id
stringlengths
11
15
title
stringlengths
7
171
problem
stringlengths
9
4.33k
solution
stringlengths
6
19k
problem_wikitext
stringlengths
9
4.42k
solution_wikitext
stringlengths
7
19.1k
proof_title
stringlengths
9
171
theorem_url
stringlengths
34
198
proof_url
stringlengths
36
198
categories
listlengths
0
9
theorem_references
listlengths
0
36
proof_references
listlengths
0
253
proofwiki-11000
Linear First Order ODE/y' + y cot x = 2 x cosec x
The linear first order ODE: :$(1): \quad y' + y \cot x = 2 x \csc x$ has the general solution: :$y = x^2 \csc x + C \csc x$
$(1)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = \cot x$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x | r = \int \cot x \rd x | c = }} {{eqn | r = \map \ln {\sin x} | c = }} {{eqn | ll= \leadsto | l = e^{\int P \rd x} | r = \sin x | c...
The [[Definition:Linear First Order ODE|linear first order ODE]]: :$(1): \quad y' + y \cot x = 2 x \csc x$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = x^2 \csc x + C \csc x$
$(1)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = \cot x$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x | r = \int \cot x \rd x | c = }} {{eqn | r = \map \ln {\sin x} | c = }} {{eqn | ll= \leadsto | l = e^{\int P \rd x} | r = \sin x | ...
Linear First Order ODE/y' + y cot x = 2 x cosec x
https://proofwiki.org/wiki/Linear_First_Order_ODE/y'_+_y_cot_x_=_2_x_cosec_x
https://proofwiki.org/wiki/Linear_First_Order_ODE/y'_+_y_cot_x_=_2_x_cosec_x
[ "Examples of Linear First Order ODEs" ]
[ "Definition:Linear First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution to Linear First Order Ordinary Differential Equation/Solution by Integrating Factor", "Definition:Differential Equation/Solution/General Solution" ]
proofwiki-11001
Linear First Order ODE/(2 y - x^3) dx = x dy
The linear first order ODE: :$(1): \quad \paren {2 y - x^3} \rd x = x \rd y$ has the general solution: :$y = -x^3 + C x^2$
Rearranging $(1)$: :$(2): \quad \dfrac {\d y} {\d x} - \dfrac 2 x y = - x^2$ $(2)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where $\map P x = -\dfrac 2 x$. Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x | r = \int -\dfrac 2 x \rd x | c = }} {{eqn | r = -2 \ln x | c = }} {...
The [[Definition:Linear First Order ODE|linear first order ODE]]: :$(1): \quad \paren {2 y - x^3} \rd x = x \rd y$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = -x^3 + C x^2$
Rearranging $(1)$: :$(2): \quad \dfrac {\d y} {\d x} - \dfrac 2 x y = - x^2$ $(2)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where $\map P x = -\dfrac 2 x$. Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x | r = \int -\dfrac 2 x \rd x | c = }} {{eqn | r = -2 \ln x | c = }}...
Linear First Order ODE/(2 y - x^3) dx = x dy
https://proofwiki.org/wiki/Linear_First_Order_ODE/(2_y_-_x^3)_dx_=_x_dy
https://proofwiki.org/wiki/Linear_First_Order_ODE/(2_y_-_x^3)_dx_=_x_dy
[ "Examples of Linear First Order ODEs" ]
[ "Definition:Linear First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution to Linear First Order Ordinary Differential Equation/Solution by Integrating Factor", "Definition:Differential Equation/Solution/General Solution" ]
proofwiki-11002
Solution to Bernoulli's Equation
'''Bernoulli's equation''': :$(1): \quad \dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$ where: :$n \ne 0, n \ne 1$ has the general solution: :$\ds \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \map \mu x \rd x + C$ where: :$\map \mu x = e^{\paren {1 - n} \int \map P x \rd x}$
Make the substitution: :$z = y^{1 - n}$ in $(1)$. Then we have: {{begin-eqn}} {{eqn | l = \frac {\d z} {\d y} | r = \paren {1 - n} y^{-n} | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto | l = \frac {\d z} {\d y} \frac {\d y} {\d x} + \map P x y \paren {1 - n} y^{-n} | r = \map Q x y^n \...
'''[[Definition:Bernoulli's Equation|Bernoulli's equation]]''': :$(1): \quad \dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$ where: :$n \ne 0, n \ne 1$ has the [[Definition:General Solution to Differential Equation|general solution]]: :$\ds \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \map \mu x \r...
Make the substitution: :$z = y^{1 - n}$ in $(1)$. Then we have: {{begin-eqn}} {{eqn | l = \frac {\d z} {\d y} | r = \paren {1 - n} y^{-n} | c = [[Power Rule for Derivatives]] }} {{eqn | ll= \leadsto | l = \frac {\d z} {\d y} \frac {\d y} {\d x} + \map P x y \paren {1 - n} y^{-n} | r = \map Q ...
Solution to Bernoulli's Equation
https://proofwiki.org/wiki/Solution_to_Bernoulli's_Equation
https://proofwiki.org/wiki/Solution_to_Bernoulli's_Equation
[ "Bernoulli's Equation" ]
[ "Definition:Bernoulli's Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Power Rule for Derivatives", "Derivative of Composite Function", "Definition:Linear First Order Ordinary Differential Equation", "Definition:Integrating Factor" ]
proofwiki-11003
Bernoulli's Equation/x y' + y = x^4 y^3
The first order ODE: :$(1): \quad x y' + y = x^4 y^3$ has the general solution: :$\dfrac 1 {y^2} = - x^4 + C x^2$
Let $(1)$ be rearranged as: :$(2): \quad \dfrac {\d y} {\d x} + \dfrac 1 x y = x^3 y^3$ It can be seen that $(2)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$ where: :$\map P x = \dfrac 1 x$ :$\map Q x = x^3$ :$n = 3$ and so is an example of Bernoulli's equation. By Solution to Bernoulli's Equati...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad x y' + y = x^4 y^3$ has the [[Definition:General Solution to Differential Equation|general solution]]: :$\dfrac 1 {y^2} = - x^4 + C x^2$
Let $(1)$ be rearranged as: :$(2): \quad \dfrac {\d y} {\d x} + \dfrac 1 x y = x^3 y^3$ It can be seen that $(2)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$ where: :$\map P x = \dfrac 1 x$ :$\map Q x = x^3$ :$n = 3$ and so is an example of [[Definition:Bernoulli's Equation|Bernoulli's equati...
Bernoulli's Equation/x y' + y = x^4 y^3
https://proofwiki.org/wiki/Bernoulli's_Equation/x_y'_+_y_=_x^4_y^3
https://proofwiki.org/wiki/Bernoulli's_Equation/x_y'_+_y_=_x^4_y^3
[ "Examples of Bernoulli's Equation" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Bernoulli's Equation", "Solution to Bernoulli's Equation", "Definition:Differential Equation/Solution/General Solution", "Definition:Differential Equation/Solution/General Solution" ]
proofwiki-11004
Bernoulli's Equation/x y^2 y' + y^3 = x cosine x
The first order ODE: :$(1): \quad x y^2 y' + y^3 = x \cos x$ has the general solution: :$y^3 = 3 \sin x + \dfrac {9 \cos x} x - \dfrac {18 \sin x} {x^2} - \dfrac {18 \cos x} {x^3} + \dfrac C {x^3}$
Let $(1)$ be rearranged as: :$(2): \quad \dfrac {\d y} {\d x} + \dfrac 1 x y = \dfrac {\cos x} {y^2}$ It can be seen that $(2)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$ where: :$\map P x = \dfrac 1 x$ :$\map Q x = \cos x$ :$n = -2$ and so is an example of Bernoulli's equation. By Solution to ...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad x y^2 y' + y^3 = x \cos x$ has the [[Definition:General Solution to Differential Equation|general solution]]: :$y^3 = 3 \sin x + \dfrac {9 \cos x} x - \dfrac {18 \sin x} {x^2} - \dfrac {18 \cos x} {x^3} + \dfrac C {x^3}$
Let $(1)$ be rearranged as: :$(2): \quad \dfrac {\d y} {\d x} + \dfrac 1 x y = \dfrac {\cos x} {y^2}$ It can be seen that $(2)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$ where: :$\map P x = \dfrac 1 x$ :$\map Q x = \cos x$ :$n = -2$ and so is an example of [[Definition:Bernoulli's Equation|...
Bernoulli's Equation/x y^2 y' + y^3 = x cosine x
https://proofwiki.org/wiki/Bernoulli's_Equation/x_y^2_y'_+_y^3_=_x_cosine_x
https://proofwiki.org/wiki/Bernoulli's_Equation/x_y^2_y'_+_y^3_=_x_cosine_x
[ "Examples of Bernoulli's Equation" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Bernoulli's Equation", "Solution to Bernoulli's Equation", "Definition:Differential Equation/Solution/General Solution", "Primitive of x cubed by Cosine of a x", "Definition:Differential Equation/Solution/General Solution" ]
proofwiki-11005
First Order ODE/(exp y - 2 x y) y' = y^2
The first order ODE: :$(1): \quad \paren {e^y - 2 x y} y' = y^2$ has the general solution: :$x y^2 = e^y + C$
Let $(1)$ be rearranged as: :$\dfrac {\d y} {\d x} = \dfrac {y^2} {e^y - 2 x y}$ Hence: :$(2): \quad \dfrac {\d x} {\d y} + \dfrac 2 y x = \dfrac {e^y} {y^2}$ It can be seen that $(2)$ is a linear first order ODE in the form: :$\dfrac {\d x} {\d y} + \map P y x = \map Q y$ where: :$\map P y = \dfrac 2 y$ :$\map Q y = \...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad \paren {e^y - 2 x y} y' = y^2$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$x y^2 = e^y + C$
Let $(1)$ be rearranged as: :$\dfrac {\d y} {\d x} = \dfrac {y^2} {e^y - 2 x y}$ Hence: :$(2): \quad \dfrac {\d x} {\d y} + \dfrac 2 y x = \dfrac {e^y} {y^2}$ It can be seen that $(2)$ is a [[Definition:Linear First Order ODE|linear first order ODE]] in the form: :$\dfrac {\d x} {\d y} + \map P y x = \map Q y$ where...
First Order ODE/(exp y - 2 x y) y' = y^2
https://proofwiki.org/wiki/First_Order_ODE/(exp_y_-_2_x_y)_y'_=_y^2
https://proofwiki.org/wiki/First_Order_ODE/(exp_y_-_2_x_y)_y'_=_y^2
[ "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Linear First Order Ordinary Differential Equation", "Solution to Linear First Order Ordinary Differential Equation/Solution by Integrating Factor", "Definition:Differential Equation/Solution/General Solution" ]
proofwiki-11006
First Order ODE/y - x y' = y' y^2 exp y
The first order ODE: :$(1): \quad y - x y' = y' y^2 e^y$ has the general solution: :$x y^2 = e^y + C$
Let $(1)$ be rearranged as: :$\dfrac {\d y} {\d x} = \dfrac y {y^2 e^y + x}$ Hence: :$(2): \quad \dfrac {\d x} {\d y} - \dfrac 1 y x = y e^y$ It can be seen that $(2)$ is a linear first order ODE in the form: :$\dfrac {\d x} {\d y} + \map P y x = \map Q y$ where: :$\map P y = -\dfrac 1 y$ :$\map Q y = y e^y$ Thus: {{be...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad y - x y' = y' y^2 e^y$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$x y^2 = e^y + C$
Let $(1)$ be rearranged as: :$\dfrac {\d y} {\d x} = \dfrac y {y^2 e^y + x}$ Hence: :$(2): \quad \dfrac {\d x} {\d y} - \dfrac 1 y x = y e^y$ It can be seen that $(2)$ is a [[Definition:Linear First Order ODE|linear first order ODE]] in the form: :$\dfrac {\d x} {\d y} + \map P y x = \map Q y$ where: :$\map P y = -\...
First Order ODE/y - x y' = y' y^2 exp y
https://proofwiki.org/wiki/First_Order_ODE/y_-_x_y'_=_y'_y^2_exp_y
https://proofwiki.org/wiki/First_Order_ODE/y_-_x_y'_=_y'_y^2_exp_y
[ "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Linear First Order Ordinary Differential Equation", "Solution to Linear First Order Ordinary Differential Equation/Solution by Integrating Factor", "Definition:Differential Equation/Solution/General Solution" ]
proofwiki-11007
Topology Defined by Neighborhood System
Let $S$ be a set. Let $\family {\NN_x}_{x \mathop \in S}$ be an indexed family where $\NN_x$ is non-empty set of subsets of $S$. Assume that :$(\text N 1): \quad \forall x \in S, U \in \NN_x: x \in U$ :$(\text N 2): \quad \forall x \in S, U \in \NN_x, y \in U:\exists V \in \NN_y: V \subseteq U$ :$(\text N 3): \quad \fo...
Define: :$\BB := \ds \bigcup_{x \mathop \in S} \NN_x$ According to Topology Defined by Basis it should be proved that :$(\text B 1): \quad \forall A_1, A_2 \in \BB: \forall x \in A_1 \cap A_2: \exists A \in \BB: x \in A \subseteq A_1 \cap A_2$ :$(\text B 2): \quad \forall x \in S: \exists A \in \BB: x \in A$ Ad $(\text...
Let $S$ be a [[Definition:Set|set]]. Let $\family {\NN_x}_{x \mathop \in S}$ be an [[Definition:Indexed Family|indexed family]] where $\NN_x$ is [[Definition:Non-Empty Set|non-empty]] [[Definition:Set of Sets|set]] of [[Definition:Subset|subsets]] of $S$. Assume that :$(\text N 1): \quad \forall x \in S, U \in \NN_x:...
Define: :$\BB := \ds \bigcup_{x \mathop \in S} \NN_x$ According to [[Topology Defined by Basis]] it should be proved that :$(\text B 1): \quad \forall A_1, A_2 \in \BB: \forall x \in A_1 \cap A_2: \exists A \in \BB: x \in A \subseteq A_1 \cap A_2$ :$(\text B 2): \quad \forall x \in S: \exists A \in \BB: x \in A$ Ad $...
Topology Defined by Neighborhood System
https://proofwiki.org/wiki/Topology_Defined_by_Neighborhood_System
https://proofwiki.org/wiki/Topology_Defined_by_Neighborhood_System
[ "Topology" ]
[ "Definition:Set", "Definition:Indexing Set/Family", "Definition:Non-Empty Set", "Definition:Set of Sets", "Definition:Subset", "Definition:Topological Space", "Definition:Neighborhood System" ]
[ "Topology Defined by Basis", "Definition:Set Intersection", "Definition:Set Union/Family of Sets", "Definition:Set Union/Family of Sets", "Set Intersection Preserves Subsets", "Subset Relation is Transitive", "Definition:Empty Set", "Definition:Set Union/Family of Sets", "Topology Defined by Basis",...
proofwiki-11008
Orthogonal Trajectories/Examples/x + C exp -x
Consider the one-parameter family of curves: :$(1): \quad y = x + C e^{-x}$ Its family of orthogonal trajectories is given by the equation: :$x = y - 2 + C e^{-y}$ :600px
We use the technique of formation of ordinary differential equation by elimination. Differentiating $(1)$ {{WRT|Differentiation}} $x$ gives: :$\dfrac {\d y} {\d x} = 1 - C e^{-x}$ Eliminating $C$: {{begin-eqn}} {{eqn | l = C | o = = | r = \frac {y - x} {e^{-x} } | c = }} {{eqn | ll= \leadsto | ...
Consider the [[Definition:One-Parameter Family of Curves|one-parameter family of curves]]: :$(1): \quad y = x + C e^{-x}$ Its [[Definition:Orthogonal Trajectories|family of orthogonal trajectories]] is given by the equation: :$x = y - 2 + C e^{-y}$ :[[File:XplusCExpMinusXOrthogonalTrajectories.png|600px]]
We use the technique of [[Definition:Formation of Ordinary Differential Equation by Elimination|formation of ordinary differential equation by elimination]]. [[Definition:Differentiation|Differentiating]] $(1)$ {{WRT|Differentiation}} $x$ gives: :$\dfrac {\d y} {\d x} = 1 - C e^{-x}$ Eliminating $C$: {{begin-eqn}} ...
Orthogonal Trajectories/Examples/x + C exp -x
https://proofwiki.org/wiki/Orthogonal_Trajectories/Examples/x_+_C_exp_-x
https://proofwiki.org/wiki/Orthogonal_Trajectories/Examples/x_+_C_exp_-x
[ "Examples of Orthogonal Trajectories", "Exponential Function" ]
[ "Definition:Family of Curves/One-Parameter", "Definition:Orthogonal Trajectories", "File:XplusCExpMinusXOrthogonalTrajectories.png" ]
[ "Definition:Formation of Ordinary Differential Equation by Elimination", "Definition:Differentiation", "Orthogonal Trajectories of One-Parameter Family of Curves", "Definition:Orthogonal Trajectories", "Definition:Integrating Factor", "Primitive of x by Exponential of a x" ]
proofwiki-11009
Differential Equation of Family of Linear Combination of Functions is Linear
Consider the one-parameter family of curves: :$(1): \quad y = C \map f x + \map g x$ The differential equation that describes $(1)$ is linear and of first order.
Differentiating $(1)$ {{WRT|Differentiation}} $x$ gives: :$(2): \quad \dfrac {\d y} {\d x} = C \map {f'} x + \map {g'} x$ Rearranging $(1)$, we have: :$C = \dfrac {y - \map g x} {\map f x}$ Substituting for $C$ in $(2)$: {{begin-eqn}} {{eqn | l = \dfrac {\d y} {\d x} | r = \dfrac {y - \map g x} {\map f x} \map {f...
Consider the [[Definition:One-Parameter Family of Curves|one-parameter family of curves]]: :$(1): \quad y = C \map f x + \map g x$ The [[Definition:Differential Equation|differential equation]] that describes $(1)$ is [[Definition:Linear First Order ODE|linear and of first order]].
[[Definition:Differentiation|Differentiating]] $(1)$ {{WRT|Differentiation}} $x$ gives: :$(2): \quad \dfrac {\d y} {\d x} = C \map {f'} x + \map {g'} x$ Rearranging $(1)$, we have: :$C = \dfrac {y - \map g x} {\map f x}$ Substituting for $C$ in $(2)$: {{begin-eqn}} {{eqn | l = \dfrac {\d y} {\d x} | r = \dfra...
Differential Equation of Family of Linear Combination of Functions is Linear
https://proofwiki.org/wiki/Differential_Equation_of_Family_of_Linear_Combination_of_Functions_is_Linear
https://proofwiki.org/wiki/Differential_Equation_of_Family_of_Linear_Combination_of_Functions_is_Linear
[ "Linear First Order ODEs" ]
[ "Definition:Family of Curves/One-Parameter", "Definition:Differential Equation", "Definition:Linear First Order Ordinary Differential Equation" ]
[ "Definition:Differentiation", "Definition:Linear First Order Ordinary Differential Equation" ]
proofwiki-11010
Solution of Second Order Differential Equation with Missing Dependent Variable
Let $\map f {x, y', y' '} = 0$ be a second order ordinary differential equation in which the dependent variable $y$ is not explicitly present. Then $f$ can be reduced to a first order ordinary differential equation, whose solution can be determined.
Consider the second order ordinary differential equation: :$(1): \quad \map f {x, y', y' '} = 0$ Let a new dependent variable $p$ be introduced: :$y' = p$ :$y' ' = \dfrac {\d p} {\d x}$ Then $(1)$ can be transformed into: :$(2): \quad \map f {x, p, \dfrac {\d p} {\d x} } = 0$ which is a first order ODE. If $(2)$ has a ...
Let $\map f {x, y', y' '} = 0$ be a [[Definition:Second Order Ordinary Differential Equation|second order ordinary differential equation]] in which the [[Definition:Dependent Variable|dependent variable]] $y$ is not explicitly present. Then $f$ can be reduced to a [[Definition:First Order Ordinary Differential Equatio...
Consider the [[Definition:Second Order Ordinary Differential Equation|second order ordinary differential equation]]: :$(1): \quad \map f {x, y', y' '} = 0$ Let a new [[Definition:Dependent Variable|dependent variable]] $p$ be introduced: :$y' = p$ :$y' ' = \dfrac {\d p} {\d x}$ Then $(1)$ can be transformed into: :$(...
Solution of Second Order Differential Equation with Missing Dependent Variable
https://proofwiki.org/wiki/Solution_of_Second_Order_Differential_Equation_with_Missing_Dependent_Variable
https://proofwiki.org/wiki/Solution_of_Second_Order_Differential_Equation_with_Missing_Dependent_Variable
[ "Second Order ODEs" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Dependent Variable", "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Dependent Variable", "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Definition:First Order Ordinary Differential Equation", "Definition:Second Order Ordinary Diffe...
proofwiki-11011
Second Order ODE/x y'' - y' = 3 x^2
The second order ODE: :$(1): \quad x y' ' - y' = 3 x^2$ has the general solution: :$y = x^3 + \dfrac {C_1 x^2} 2 + C^2$
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable. Substitute $p$ for $y'$ in $(1)$: :$x \dfrac {\d p} {\d x} - p = 3 x^2$ and divide through by $x$: :$\dfrac {\d p} {\d x} - \dfrac p x = 3 x$ From: :Linear First Order ODE: $y' - \dfrac y x = 3 x$ its solution is...
The [[Definition:Second Order ODE|second order ODE]]: :$(1): \quad x y' ' - y' = 3 x^2$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = x^3 + \dfrac {C_1 x^2} 2 + C^2$
The proof proceeds by using [[Solution of Second Order Differential Equation with Missing Dependent Variable]]. Substitute $p$ for $y'$ in $(1)$: :$x \dfrac {\d p} {\d x} - p = 3 x^2$ and divide through by $x$: :$\dfrac {\d p} {\d x} - \dfrac p x = 3 x$ From: :[[Linear First Order ODE/y' - (y over x) = 3 x|Linear F...
Second Order ODE/x y'' - y' = 3 x^2
https://proofwiki.org/wiki/Second_Order_ODE/x_y''_-_y'_=_3_x^2
https://proofwiki.org/wiki/Second_Order_ODE/x_y''_-_y'_=_3_x^2
[ "Examples of Homogeneous LSOODEs" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution of Second Order Differential Equation with Missing Dependent Variable", "Linear First Order ODE/y' - (y over x) = 3 x", "Definition:Differential Equation/Solution/General Solution", "Definition:Separable Differential Equation" ]
proofwiki-11012
Niemytzki Plane is Topology
Niemytzki plane is a topological space.
By definition $T = \struct {S, \tau}$ is the Niemytzki plane {{iff}}: {{begin-eqn}} {{eqn | n = 1 | l = S | r = \set {\tuple {x, y}: y \ge 0} }} {{eqn | n = 2 | l = \map \BB {x, y} | r = \set {\map {B_r} {x, y} \cap S: r > 0} | c = if $x, y \in \R, y > 0$ }} {{eqn | n = 3 | l = \map ...
[[Definition:Niemytzki Plane|Niemytzki plane]] is a [[Definition:Topological Space|topological space]].
By definition $T = \struct {S, \tau}$ is the [[Definition:Niemytzki Plane|Niemytzki plane]] {{iff}}: {{begin-eqn}} {{eqn | n = 1 | l = S | r = \set {\tuple {x, y}: y \ge 0} }} {{eqn | n = 2 | l = \map \BB {x, y} | r = \set {\map {B_r} {x, y} \cap S: r > 0} | c = if $x, y \in \R, y > 0$ }}...
Niemytzki Plane is Topology
https://proofwiki.org/wiki/Niemytzki_Plane_is_Topology
https://proofwiki.org/wiki/Niemytzki_Plane_is_Topology
[ "Niemytzki Plane" ]
[ "Definition:Niemytzki Plane", "Definition:Topological Space" ]
[ "Definition:Niemytzki Plane", "Definition:Open Ball", "Definition:Euclidean Metric/Real Number Plane", "Topology Defined by Neighborhood System", "Definition:Empty Set", "Definition:Set of Sets", "Definition:Subset", "Definition:Non-Empty Set", "Intersection is Subset", "Definition:Subset", "Def...
proofwiki-11013
Solution of Second Order Differential Equation with Missing Independent Variable
Let $\map g {y, \dfrac {\d y} {\d x}, \dfrac {\d^2 y} {\d x^2} } = 0$ be a second order ordinary differential equation in which the independent variable $x$ is not explicitly present. Then $g$ can be reduced to a first order ordinary differential equation, whose solution can be determined.
Consider the second order ordinary differential equation: :$(1): \quad \map g {y, \dfrac {\d y} {\d x}, \dfrac {\d^2 y} {\d x^2} } = 0$ Let a new dependent variable $p$ be introduced: :$y' = p$ Hence: :$y' ' = \dfrac {\d p} {\d x} = \dfrac {\d p} {\d y} \dfrac {\d y} {\d x} = p \dfrac {\d p} {\d y}$ Then $(1)$ can be t...
Let $\map g {y, \dfrac {\d y} {\d x}, \dfrac {\d^2 y} {\d x^2} } = 0$ be a [[Definition:Second Order Ordinary Differential Equation|second order ordinary differential equation]] in which the [[Definition:Independent Variable|independent variable]] $x$ is not explicitly present. Then $g$ can be reduced to a [[Definitio...
Consider the [[Definition:Second Order Ordinary Differential Equation|second order ordinary differential equation]]: :$(1): \quad \map g {y, \dfrac {\d y} {\d x}, \dfrac {\d^2 y} {\d x^2} } = 0$ Let a new [[Definition:Dependent Variable|dependent variable]] $p$ be introduced: :$y' = p$ Hence: :$y' ' = \dfrac {\d p} {...
Solution of Second Order Differential Equation with Missing Independent Variable
https://proofwiki.org/wiki/Solution_of_Second_Order_Differential_Equation_with_Missing_Independent_Variable
https://proofwiki.org/wiki/Solution_of_Second_Order_Differential_Equation_with_Missing_Independent_Variable
[ "Second Order ODEs" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Independent Variable", "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Dependent Variable", "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Definition:First Order Ordinary Differential Equation", "Definition:Second Order Ordinary Diffe...
proofwiki-11014
Linear Second Order ODE/y'' + k^2 y = 0
The second order ODE: :$(1): \quad y' ' + k^2 y = 0$ has the general solution: :$y = A \, \map \sin {k x + B}$ or can be expressed as: :$y = C_1 \sin k x + C_2 \cos k x$
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as: :$p \dfrac {\d p} {\d y} = -k^2 y$ where $p = \dfrac {\d y} {\d x}$. From: :First Order ODE: $y \rd y = k x \rd x$ this has the general solution: :$p^2 = -k^2 y^2 + C$ or: :$p^2 + k^2 y^2 = C$ As the {{LH...
The [[Definition:Second Order ODE|second order ODE]]: :$(1): \quad y' ' + k^2 y = 0$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = A \, \map \sin {k x + B}$ or can be expressed as: :$y = C_1 \sin k x + C_2 \cos k x$
Using [[Solution of Second Order Differential Equation with Missing Independent Variable]], $(1)$ can be expressed as: :$p \dfrac {\d p} {\d y} = -k^2 y$ where $p = \dfrac {\d y} {\d x}$. From: :[[First Order ODE/y dy = k x dx|First Order ODE: $y \rd y = k x \rd x$]] this has the [[Definition:General Solution to Diff...
Linear Second Order ODE/y'' + k^2 y = 0/Proof 1
https://proofwiki.org/wiki/Linear_Second_Order_ODE/y''_+_k^2_y_=_0
https://proofwiki.org/wiki/Linear_Second_Order_ODE/y''_+_k^2_y_=_0/Proof_1
[ "Linear Second Order ODE/y'' + k^2 y = 0", "Examples of Constant Coefficient Homogeneous LSOODEs" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution of Second Order Differential Equation with Missing Independent Variable", "First Order ODE/y dy = k x dx", "Definition:Differential Equation/Solution/General Solution", "Definition:Square/Function", "Definition:Positive/Real Number", "Solution to Separable Differential Equation", "Primitive of...
proofwiki-11015
Linear Second Order ODE/y'' + k^2 y = 0
The second order ODE: :$(1): \quad y' ' + k^2 y = 0$ has the general solution: :$y = A \, \map \sin {k x + B}$ or can be expressed as: :$y = C_1 \sin k x + C_2 \cos k x$
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE. Its auxiliary equation is: :$(2): \quad: m^2 + k^2 = 0$ From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are: :$m_1 = k i$ :$m_2 = -k i$ These are complex and unequal. So from Solution of Constant Coef...
The [[Definition:Second Order ODE|second order ODE]]: :$(1): \quad y' ' + k^2 y = 0$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = A \, \map \sin {k x + B}$ or can be expressed as: :$y = C_1 \sin k x + C_2 \cos k x$
It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]. Its [[Definition:Auxiliary Equation|auxiliary equation]] is: :$(2): \quad: m^2 + k^2 = 0$ From [[Solution to Quadratic Equation with Real Coefficients]], the...
Linear Second Order ODE/y'' + k^2 y = 0/Proof 2
https://proofwiki.org/wiki/Linear_Second_Order_ODE/y''_+_k^2_y_=_0
https://proofwiki.org/wiki/Linear_Second_Order_ODE/y''_+_k^2_y_=_0/Proof_2
[ "Linear Second Order ODE/y'' + k^2 y = 0", "Examples of Constant Coefficient Homogeneous LSOODEs" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Homogeneous Linear Second Order ODE with Constant Coefficients", "Definition:Auxiliary Equation", "Solution to Quadratic Equation/Real Coefficients", "Definition:Root of Polynomial", "Definition:Complex Number", "Solution of Constant Coefficient Homogeneous LSOODE", "Definition:Differential ...
proofwiki-11016
Second Order ODE/y y'' + (y')^2 = 0
The second order ODE: :$(1): \quad y y' ' + \paren {y'}^2 = 0$ has the general solution: :$y^2 = C_1 x + C_2$
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = y p \frac {\d p} {\d y} + p^2 | r = 0 | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto | l = y \rd p + p \rd y | r = 0 | c = multiplyin...
The [[Definition:Second Order ODE|second order ODE]]: :$(1): \quad y y' ' + \paren {y'}^2 = 0$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y^2 = C_1 x + C_2$
Using [[Solution of Second Order Differential Equation with Missing Independent Variable]], $(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = y p \frac {\d p} {\d y} + p^2 | r = 0 | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto | l = y \rd p + p \rd y | r = 0 | c = multi...
Second Order ODE/y y'' + (y')^2 = 0
https://proofwiki.org/wiki/Second_Order_ODE/y_y''_+_(y')^2_=_0
https://proofwiki.org/wiki/Second_Order_ODE/y_y''_+_(y')^2_=_0
[ "Examples of Second Order ODEs" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution of Second Order Differential Equation with Missing Independent Variable", "First Order ODE/y dx + x dy = 0", "First Order ODE/y dy = k dx" ]
proofwiki-11017
Second Order ODE/x y'' = y' + (y')^3
The second order ODE: :$(1): \quad x y' ' = y' + \paren {y'}^3$ has the general solution: :$x^2 + \paren {y - C_2}^2 = C_1^2$
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable. Substitute $p$ for $y'$ in $(1)$: {{begin-eqn}} {{eqn | l = x \dfrac {\d p} {\d x} | r = p + p^3 | c = }} {{eqn | ll= \leadsto | l = p = \frac {\d y} {\d x} | r = \frac x {\sqrt {C_1^2 - ...
The [[Definition:Second Order ODE|second order ODE]]: :$(1): \quad x y' ' = y' + \paren {y'}^3$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$x^2 + \paren {y - C_2}^2 = C_1^2$
The proof proceeds by using [[Solution of Second Order Differential Equation with Missing Dependent Variable]]. Substitute $p$ for $y'$ in $(1)$: {{begin-eqn}} {{eqn | l = x \dfrac {\d p} {\d x} | r = p + p^3 | c = }} {{eqn | ll= \leadsto | l = p = \frac {\d y} {\d x} | r = \frac x {\sqrt {C_...
Second Order ODE/x y'' = y' + (y')^3
https://proofwiki.org/wiki/Second_Order_ODE/x_y''_=_y'_+_(y')^3
https://proofwiki.org/wiki/Second_Order_ODE/x_y''_=_y'_+_(y')^3
[ "Examples of Homogeneous LSOODEs" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution of Second Order Differential Equation with Missing Dependent Variable", "First Order ODE/x dy = (y + y^3) dx", "Solution to Separable Differential Equation", "Primitive of x over Root of a squared minus x squared" ]
proofwiki-11018
Linear Second Order ODE/y'' - k^2 y = 0
The second order ODE: :$(1): \quad y' ' - k^2 y = 0$ has the general solution: :$y = C_1 e^{k x} + C_2 e^{-k x}$
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = p \frac {\d p} {\d y} | r = k^2 y | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto | l = p^2 | r = k^2 y^2 + k^2 \alpha | c = First Or...
The [[Definition:Second Order ODE|second order ODE]]: :$(1): \quad y' ' - k^2 y = 0$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = C_1 e^{k x} + C_2 e^{-k x}$
Using [[Solution of Second Order Differential Equation with Missing Independent Variable]], $(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = p \frac {\d p} {\d y} | r = k^2 y | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto | l = p^2 | r = k^2 y^2 + k^2 \alpha | c = [[F...
Linear Second Order ODE/y'' - k^2 y = 0/Proof 1
https://proofwiki.org/wiki/Linear_Second_Order_ODE/y''_-_k^2_y_=_0
https://proofwiki.org/wiki/Linear_Second_Order_ODE/y''_-_k^2_y_=_0/Proof_1
[ "Linear Second Order ODE/y'' - k^2 y = 0", "Examples of Constant Coefficient Homogeneous LSOODEs" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution of Second Order Differential Equation with Missing Independent Variable", "First Order ODE/y dy = k x dx", "Solution to Separable Differential Equation", "Primitive of Reciprocal of Root of x squared plus k", "Solution to Quadratic Equation", "Solution to Quadratic Equation" ]
proofwiki-11019
Linear Second Order ODE/y'' - k^2 y = 0
The second order ODE: :$(1): \quad y' ' - k^2 y = 0$ has the general solution: :$y = C_1 e^{k x} + C_2 e^{-k x}$
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE. Its auxiliary equation is: :$(2): \quad m^2 - k^2 = 0$ From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are: :$m_1 = k$ :$m_2 = -k$ These are real and unequal. So from Solution of Constant Coefficient ...
The [[Definition:Second Order ODE|second order ODE]]: :$(1): \quad y' ' - k^2 y = 0$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = C_1 e^{k x} + C_2 e^{-k x}$
It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]. Its [[Definition:Auxiliary Equation|auxiliary equation]] is: :$(2): \quad m^2 - k^2 = 0$ From [[Solution to Quadratic Equation with Real Coefficients]], the ...
Linear Second Order ODE/y'' - k^2 y = 0/Proof 2
https://proofwiki.org/wiki/Linear_Second_Order_ODE/y''_-_k^2_y_=_0
https://proofwiki.org/wiki/Linear_Second_Order_ODE/y''_-_k^2_y_=_0/Proof_2
[ "Linear Second Order ODE/y'' - k^2 y = 0", "Examples of Constant Coefficient Homogeneous LSOODEs" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Homogeneous Linear Second Order ODE with Constant Coefficients", "Definition:Auxiliary Equation", "Solution to Quadratic Equation/Real Coefficients", "Definition:Root of Polynomial", "Definition:Real Number", "Solution of Constant Coefficient Homogeneous LSOODE", "Definition:Differential Equ...
proofwiki-11020
Second Order ODE/x^2 y'' = 2 x y' + (y')^2
The second order ODE: :$x^2 y' ' = 2 x y' + \paren {y'}^2$ has the general solution: :$y = -\dfrac {x^2} 2 - C_1 x - {C_1}^2 \, \map \ln {x - C_1} + C_2$
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable. Substitute $p$ for $y'$: {{begin-eqn}} {{eqn | l = x^2 \dfrac {\d p} {\d x} | r = 2 x p + p^2 | c = }} {{eqn | ll= \leadsto | l = p = \frac {\d y}{\d x} | r = -\frac {x^2} {x - C_1} ...
The [[Definition:Second Order ODE|second order ODE]]: :$x^2 y' ' = 2 x y' + \paren {y'}^2$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = -\dfrac {x^2} 2 - C_1 x - {C_1}^2 \, \map \ln {x - C_1} + C_2$
The proof proceeds by using [[Solution of Second Order Differential Equation with Missing Dependent Variable]]. Substitute $p$ for $y'$: {{begin-eqn}} {{eqn | l = x^2 \dfrac {\d p} {\d x} | r = 2 x p + p^2 | c = }} {{eqn | ll= \leadsto | l = p = \frac {\d y}{\d x} | r = -\frac {x^2} {x - C_1...
Second Order ODE/x^2 y'' = 2 x y' + (y')^2
https://proofwiki.org/wiki/Second_Order_ODE/x^2_y''_=_2_x_y'_+_(y')^2
https://proofwiki.org/wiki/Second_Order_ODE/x^2_y''_=_2_x_y'_+_(y')^2
[ "Examples of Second Order ODEs" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution of Second Order Differential Equation with Missing Dependent Variable", "Bernoulli's Equation/x^2 dy = (2 x y + y^2) dx", "Primitive of x squared over a x + b" ]
proofwiki-11021
Second Order ODE/(x^2 + 2 y') y'' + 2 x y' = 0
The second order ODE: :$(1): \quad \paren {x^2 + 2 y'} y' ' + 2 x y' = 0$ subject to the initial conditions: :$y = 1$ and $y' = 0$ when $x = 0$ has the particular solution: :$y = 1$ or: :$3 y + x^3 = 3$
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable. Substitute $p$ for $y'$ in $(1)$: {{begin-eqn}} {{eqn | l = \paren {x^2 + 2 p} \dfrac {\d p} {\d x} + 2 x p | r = 0 | c = }} {{eqn | ll= \leadsto | l = 2 x p \rd x + \paren {x^2 + 2 p} \rd p ...
The [[Definition:Second Order ODE|second order ODE]]: :$(1): \quad \paren {x^2 + 2 y'} y' ' + 2 x y' = 0$ subject to the [[Definition:Initial Condition|initial conditions]]: :$y = 1$ and $y' = 0$ when $x = 0$ has the [[Definition:Particular Solution of Differential Equation|particular solution]]: :$y = 1$ or: :$3 y + ...
The proof proceeds by using [[Solution of Second Order Differential Equation with Missing Dependent Variable]]. Substitute $p$ for $y'$ in $(1)$: {{begin-eqn}} {{eqn | l = \paren {x^2 + 2 p} \dfrac {\d p} {\d x} + 2 x p | r = 0 | c = }} {{eqn | ll= \leadsto | l = 2 x p \rd x + \paren {x^2 + 2 p} \r...
Second Order ODE/(x^2 + 2 y') y'' + 2 x y' = 0
https://proofwiki.org/wiki/Second_Order_ODE/(x^2_+_2_y')_y''_+_2_x_y'_=_0
https://proofwiki.org/wiki/Second_Order_ODE/(x^2_+_2_y')_y''_+_2_x_y'_=_0
[ "Examples of Homogeneous LSOODEs" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Initial Condition", "Definition:Differential Equation/Solution/Particular Solution" ]
[ "Solution of Second Order Differential Equation with Missing Dependent Variable", "Bernoulli's Equation/2 x y dx + (x^2 + 2 y) dy = 0", "Definition:Initial Condition", "Definition:Initial Condition", "Definition:Differential Equation/Solution/Particular Solution", "Definition:Initial Condition" ]
proofwiki-11022
Second Order ODE/y y'' = y^2 y' + (y')^2
The second order ODE: :$(1): \quad y y' ' = y^2 y' + \paren {y'}^2$ subject to the initial conditions: :$y = -\dfrac 1 2$ and $y' = 1$ when $x = 0$ has the particular solution: :$2 y - 3 = 8 y \, \map \exp {\dfrac {3 x} 2}$
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = y p \frac {\d p} {\d y} | r = y^2 p + p^2 | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto | l = \frac {\d p} {\d y} - \frac p y | r = y ...
The [[Definition:Second Order ODE|second order ODE]]: :$(1): \quad y y' ' = y^2 y' + \paren {y'}^2$ subject to the [[Definition:Initial Condition|initial conditions]]: :$y = -\dfrac 1 2$ and $y' = 1$ when $x = 0$ has the [[Definition:Particular Solution of Differential Equation|particular solution]]: :$2 y - 3 = 8 y \...
Using [[Solution of Second Order Differential Equation with Missing Independent Variable]], $(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = y p \frac {\d p} {\d y} | r = y^2 p + p^2 | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto | l = \frac {\d p} {\d y} - \frac p y | r = y...
Second Order ODE/y y'' = y^2 y' + (y')^2
https://proofwiki.org/wiki/Second_Order_ODE/y_y''_=_y^2_y'_+_(y')^2
https://proofwiki.org/wiki/Second_Order_ODE/y_y''_=_y^2_y'_+_(y')^2
[ "Examples of Second Order ODEs" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Initial Condition", "Definition:Differential Equation/Solution/Particular Solution" ]
[ "Solution of Second Order Differential Equation with Missing Independent Variable", "Linear First Order ODE/y' - (y over x) = k x", "Solution to Separable Differential Equation", "Primitive of Reciprocal of x by a x + b", "Definition:Initial Condition" ]
proofwiki-11023
Second Order ODE/y'' = 1 + (y')^2
The second order ODE: :$(1): \quad y' ' = 1 + \paren {y'}^2$ has the general solution: :$y = \map {\ln \sec} {x + C_1} + C_2$
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = p \frac {\d p} {\d y} | r = p^2 + 1 | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto | l = \int \rd y | r = \int \frac {p \rd p} {p^2 + 1} ...
The [[Definition:Second Order ODE|second order ODE]]: :$(1): \quad y' ' = 1 + \paren {y'}^2$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = \map {\ln \sec} {x + C_1} + C_2$
Using [[Solution of Second Order Differential Equation with Missing Independent Variable]], $(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = p \frac {\d p} {\d y} | r = p^2 + 1 | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto | l = \int \rd y | r = \int \frac {p \rd p} {p^2 + ...
Second Order ODE/y'' = 1 + (y')^2/Proof 1
https://proofwiki.org/wiki/Second_Order_ODE/y''_=_1_+_(y')^2
https://proofwiki.org/wiki/Second_Order_ODE/y''_=_1_+_(y')^2/Proof_1
[ "Examples of Second Order ODEs", "Second Order ODE/y'' = 1 + (y')^2" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution of Second Order Differential Equation with Missing Independent Variable", "Solution to Separable Differential Equation", "Primitive of x over x squared plus a squared", "Solution to Separable Differential Equation", "Primitive of Reciprocal of x by Root of x squared minus a squared" ]
proofwiki-11024
Second Order ODE/y'' = 1 + (y')^2
The second order ODE: :$(1): \quad y' ' = 1 + \paren {y'}^2$ has the general solution: :$y = \map {\ln \sec} {x + C_1} + C_2$
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable. Substitute $p$ for $y'$ in $(1)$: {{begin-eqn}} {{eqn | l = \dfrac {\d p} {\d x} | r = p^2 + 1 | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto | l = \int \rd x | r = \int \f...
The [[Definition:Second Order ODE|second order ODE]]: :$(1): \quad y' ' = 1 + \paren {y'}^2$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = \map {\ln \sec} {x + C_1} + C_2$
The proof proceeds by using [[Solution of Second Order Differential Equation with Missing Dependent Variable]]. Substitute $p$ for $y'$ in $(1)$: {{begin-eqn}} {{eqn | l = \dfrac {\d p} {\d x} | r = p^2 + 1 | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto | l = \int \rd x | r = \...
Second Order ODE/y'' = 1 + (y')^2/Proof 2
https://proofwiki.org/wiki/Second_Order_ODE/y''_=_1_+_(y')^2
https://proofwiki.org/wiki/Second_Order_ODE/y''_=_1_+_(y')^2/Proof_2
[ "Examples of Second Order ODEs", "Second Order ODE/y'' = 1 + (y')^2" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution of Second Order Differential Equation with Missing Dependent Variable", "Solution to Separable Differential Equation", "Primitive of Reciprocal of x squared plus a squared/Arctangent Form", "Solution to Separable Differential Equation", "Primitive of Tangent of a x" ]
proofwiki-11025
Cycloid has Tautochrone Property
Let a wire $AB$ be curved into the shape of an arc of a cycloid such that: :$A$ is at the cusp :$B$ is the highest point of the arc and inverted so that its cusps are uppermost and on the same horizontal line. Thus $B$ is the lowest point of the arc. Let $AB$ be embedded in a constant and uniform gravitational field wh...
:500px By the Principle of Conservation of Energy, the speed of the bead at a particular height is determined by its loss in potential energy in getting there. Thus, at the point $\tuple {x, y}$, we have: :$(1): \quad v = \dfrac {\d s} {\d t} = \sqrt {2 g y}$ This can be written: {{begin-eqn}} {{eqn | l = \d t | ...
Let a [[Definition:Wire|wire]] $AB$ be curved into the shape of an [[Definition:Arc of Cycloid|arc]] of a [[Definition:Cycloid|cycloid]] such that: :$A$ is at the [[Definition:Cusp of Cycloid|cusp]] :$B$ is the highest point of the [[Definition:Arc of Cycloid|arc]] and inverted so that its [[Definition:Cusp of Cycloid|...
:[[File:Brachistochrone.png|500px]] By the [[Principle of Conservation of Energy]], the [[Definition:Speed|speed]] of the bead at a particular height is determined by its loss in [[Definition:Potential Energy|potential energy]] in getting there. Thus, at the [[Definition:Point|point]] $\tuple {x, y}$, we have: :$(1)...
Cycloid has Tautochrone Property
https://proofwiki.org/wiki/Cycloid_has_Tautochrone_Property
https://proofwiki.org/wiki/Cycloid_has_Tautochrone_Property
[ "Cycloid has Tautochrone Property", "Tautochrone Problem", "Cycloids" ]
[ "Definition:Wire", "Definition:Cycloid/Arc", "Definition:Cycloid", "Definition:Cycloid/Cusp", "Definition:Cycloid/Arc", "Definition:Cycloid/Cusp", "Definition:Horizontal Line", "Definition:Point", "Definition:Cycloid/Arc", "Definition:Constant", "Definition:Uniform", "Definition:Gravitational ...
[ "File:Brachistochrone.png", "Principle of Conservation of Energy", "Definition:Speed", "Definition:Potential Energy", "Definition:Point", "Definition:Time", "Equation of Cycloid", "Half Angle Formulas/Cosine", "Half Angle Formulas/Sine" ]
proofwiki-11026
Velocity of Bead on Brachistochrone
Consider a wire bent into the shape of an arc of a cycloid $C$ and inverted so that its cusps are uppermost and on the same horizontal line. Let $C$ be defined by Equation of Cycloid embedded in a cartesian plane: :$x = a \paren {\theta - \sin \theta}$ :$y = a \paren {1 - \cos \theta}$ Let a bead $B$ be released from s...
By Brachistochrone is Cycloid, $C$ is a brachistochrone. {{ProofWanted}}
Consider a [[Definition:Wire|wire]] bent into the shape of an [[Definition:Arc of Cycloid|arc]] of a [[Definition:Cycloid|cycloid]] $C$ and inverted so that its [[Definition:Cusp of Cycloid|cusps]] are uppermost and on the same [[Definition:Horizontal Line|horizontal line]]. Let $C$ be defined by [[Equation of Cycloid...
By [[Brachistochrone is Cycloid]], $C$ is a [[Definition:Brachistochrone|brachistochrone]]. {{ProofWanted}}
Velocity of Bead on Brachistochrone
https://proofwiki.org/wiki/Velocity_of_Bead_on_Brachistochrone
https://proofwiki.org/wiki/Velocity_of_Bead_on_Brachistochrone
[ "Cycloids" ]
[ "Definition:Wire", "Definition:Cycloid/Arc", "Definition:Cycloid", "Definition:Cycloid/Cusp", "Definition:Horizontal Line", "Equation of Cycloid", "Definition:Cartesian Plane", "Definition:Bead", "Definition:Point", "Definition:Wire", "Definition:Friction", "Definition:Gravitational Field", ...
[ "Brachistochrone is Cycloid", "Definition:Brachistochrone" ]
proofwiki-11027
Open Ball is Subset of Open Ball
Let $M = \struct {A, d}$ be a metric space. Let $x, y$ be points of $A$. Then: :$\epsilon-\delta \ge \map d {x, y} \implies \map {B_\delta} y \subseteq \map {B_\epsilon} x$ where $\map {B_\epsilon} x$ denotes the open $\epsilon$-ball in $M = \struct {A, d}$.
Let $\epsilon - \delta \ge \map d {x, y}$. Then $\epsilon \ge \map d {x, y} + \delta$. If $z \in \map {B_\delta} y$, then $\map d {y, z} < \delta$. So: {{begin-eqn}} {{eqn | l = \map d {x, z} | o = \le | r = \map d {x, y} + \map d {y, z} | c = {{Metric-space-axiom|2}} }} {{eqn | o = < | r = \map...
Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]]. Let $x, y$ be [[Definition:Element|points]] of $A$. Then: :$\epsilon-\delta \ge \map d {x, y} \implies \map {B_\delta} y \subseteq \map {B_\epsilon} x$ where $\map {B_\epsilon} x$ denotes the [[Definition:Open Ball of Metric Space|open $\epsilon...
Let $\epsilon - \delta \ge \map d {x, y}$. Then $\epsilon \ge \map d {x, y} + \delta$. If $z \in \map {B_\delta} y$, then $\map d {y, z} < \delta$. So: {{begin-eqn}} {{eqn | l = \map d {x, z} | o = \le | r = \map d {x, y} + \map d {y, z} | c = {{Metric-space-axiom|2}} }} {{eqn | o = < | r = \...
Open Ball is Subset of Open Ball
https://proofwiki.org/wiki/Open_Ball_is_Subset_of_Open_Ball
https://proofwiki.org/wiki/Open_Ball_is_Subset_of_Open_Ball
[ "Open Balls" ]
[ "Definition:Metric Space", "Definition:Element", "Definition:Open Ball" ]
[]
proofwiki-11028
Pursuit Curve of Boat in River
Consider a straight river $R$ whose parallel banks are aligned with the $y$-axis and the line $x = c$ of a cartesian plane. Let the current of $R$ have a constant and uniform speed $a$ in the negative $y$ direction. Let a boat $B$ be launched from the point $\tuple {c, 0}$ and headed directly towards the origin with sp...
The components of the velocity of $b$ are: :$\dfrac {\d x} {\d t} = - b \cos \theta$ :$\dfrac {\d y} {\d t} = - a + b \sin \theta$ Hence: {{begin-eqn}} {{eqn | l = \frac {\d y} {\d x} | r = \frac {-a + b \sin \theta} {-b \cos \theta} | c = }} {{eqn | r = \frac {-a + b \paren {-\dfrac y {\sqrt {x^2 + y^2} }...
Consider a straight river $R$ whose [[Definition:Parallel Lines|parallel]] banks are aligned with the [[Definition:Y-Axis|$y$-axis]] and the [[Definition:Straight Line|line]] $x = c$ of a [[Definition:Cartesian Plane|cartesian plane]]. Let the current of $R$ have a constant and uniform [[Definition:Speed|speed]] $a$ i...
The [[Definition:Vector Component|components]] of the [[Definition:Velocity|velocity]] of $b$ are: :$\dfrac {\d x} {\d t} = - b \cos \theta$ :$\dfrac {\d y} {\d t} = - a + b \sin \theta$ Hence: {{begin-eqn}} {{eqn | l = \frac {\d y} {\d x} | r = \frac {-a + b \sin \theta} {-b \cos \theta} | c = }} {{eqn ...
Pursuit Curve of Boat in River
https://proofwiki.org/wiki/Pursuit_Curve_of_Boat_in_River
https://proofwiki.org/wiki/Pursuit_Curve_of_Boat_in_River
[ "Pursuit Curves" ]
[ "Definition:Parallel (Geometry)/Lines", "Definition:Axis/Y-Axis", "Definition:Line/Straight Line", "Definition:Cartesian Plane", "Definition:Speed", "Definition:Negative/Real Number", "Definition:Coordinate System/Origin", "Definition:Speed" ]
[ "Definition:Vector Quantity/Component", "Definition:Velocity", "Definition:Homogeneous Differential Equation", "Solution to Homogeneous Differential Equation", "Primitive of Reciprocal of Root of x squared plus a squared" ]
proofwiki-11029
Current in Electric Circuit/L, R, C in Series
Consider the electronic circuit $K$ consisting of: :a resistance $R$ :an inductance $L$ :a capacitance $C$ in series with a source of electromotive force $E$ which is a function of time $t$. :File:CircuitRLCseries.png The electric current $I$ in $K$ is given by the linear second order ODE: :$L \dfrac {\d^2 I} {\d t^2} ...
Let: :$E_L$ be the drop in electromotive force across $L$ :$E_R$ be the drop in electromotive force across $R$ :$E_C$ be the drop in electromotive force across $C$. From Kirchhoff's Voltage Law: :$E - E_L - E_R - E_C = 0$ From Ohm's Law: :$E_R = R I$ From Drop in EMF caused by Inductance is proportional to Rate of Chan...
Consider the [[Definition:Electronic Circuit|electronic circuit]] $K$ consisting of: :a [[Definition:Resistance|resistance]] $R$ :an [[Definition:Inductance|inductance]] $L$ :a [[Definition:Capacitance|capacitance]] $C$ in [[Definition:Series (Electronics)|series]] with a source of [[Definition:Electromotive Force|ele...
Let: :$E_L$ be the drop in [[Definition:Electromotive Force|electromotive force]] across $L$ :$E_R$ be the drop in [[Definition:Electromotive Force|electromotive force]] across $R$ :$E_C$ be the drop in [[Definition:Electromotive Force|electromotive force]] across $C$. From [[Kirchhoff's Voltage Law]]: :$E - E_L - E_R...
Current in Electric Circuit/L, R, C in Series
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R,_C_in_Series
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R,_C_in_Series
[ "Electronics" ]
[ "Definition:Electronic Circuit", "Definition:Resistance", "Definition:Inductance", "Definition:Capacitance", "Definition:Series (Electronics)", "Definition:Electromotive Force", "Definition:Function", "Definition:Time", "File:CircuitRLCseries.png", "Definition:Electric Current", "Definition:Line...
[ "Definition:Electromotive Force", "Definition:Electromotive Force", "Definition:Electromotive Force", "Kirchhoff's Voltage Law", "Ohm's Law", "Drop in EMF caused by Inductance is proportional to Rate of Change of Current", "Drop in EMF caused by Capacitance is proportional to Accumulated Charge", "Def...
proofwiki-11030
Electric Charge in Electric Circuit/L, R, C in Series
Consider the electronic circuit $K$ consisting of: :a resistance $R$ :an inductance $L$ :a capacitance $C$ in series with a source of electromotive force $E$ which is a function of time $t$. :File:CircuitRLCseries.png The electric charge $Q$ in $K$ is given by the linear second order ODE: :$L \dfrac {\d^2 Q} {\d t^2} +...
Let: :$E_L$ be the drop in electromotive force across $L$ :$E_R$ be the drop in electromotive force across $R$ :$E_C$ be the drop in electromotive force across $C$. From Kirchhoff's Voltage Law: :$E - E_L - E_R - E_C = 0$ From Ohm's Law: :$E_R = R I$ From Drop in EMF caused by Inductance is proportional to Rate of Chan...
Consider the [[Definition:Electronic Circuit|electronic circuit]] $K$ consisting of: :a [[Definition:Resistance|resistance]] $R$ :an [[Definition:Inductance|inductance]] $L$ :a [[Definition:Capacitance|capacitance]] $C$ in [[Definition:Series (Electronics)|series]] with a source of [[Definition:Electromotive Force|ele...
Let: :$E_L$ be the drop in [[Definition:Electromotive Force|electromotive force]] across $L$ :$E_R$ be the drop in [[Definition:Electromotive Force|electromotive force]] across $R$ :$E_C$ be the drop in [[Definition:Electromotive Force|electromotive force]] across $C$. From [[Kirchhoff's Voltage Law]]: :$E - E_L - E_R...
Electric Charge in Electric Circuit/L, R, C in Series
https://proofwiki.org/wiki/Electric_Charge_in_Electric_Circuit/L,_R,_C_in_Series
https://proofwiki.org/wiki/Electric_Charge_in_Electric_Circuit/L,_R,_C_in_Series
[ "Electronics" ]
[ "Definition:Electronic Circuit", "Definition:Resistance", "Definition:Inductance", "Definition:Capacitance", "Definition:Series (Electronics)", "Definition:Electromotive Force", "Definition:Function", "Definition:Time", "File:CircuitRLCseries.png", "Definition:Electric Charge", "Definition:Linea...
[ "Definition:Electromotive Force", "Definition:Electromotive Force", "Definition:Electromotive Force", "Kirchhoff's Voltage Law", "Ohm's Law", "Drop in EMF caused by Inductance is proportional to Rate of Change of Current", "Drop in EMF caused by Capacitance is proportional to Accumulated Charge", "Def...
proofwiki-11031
Current in Electric Circuit/L, R in Series
Consider the electronic circuit $K$ consisting of: :a resistance $R$ :an inductance $L$ in series with a source of electromotive force $E$ which is a function of time $t$. :File:CircuitRLseries.png The electric current $I$ in $K$ is given by the linear first order ODE: :$L \dfrac {\d I} {\d t} + R I = E$
Let: :$E_L$ be the drop in electromotive force across $L$ :$E_R$ be the drop in electromotive force across $R$ From Kirchhoff's Voltage Law: :$E - E_L - E_R = 0$ From Ohm's Law: :$E_R = R I$ From Drop in EMF caused by Inductance is proportional to Rate of Change of Current: :$E_L = L \dfrac {\d I} {\d t}$ Thus: :$E - L...
Consider the [[Definition:Electronic Circuit|electronic circuit]] $K$ consisting of: :a [[Definition:Resistance|resistance]] $R$ :an [[Definition:Inductance|inductance]] $L$ in [[Definition:Series (Electronics)|series]] with a source of [[Definition:Electromotive Force|electromotive force]] $E$ which is a [[Definition...
Let: :$E_L$ be the drop in [[Definition:Electromotive Force|electromotive force]] across $L$ :$E_R$ be the drop in [[Definition:Electromotive Force|electromotive force]] across $R$ From [[Kirchhoff's Voltage Law]]: :$E - E_L - E_R = 0$ From [[Ohm's Law]]: :$E_R = R I$ From [[Drop in EMF caused by Inductance is propo...
Current in Electric Circuit/L, R in Series
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series
[ "Electronics", "Decay Equation" ]
[ "Definition:Electronic Circuit", "Definition:Resistance", "Definition:Inductance", "Definition:Series (Electronics)", "Definition:Electromotive Force", "Definition:Function", "Definition:Time", "File:CircuitRLseries.png", "Definition:Electric Current", "Definition:Linear First Order Ordinary Diffe...
[ "Definition:Electromotive Force", "Definition:Electromotive Force", "Kirchhoff's Voltage Law", "Ohm's Law", "Drop in EMF caused by Inductance is proportional to Rate of Change of Current" ]
proofwiki-11032
Current in Electric Circuit/L, R in Series/Constant EMF at t = 0
Let the electric current flowing in $K$ at time $t = 0$ be $I_0$. Let a constant EMF $E_0$ be imposed upon $K$ at time $t = 0$. The electric current $I$ in $K$ is given by the equation: :$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$
From Electric Current in Electric Circuit: L, R in Series: :$L \dfrac {\d I} {\d t} + R I = E_0$ defines the behaviour of $I$. This can be written as: :$(1): \quad \dfrac {\d I} {\d t} = \dfrac R L \paren {\dfrac {E_0} R - I}$ $(1)$ is in the form: :$\dfrac {\d y} {\d x} = k \paren {y_a - y}$ where: :$k \in \R: k > 0$ ...
Let the [[Definition:Electric Current|electric current]] flowing in $K$ at [[Definition:Time|time]] $t = 0$ be $I_0$. Let a constant [[Definition:Electromotive Force|EMF]] $E_0$ be imposed upon $K$ at [[Definition:Time|time]] $t = 0$. The [[Definition:Electric Current|electric current]] $I$ in $K$ is given by the equ...
From [[Current in Electric Circuit/L, R in Series|Electric Current in Electric Circuit: L, R in Series]]: :$L \dfrac {\d I} {\d t} + R I = E_0$ defines the behaviour of $I$. This can be written as: :$(1): \quad \dfrac {\d I} {\d t} = \dfrac R L \paren {\dfrac {E_0} R - I}$ $(1)$ is in the form: :$\dfrac {\d y} {\d x}...
Current in Electric Circuit/L, R in Series/Constant EMF at t = 0
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Constant_EMF_at_t_=_0
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Constant_EMF_at_t_=_0
[ "Electronics" ]
[ "Definition:Electric Current", "Definition:Time", "Definition:Electromotive Force", "Definition:Time", "Definition:Electric Current" ]
[ "Current in Electric Circuit/L, R in Series", "Decay Equation", "Definition:Differential Equation/Solution/Particular Solution", "Definition:Differential Equation/Solution/Particular Solution" ]
proofwiki-11033
Current in Electric Circuit/L, R in Series/Constant EMF at t = 0/Corollary 1
Let the electric current flowing in $K$ at time $t = 0$ be $I_0$. Let a constant EMF $E_0$ be imposed upon $K$ at time $t = 0$. After a sufficiently long time, the electric current $I$ in $K$ is given by the equation: :$E_0 = R I$
From Electric Current in Electric Circuit: L, R in Series: Constant EMF at $t = 0$: :$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$ We have that: :$\ds \lim_{t \mathop \to \infty} e^{-R t / L} \to 0$ and so: :$\ds \lim_{t \mathop \to \infty} I \to \dfrac {E_0} R$ Hence the result. {{qed}}
Let the [[Definition:Electric Current|electric current]] flowing in $K$ at [[Definition:Time|time]] $t = 0$ be $I_0$. Let a constant [[Definition:Electromotive Force|EMF]] $E_0$ be imposed upon $K$ at [[Definition:Time|time]] $t = 0$. After a sufficiently long time, the [[Definition:Electric Current|electric current]...
From [[Current in Electric Circuit/L, R in Series/Constant EMF at t = 0|Electric Current in Electric Circuit: L, R in Series: Constant EMF at $t = 0$]]: :$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$ We have that: :$\ds \lim_{t \mathop \to \infty} e^{-R t / L} \to 0$ and so: :$\ds \lim_{t \mathop ...
Current in Electric Circuit/L, R in Series/Constant EMF at t = 0/Corollary 1
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Constant_EMF_at_t_=_0/Corollary_1
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Constant_EMF_at_t_=_0/Corollary_1
[ "Electronics" ]
[ "Definition:Electric Current", "Definition:Time", "Definition:Electromotive Force", "Definition:Time", "Definition:Electric Current" ]
[ "Current in Electric Circuit/L, R in Series/Constant EMF at t = 0" ]
proofwiki-11034
Current in Electric Circuit/L, R in Series/Constant EMF at t = 0/Corollary 2
Let the electric current flowing in $K$ at time $t = 0$ be $0$. Let a constant EMF $E_0$ be imposed upon $K$ at time $t = 0$. The electric current $I$ in $K$ is given by the equation: :$I = \dfrac {E_0} R \paren {1 - e^{-R t / L} }$
From Electric Current in Electric Circuit: L, R in Series: Constant EMF at $t = 0$: :$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$ Setting $I_0 = 0$: {{begin-eqn}} {{eqn | l = I | r = \dfrac {E_0} R + \paren {0 - \dfrac {E_0} R} e^{-R t / L} | c = }} {{eqn | r = \dfrac {E_0} R \paren {1...
Let the [[Definition:Electric Current|electric current]] flowing in $K$ at [[Definition:Time|time]] $t = 0$ be $0$. Let a constant [[Definition:Electromotive Force|EMF]] $E_0$ be imposed upon $K$ at [[Definition:Time|time]] $t = 0$. The [[Definition:Electric Current|electric current]] $I$ in $K$ is given by the equat...
From [[Current in Electric Circuit/L, R in Series/Constant EMF at t = 0|Electric Current in Electric Circuit: L, R in Series: Constant EMF at $t = 0$]]: :$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$ Setting $I_0 = 0$: {{begin-eqn}} {{eqn | l = I | r = \dfrac {E_0} R + \paren {0 - \dfrac {E_...
Current in Electric Circuit/L, R in Series/Constant EMF at t = 0/Corollary 2
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Constant_EMF_at_t_=_0/Corollary_2
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Constant_EMF_at_t_=_0/Corollary_2
[ "Electronics" ]
[ "Definition:Electric Current", "Definition:Time", "Definition:Electromotive Force", "Definition:Time", "Definition:Electric Current" ]
[ "Current in Electric Circuit/L, R in Series/Constant EMF at t = 0" ]
proofwiki-11035
Current in Electric Circuit/L, R in Series/Constant EMF at t = 0/Corollary 3
Let the electric current flowing in $K$ at time $t = 0$ be $I_0$. Let EMF imposed upon $K$ be zero. The electric current $I$ in $K$ is given by the equation: :$I = I_0 e^{-R t / L}$
From Electric Current in Electric Circuit: L, R in Series: Constant EMF at $t = 0$: :$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$ Setting $E_0 = 0$: {{begin-eqn}} {{eqn | l = I | r = \dfrac 0 R + \paren {I_0 - \dfrac 0 R} e^{-R t / L} | c = }} {{eqn | r = I_0 e^{-R t / L} | c = ...
Let the [[Definition:Electric Current|electric current]] flowing in $K$ at [[Definition:Time|time]] $t = 0$ be $I_0$. Let [[Definition:Electromotive Force|EMF]] imposed upon $K$ be zero. The [[Definition:Electric Current|electric current]] $I$ in $K$ is given by the equation: :$I = I_0 e^{-R t / L}$
From [[Current in Electric Circuit/L, R in Series/Constant EMF at t = 0|Electric Current in Electric Circuit: L, R in Series: Constant EMF at $t = 0$]]: :$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$ Setting $E_0 = 0$: {{begin-eqn}} {{eqn | l = I | r = \dfrac 0 R + \paren {I_0 - \dfrac 0 R} ...
Current in Electric Circuit/L, R in Series/Constant EMF at t = 0/Corollary 3
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Constant_EMF_at_t_=_0/Corollary_3
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Constant_EMF_at_t_=_0/Corollary_3
[ "Electronics" ]
[ "Definition:Electric Current", "Definition:Time", "Definition:Electromotive Force", "Definition:Electric Current" ]
[ "Current in Electric Circuit/L, R in Series/Constant EMF at t = 0" ]
proofwiki-11036
Current in Electric Circuit/L, R in Series/Exponentially Decaying EMF at t = 0
Let the electric current flowing in $K$ at time $t = 0$ be $I_0$. Let an EMF $E$ be imposed upon $K$ at time $t = 0$ defined by the equation: :$E = E_0 e^{-k t}$ The electric current $I$ in $K$ is given by the equation: :$I = \dfrac {E_0} {R - k L} e^{-k t} + \paren {I_0 - \dfrac {E_0} {R - k L} } e^{-R t / L}$
From Electric Current in Electric Circuit: L, R in Series: :$L \dfrac {\d I} {\d t} + R I = E_0 e^{-k t}$ defines the behaviour of $I$. This can be written as: :$(1): \quad \dfrac {\d I} {\d t} + \dfrac R L I = \dfrac {E_0} L e^{-k t}$ $(1)$ is a linear first order ODE in the form: :$\dfrac {\d I} {\d t} + \map P t I ...
Let the [[Definition:Electric Current|electric current]] flowing in $K$ at [[Definition:Time|time]] $t = 0$ be $I_0$. Let an [[Definition:Electromotive Force|EMF]] $E$ be imposed upon $K$ at [[Definition:Time|time]] $t = 0$ defined by the equation: :$E = E_0 e^{-k t}$ The [[Definition:Electric Current|electric curren...
From [[Current in Electric Circuit/L, R in Series|Electric Current in Electric Circuit: L, R in Series]]: :$L \dfrac {\d I} {\d t} + R I = E_0 e^{-k t}$ defines the behaviour of $I$. This can be written as: :$(1): \quad \dfrac {\d I} {\d t} + \dfrac R L I = \dfrac {E_0} L e^{-k t}$ $(1)$ is a [[Definition:Linear Fir...
Current in Electric Circuit/L, R in Series/Exponentially Decaying EMF at t = 0
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Exponentially_Decaying_EMF_at_t_=_0
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Exponentially_Decaying_EMF_at_t_=_0
[ "Electronics" ]
[ "Definition:Electric Current", "Definition:Time", "Definition:Electromotive Force", "Definition:Time", "Definition:Electric Current" ]
[ "Current in Electric Circuit/L, R in Series", "Definition:Linear First Order Ordinary Differential Equation", "Solution to Linear First Order Ordinary Differential Equation/Solution by Integrating Factor", "Definition:Differential Equation/Solution/General Solution" ]
proofwiki-11037
Current in Electric Circuit/L, R in Series/Sinusoidal EMF
Let the electric current flowing in $K$ at time $t = 0$ be $I_0$. Let an EMF $E$ be imposed upon $K$ at time $t = 0$ defined by the equation: :$E = E_0 \sin \omega t$ The electric current $I$ in $K$ is given by the equation: :$I = \dfrac {E_0} {\sqrt {R^2 - L^2 \omega^2} } \map \sin {\omega t - \alpha} + \paren {I_0 - ...
From Electric Current in Electric Circuit: L, R in Series: :$L \dfrac {\d I} {\d t} + R I = E_0 \sin \omega t$ defines the behaviour of $I$. This can be written as: :$(1): \quad \dfrac {\d I} {\d t} + \dfrac R L I = \dfrac {E_0} L \sin \omega t$ $(1)$ is a linear first order ODE in the form: :$\dfrac {\d I} {\d t} + \m...
Let the [[Definition:Electric Current|electric current]] flowing in $K$ at [[Definition:Time|time]] $t = 0$ be $I_0$. Let an [[Definition:Electromotive Force|EMF]] $E$ be imposed upon $K$ at [[Definition:Time|time]] $t = 0$ defined by the equation: :$E = E_0 \sin \omega t$ The [[Definition:Electric Current|electric c...
From [[Current in Electric Circuit/L, R in Series|Electric Current in Electric Circuit: L, R in Series]]: :$L \dfrac {\d I} {\d t} + R I = E_0 \sin \omega t$ defines the behaviour of $I$. This can be written as: :$(1): \quad \dfrac {\d I} {\d t} + \dfrac R L I = \dfrac {E_0} L \sin \omega t$ $(1)$ is a [[Definition:L...
Current in Electric Circuit/L, R in Series/Sinusoidal EMF
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Sinusoidal_EMF
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Sinusoidal_EMF
[ "Electronics" ]
[ "Definition:Electric Current", "Definition:Time", "Definition:Electromotive Force", "Definition:Time", "Definition:Electric Current" ]
[ "Current in Electric Circuit/L, R in Series", "Definition:Linear First Order Ordinary Differential Equation", "Solution to Linear First Order Ordinary Differential Equation/Solution by Integrating Factor", "Definition:Differential Equation/Solution/General Solution", "Primitive of Exponential of a x by Sine...
proofwiki-11038
Current in Electric Circuit/L, R in Series/Condition for Ohm's Law
Ohm's Law is satisfied by $K$ whenever the current $I$ is at a maximum or a minimum.
From Electric Current in Electric Circuit: L, R in Series: :$L \dfrac {\d I} {\d t} + R I = E$ defines the behaviour of $I$. Let $I$ be at a maximum or a minimum. Then from the Interior Extremum Theorem: :$\dfrac {\d I} {\d t} = 0$ and so: {{begin-eqn}} {{eqn | l = E | r = 0 + R I | c = }} {{eqn | r = R I ...
[[Ohm's Law]] is satisfied by $K$ whenever the [[Definition:Electric Current|current]] $I$ is at a [[Definition:Local Maximum|maximum]] or a [[Definition:Local Minimum|minimum]].
From [[Current in Electric Circuit/L, R in Series|Electric Current in Electric Circuit: L, R in Series]]: :$L \dfrac {\d I} {\d t} + R I = E$ defines the behaviour of $I$. Let $I$ be at a [[Definition:Local Maximum|maximum]] or a [[Definition:Local Minimum|minimum]]. Then from the [[Interior Extremum Theorem]]: :$\d...
Current in Electric Circuit/L, R in Series/Condition for Ohm's Law
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Condition_for_Ohm's_Law
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Condition_for_Ohm's_Law
[ "Electronics" ]
[ "Ohm's Law", "Definition:Electric Current", "Definition:Maximum Value of Real Function/Local", "Definition:Minimum Value of Real Function/Local" ]
[ "Current in Electric Circuit/L, R in Series", "Definition:Maximum Value of Real Function/Local", "Definition:Minimum Value of Real Function/Local", "Interior Extremum Theorem", "Ohm's Law" ]
proofwiki-11039
Current in Electric Circuit/L, R in Series/Minimum Current implies Increasing EMF
Let the current $I$ be at a minimum. Then the EMF $E$ is increasing.
From Electric Current in Electric Circuit: L, R in Series: :$L \dfrac {\d I} {\d t} + R I = E$ defines the behaviour of $I$. Taking the derivative: :$L \dfrac {\d^2 I} {\d t^2} + R \dfrac {\d I} {\d t} = \dfrac {\d E} {\d t}$ From Second Derivative of Real Function at Minimum: :$L \dfrac {\d^2 I} {\d t^2} \ge 0$ while ...
Let the [[Definition:Electric Current|current]] $I$ be at a [[Definition:Local Minimum|minimum]]. Then the [[Definition:Electromotive Force|EMF]] $E$ is [[Definition:Increasing|increasing]].
From [[Current in Electric Circuit/L, R in Series|Electric Current in Electric Circuit: L, R in Series]]: :$L \dfrac {\d I} {\d t} + R I = E$ defines the behaviour of $I$. Taking the [[Definition:Derivative|derivative]]: :$L \dfrac {\d^2 I} {\d t^2} + R \dfrac {\d I} {\d t} = \dfrac {\d E} {\d t}$ From [[Second Deriv...
Current in Electric Circuit/L, R in Series/Minimum Current implies Increasing EMF
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Minimum_Current_implies_Increasing_EMF
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Minimum_Current_implies_Increasing_EMF
[ "Electronics" ]
[ "Definition:Electric Current", "Definition:Minimum Value of Real Function/Local", "Definition:Electromotive Force", "Definition:Increasing" ]
[ "Current in Electric Circuit/L, R in Series", "Definition:Derivative", "Second Derivative of Real Function at Minimum", "Interior Extremum Theorem", "Increasing Function has Positive Derivative" ]
proofwiki-11040
Current in Electric Circuit/L, R in Series/Maximum Current implies Decreasing EMF
Let the current $I$ be at a maximum. Then the EMF $E$ is decreasing.
From Electric Current in Electric Circuit: L, R in Series: :$L \dfrac {\d I} {\d t} + R I = E$ defines the behaviour of $I$. Taking the derivative: :$L \dfrac {\d^2 I} {\d t^2} + R \dfrac {\d I} {\d t} = \dfrac {\d E} {\d t}$ From Second Derivative of Real Function at Minimum: :$L \dfrac {\d^2 I} {\d t^2} \le 0$ while ...
Let the [[Definition:Electric Current|current]] $I$ be at a [[Definition:Local Maximum|maximum]]. Then the [[Definition:Electromotive Force|EMF]] $E$ is [[Definition:Decreasing|decreasing]].
From [[Current in Electric Circuit/L, R in Series|Electric Current in Electric Circuit: L, R in Series]]: :$L \dfrac {\d I} {\d t} + R I = E$ defines the behaviour of $I$. Taking the [[Definition:Derivative|derivative]]: :$L \dfrac {\d^2 I} {\d t^2} + R \dfrac {\d I} {\d t} = \dfrac {\d E} {\d t}$ From [[Second Deriv...
Current in Electric Circuit/L, R in Series/Maximum Current implies Decreasing EMF
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Maximum_Current_implies_Decreasing_EMF
https://proofwiki.org/wiki/Current_in_Electric_Circuit/L,_R_in_Series/Maximum_Current_implies_Decreasing_EMF
[ "Electronics" ]
[ "Definition:Electric Current", "Definition:Maximum Value of Real Function/Local", "Definition:Electromotive Force", "Definition:Decreasing" ]
[ "Current in Electric Circuit/L, R in Series", "Definition:Derivative", "Second Derivative of Real Function at Minimum", "Interior Extremum Theorem", "Decreasing Function has Negative Derivative" ]
proofwiki-11041
Second Order ODE/y y'' = (y')^2
The second order ODE: :$(1): \quad y y' ' = \paren {y'}^2$ has the general solution: :$y = C_2 e^{C_1 x}$
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = y p \frac {\d p} {\d y} | r = p^2 | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto | l = y \frac {\d p} {\d y} | r = p | c = }} {{eqn ...
The [[Definition:Second Order ODE|second order ODE]]: :$(1): \quad y y' ' = \paren {y'}^2$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = C_2 e^{C_1 x}$
Using [[Solution of Second Order Differential Equation with Missing Independent Variable]], $(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = y p \frac {\d p} {\d y} | r = p^2 | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto | l = y \frac {\d p} {\d y} | r = p | c = }} {...
Second Order ODE/y y'' = (y')^2
https://proofwiki.org/wiki/Second_Order_ODE/y_y''_=_(y')^2
https://proofwiki.org/wiki/Second_Order_ODE/y_y''_=_(y')^2
[ "Examples of Second Order ODEs" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution of Second Order Differential Equation with Missing Independent Variable", "First Order ODE/x dy = k y dx", "First Order ODE/dy = k y dx" ]
proofwiki-11042
First Order ODE/(1 - x y) y' = y^2
The first order ODE: :$(1): \quad \paren {1 - x y} y' = y^2$ has the general solution: :$x y = \ln y + C$
Let $(1)$ be rearranged as: {{begin-eqn}} {{eqn | l = \frac {\d x} {\d y} | r = \frac {1 - x y} {y^2} | c = }} {{eqn | n = 2 | ll= \leadsto | l = \frac {\d x} {\d y} + \frac x y | r = \frac 1 {y^2} | c = }} {{end-eqn}} It can be seen that $(2)$ is a linear first order ODE in the fo...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad \paren {1 - x y} y' = y^2$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$x y = \ln y + C$
Let $(1)$ be rearranged as: {{begin-eqn}} {{eqn | l = \frac {\d x} {\d y} | r = \frac {1 - x y} {y^2} | c = }} {{eqn | n = 2 | ll= \leadsto | l = \frac {\d x} {\d y} + \frac x y | r = \frac 1 {y^2} | c = }} {{end-eqn}} It can be seen that $(2)$ is a [[Definition:Linear First Ord...
First Order ODE/(1 - x y) y' = y^2
https://proofwiki.org/wiki/First_Order_ODE/(1_-_x_y)_y'_=_y^2
https://proofwiki.org/wiki/First_Order_ODE/(1_-_x_y)_y'_=_y^2
[ "Examples of Linear First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Linear First Order Ordinary Differential Equation", "Solution to Linear First Order Ordinary Differential Equation/Solution by Integrating Factor" ]
proofwiki-11043
First Order ODE/(2 x + 3 y + 1) dx + (2 y - 3 x + 5) dy = 0
The first order ODE: :$(1): \quad \paren {2 x + 3 y + 1} \rd x + \paren {2 y - 3 x + 5} \rd y = 0$ has the general solution: :$3 \, \map \arctan {\dfrac {y + 1} {x - 1} } = \map \ln {\paren {y + 1}^2 + \paren {x - 1}^2} + C$
Rewriting $(1)$ as: :$\dfrac {\d y} {\d x} = -\dfrac {2 x + 3 y + 1} {-3 x + 2 y + 5}$ we note that it is in the form: :$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$ where: :$a e = -4 \ne b d = 9$ Hence we can use: :First Order ODE in form $y' = \map F {\dfrac {a x + b y + c} {d x + e y + f}...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad \paren {2 x + 3 y + 1} \rd x + \paren {2 y - 3 x + 5} \rd y = 0$ has the [[Definition:General Solution to Differential Equation|general solution]]: :$3 \, \map \arctan {\dfrac {y + 1} {x - 1} } = \map \ln {\paren {y + 1}^2 + \paren {x - 1}^2} + C$
Rewriting $(1)$ as: :$\dfrac {\d y} {\d x} = -\dfrac {2 x + 3 y + 1} {-3 x + 2 y + 5}$ we note that it is in the form: :$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$ where: :$a e = -4 \ne b d = 9$ Hence we can use: :[[First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f))|F...
First Order ODE/(2 x + 3 y + 1) dx + (2 y - 3 x + 5) dy = 0
https://proofwiki.org/wiki/First_Order_ODE/(2_x_+_3_y_+_1)_dx_+_(2_y_-_3_x_+_5)_dy_=_0
https://proofwiki.org/wiki/First_Order_ODE/(2_x_+_3_y_+_1)_dx_+_(2_y_-_3_x_+_5)_dy_=_0
[ "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f))", "Definition:Homogeneous Function/Real Space", "Definition:Homogeneous Function/Real Space/Degree", "Definition:Homogeneous Differential Equation", "Definition:Fraction/Numerator", "Definition:Fraction/Denominator", "Solution to Hom...
proofwiki-11044
First Order ODE/x y' = Root of (x^2 + y^2)
is a homogeneous differential equation with solution: :$3 x^2 \ln x = y \sqrt {x^2 + y^2} + x^2 \map \ln {y + \sqrt {x^2 + y^2} } + y^2 + C x^2$
We divide through by $x$ to show that $(1)$ is homogeneous: {{begin-eqn}} {{eqn | l = \frac {\d x} {\d y} | r = \frac {\sqrt {x^2 + y^2} } x | c = }} {{eqn | r = \sqrt {\frac {x^2 + y^2} {x^2} } | c = }} {{eqn | r = \sqrt {1 + \paren {\frac y x}^2} | c = }} {{end-eqn}} By Solution to Homogene...
is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]] with [[Definition:General Solution to Differential Equation|solution]]: :$3 x^2 \ln x = y \sqrt {x^2 + y^2} + x^2 \map \ln {y + \sqrt {x^2 + y^2} } + y^2 + C x^2$
We divide through by $x$ to show that $(1)$ is [[Definition:Homogeneous Differential Equation|homogeneous]]: {{begin-eqn}} {{eqn | l = \frac {\d x} {\d y} | r = \frac {\sqrt {x^2 + y^2} } x | c = }} {{eqn | r = \sqrt {\frac {x^2 + y^2} {x^2} } | c = }} {{eqn | r = \sqrt {1 + \paren {\frac y x}^2} ...
First Order ODE/x y' = Root of (x^2 + y^2)
https://proofwiki.org/wiki/First_Order_ODE/x_y'_=_Root_of_(x^2_+_y^2)
https://proofwiki.org/wiki/First_Order_ODE/x_y'_=_Root_of_(x^2_+_y^2)
[ "Examples of Homogeneous Differential Equation", "Examples of First Order ODEs" ]
[ "Definition:Homogeneous Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Homogeneous Differential Equation", "Solution to Homogeneous Differential Equation", "Primitive of Root of x squared plus a squared" ]
proofwiki-11045
Bernoulli's Equation/y^2 dx = (x^3 - x y) dy
The first order ODE: :$(1): \quad y^2 \rd x = \paren {x^3 - x y} \rd y$ has the general solution: :$3 y = 2 x^2 + C x^2 y^2$
Dividing $(1)$ by $y^2$ and rearranging: :$(2): \quad \dfrac {\d x} {\d y} + \dfrac x y = \dfrac {x^3} {y^2}$ It can be seen that $(2)$ is in the form: :$\dfrac {\d x} {\d y} + \map P y x = \map Q y x^n$ where: :$\map P y = \dfrac 1 y$ :$\map Q y = \dfrac 1 {y^2}$ :$n = 3$ and so is an example of Bernoulli's equation. ...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad y^2 \rd x = \paren {x^3 - x y} \rd y$ has the [[Definition:General Solution to Differential Equation|general solution]]: :$3 y = 2 x^2 + C x^2 y^2$
Dividing $(1)$ by $y^2$ and rearranging: :$(2): \quad \dfrac {\d x} {\d y} + \dfrac x y = \dfrac {x^3} {y^2}$ It can be seen that $(2)$ is in the form: :$\dfrac {\d x} {\d y} + \map P y x = \map Q y x^n$ where: :$\map P y = \dfrac 1 y$ :$\map Q y = \dfrac 1 {y^2}$ :$n = 3$ and so is an example of [[Definition:Bernou...
Bernoulli's Equation/y^2 dx = (x^3 - x y) dy
https://proofwiki.org/wiki/Bernoulli's_Equation/y^2_dx_=_(x^3_-_x_y)_dy
https://proofwiki.org/wiki/Bernoulli's_Equation/y^2_dx_=_(x^3_-_x_y)_dy
[ "Examples of Bernoulli's Equation" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Bernoulli's Equation", "Solution to Bernoulli's Equation", "Definition:Differential Equation/Solution/General Solution" ]
proofwiki-11046
First Order ODE/(x^2 y^3 + y) dx = (x^3 y^2 - x) dy
The first order ODE: :$(1): \quad \paren {x^2 y^3 + y} \rd x = \paren {x^3 y^2 - x} \rd y$ has the general solution: :$-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$
Let $(1)$ be expressed as: :$(2): \quad \paren {y + x^2 y^3} \rd x + \paren {x - x^3 y^2} \rd y = 0$ We note that $(2)$ is in the form: :$\map M {x, y} \rd x + \map N {x, y} \rd y = 0$ but is not exact. So, let: :$\map M {x, y} = y + x^2 y^3$ :$\map N {x, y} = x - x^3 y^2$ Let: :$\map P {x, y} = \dfrac {\partial M} {\p...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad \paren {x^2 y^3 + y} \rd x = \paren {x^3 y^2 - x} \rd y$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$
Let $(1)$ be expressed as: :$(2): \quad \paren {y + x^2 y^3} \rd x + \paren {x - x^3 y^2} \rd y = 0$ We note that $(2)$ is in the form: :$\map M {x, y} \rd x + \map N {x, y} \rd y = 0$ but is not [[Definition:Exact Differential Equation|exact]]. So, let: :$\map M {x, y} = y + x^2 y^3$ :$\map N {x, y} = x - x^3 y^2...
First Order ODE/(x^2 y^3 + y) dx = (x^3 y^2 - x) dy
https://proofwiki.org/wiki/First_Order_ODE/(x^2_y^3_+_y)_dx_=_(x^3_y^2_-_x)_dy
https://proofwiki.org/wiki/First_Order_ODE/(x^2_y^3_+_y)_dx_=_(x^3_y^2_-_x)_dy
[ "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Exact Differential Equation", "Integrating Factor for First Order ODE/Function of Product of Variables", "Definition:Integrating Factor", "Definition:Exact Differential Equation", "First Order ODE/(1 over x^3 y^2 + 1 over x) dx + (1 over x^2 y^3 - 1 over y) dy = 0", "Definition:Differential Eq...
proofwiki-11047
Second Order ODE/y y'' + (y')^2 - 2 y y' = 0
The second order ODE: :$y y' ' + \paren {y'}^2 - 2 y y' = 0$ has the general solution: :$y^2 = C_2 e^{2 x} + C_1$
Using Solution of Second Order Differential Equation with Missing Independent Variable: {{begin-eqn}} {{eqn | l = y p \frac {\d p} {\d y} + p^2 - 2 y p | r = 0 | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto | l = \frac {\d p} {\d y} + \frac p y | r = 2 | c = }} {{eqn | ll=...
The [[Definition:Second Order ODE|second order ODE]]: :$y y' ' + \paren {y'}^2 - 2 y y' = 0$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y^2 = C_2 e^{2 x} + C_1$
Using [[Solution of Second Order Differential Equation with Missing Independent Variable]]: {{begin-eqn}} {{eqn | l = y p \frac {\d p} {\d y} + p^2 - 2 y p | r = 0 | c = where $p = \dfrac {\d y} {\d x}$ }} {{eqn | ll= \leadsto | l = \frac {\d p} {\d y} + \frac p y | r = 2 | c = }} {{eqn ...
Second Order ODE/y y'' + (y')^2 - 2 y y' = 0
https://proofwiki.org/wiki/Second_Order_ODE/y_y''_+_(y')^2_-_2_y_y'_=_0
https://proofwiki.org/wiki/Second_Order_ODE/y_y''_+_(y')^2_-_2_y_y'_=_0
[ "Examples of Second Order ODEs" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution of Second Order Differential Equation with Missing Independent Variable", "Linear First Order ODE/y' + (y over x) = k x^n", "Solution to Separable Differential Equation", "Primitive of Function under its Derivative" ]
proofwiki-11048
Rationals are Everywhere Dense in Sorgenfrey Line
$\Q$ is everywhere dense in the Sorgenfrey line.
Let $T = \struct {\R, \tau}$ be the Sorgenfrey line. Define: :$\BB := \set {\hointr x y: x, y \in \R}$ where $\hointr x y$ denotes the right half-open real interval between $x$ and $y$. By definition of Sorgenfrey line: :$\BB$ is basis of $T$. By definition of subset: :$\Q^- \subseteq \R$ where $\Q^-$ denotes the topol...
$\Q$ is [[Definition:Everywhere Dense|everywhere dense]] in the [[Definition:Sorgenfrey Line|Sorgenfrey line]].
Let $T = \struct {\R, \tau}$ be the [[Definition:Sorgenfrey Line|Sorgenfrey line]]. Define: :$\BB := \set {\hointr x y: x, y \in \R}$ where $\hointr x y$ denotes the [[Definition:Right Half-Open Real Interval|right half-open real interval]] between $x$ and $y$. By definition of [[Definition:Sorgenfrey Line|Sorgenfrey...
Rationals are Everywhere Dense in Sorgenfrey Line
https://proofwiki.org/wiki/Rationals_are_Everywhere_Dense_in_Sorgenfrey_Line
https://proofwiki.org/wiki/Rationals_are_Everywhere_Dense_in_Sorgenfrey_Line
[ "Sorgenfrey Line" ]
[ "Definition:Everywhere Dense", "Definition:Sorgenfrey Line" ]
[ "Definition:Sorgenfrey Line", "Definition:Real Interval/Half-Open/Right", "Definition:Sorgenfrey Line", "Definition:Basis (Topology)/Analytic Basis", "Definition:Subset", "Definition:Closure (Topology)", "Definition:Set Equality", "Characterization of Closure by Basis", "Definition:Real Interval/Hal...
proofwiki-11049
Linear First Order ODE/x dy + y dx = x cosine x dx
The linear first order ODE: :$x \rd y + y \rd x = x \cos x \rd x$ has the general solution: :$x y = x \sin x + \cos x + C$
Rearranging: :$\dfrac {d y} {\d x} + \dfrac y x = \cos x$ This is a linear first order ODE in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x y = \dfrac 1 x$ :$\map Q x = \cos x$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x | r = \int \frac 1 x \rd x | c = }} {{eqn | r = ...
The [[Definition:Linear First Order ODE|linear first order ODE]]: :$x \rd y + y \rd x = x \cos x \rd x$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$x y = x \sin x + \cos x + C$
Rearranging: :$\dfrac {d y} {\d x} + \dfrac y x = \cos x$ This is a [[Definition:Linear First Order ODE|linear first order ODE]] in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x y = \dfrac 1 x$ :$\map Q x = \cos x$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x | r = \int \f...
Linear First Order ODE/x dy + y dx = x cosine x dx/Proof 1
https://proofwiki.org/wiki/Linear_First_Order_ODE/x_dy_+_y_dx_=_x_cosine_x_dx
https://proofwiki.org/wiki/Linear_First_Order_ODE/x_dy_+_y_dx_=_x_cosine_x_dx/Proof_1
[ "Examples of Linear First Order ODEs", "Linear First Order ODE/x dy + y dx = x cosine x dx" ]
[ "Definition:Linear First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Linear First Order Ordinary Differential Equation", "Solution to Linear First Order Ordinary Differential Equation/Solution by Integrating Factor", "Primitive of x by Cosine of a x" ]
proofwiki-11050
Linear First Order ODE/x dy + y dx = x cosine x dx
The linear first order ODE: :$x \rd y + y \rd x = x \cos x \rd x$ has the general solution: :$x y = x \sin x + \cos x + C$
{{begin-eqn}} {{eqn | l = x \dfrac {\d y} {\d x} + y | r = x \cos x | c = }} {{eqn | ll= \leadsto | l = x y | r = \int x \cos x \rd x + C | c = Linear First Order ODE: $x y' + y = \map f x$ }} {{eqn | r = x \sin x + \cos x + C | c = Primitive of $x \cos a x$ }} {{end-eqn}} {{qed}}
The [[Definition:Linear First Order ODE|linear first order ODE]]: :$x \rd y + y \rd x = x \cos x \rd x$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$x y = x \sin x + \cos x + C$
{{begin-eqn}} {{eqn | l = x \dfrac {\d y} {\d x} + y | r = x \cos x | c = }} {{eqn | ll= \leadsto | l = x y | r = \int x \cos x \rd x + C | c = [[Linear First Order ODE/x y' + y = f (x)|Linear First Order ODE: $x y' + y = \map f x$]] }} {{eqn | r = x \sin x + \cos x + C | c = [[Prim...
Linear First Order ODE/x dy + y dx = x cosine x dx/Proof 2
https://proofwiki.org/wiki/Linear_First_Order_ODE/x_dy_+_y_dx_=_x_cosine_x_dx
https://proofwiki.org/wiki/Linear_First_Order_ODE/x_dy_+_y_dx_=_x_cosine_x_dx/Proof_2
[ "Examples of Linear First Order ODEs", "Linear First Order ODE/x dy + y dx = x cosine x dx" ]
[ "Definition:Linear First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Linear First Order ODE/x y' + y = f (x)", "Primitive of x by Cosine of a x" ]
proofwiki-11051
First Order ODE/x y dy = x^2 dy + y^2 dx
The first order ODE: :$(1): \quad x y \rd y = x^2 \rd y + y^2 \rd x$ has the general solution: :$y = x \ln y + C x$
Let $(1)$ be rearranged as: :$(2): \quad \dfrac {\d x} {\d y} - \dfrac x y = -\dfrac {x^2} {y^2}$ It can be seen that $(2)$ is in the form: :$\dfrac {\d x} {\d y} + \map P y x = \map Q y x^n$ where: :$\map P y = -\dfrac 1 y$ :$\map Q y = -\dfrac 1 {y^2}$ :$n = 2$ and so is an example of Bernoulli's equation. By Solutio...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad x y \rd y = x^2 \rd y + y^2 \rd x$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = x \ln y + C x$
Let $(1)$ be rearranged as: :$(2): \quad \dfrac {\d x} {\d y} - \dfrac x y = -\dfrac {x^2} {y^2}$ It can be seen that $(2)$ is in the form: :$\dfrac {\d x} {\d y} + \map P y x = \map Q y x^n$ where: :$\map P y = -\dfrac 1 y$ :$\map Q y = -\dfrac 1 {y^2}$ :$n = 2$ and so is an example of [[Definition:Bernoulli's Equa...
First Order ODE/x y dy = x^2 dy + y^2 dx/Proof 1
https://proofwiki.org/wiki/First_Order_ODE/x_y_dy_=_x^2_dy_+_y^2_dx
https://proofwiki.org/wiki/First_Order_ODE/x_y_dy_=_x^2_dy_+_y^2_dx/Proof_1
[ "Examples of First Order ODEs", "Examples of Bernoulli's Equation", "Examples of Homogeneous Differential Equation", "First Order ODE/x y dy = x^2 dy + y^2 dx" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Bernoulli's Equation", "Solution to Bernoulli's Equation", "Definition:Differential Equation/Solution/General Solution" ]
proofwiki-11052
First Order ODE/x y dy = x^2 dy + y^2 dx
The first order ODE: :$(1): \quad x y \rd y = x^2 \rd y + y^2 \rd x$ has the general solution: :$y = x \ln y + C x$
Let $(1)$ be rearranged as: :$(2): \quad y^2 \rd x = \paren {x y - x^2} \rd y$ Let: :$\map M {x, y} = y^2$ :$\map N {x, y} = x y - x^2$ Put $t x, t y$ for $x, y$: {{begin-eqn}} {{eqn | l = \map M {t x, t y} | r = \paren {t y}^2 | c = }} {{eqn | r = t^2 \paren {y^2} | c = }} {{eqn | r = t^2 \map M {x...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad x y \rd y = x^2 \rd y + y^2 \rd x$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = x \ln y + C x$
Let $(1)$ be rearranged as: :$(2): \quad y^2 \rd x = \paren {x y - x^2} \rd y$ Let: :$\map M {x, y} = y^2$ :$\map N {x, y} = x y - x^2$ Put $t x, t y$ for $x, y$: {{begin-eqn}} {{eqn | l = \map M {t x, t y} | r = \paren {t y}^2 | c = }} {{eqn | r = t^2 \paren {y^2} | c = }} {{eqn | r = t^2 \map...
First Order ODE/x y dy = x^2 dy + y^2 dx/Proof 2
https://proofwiki.org/wiki/First_Order_ODE/x_y_dy_=_x^2_dy_+_y^2_dx
https://proofwiki.org/wiki/First_Order_ODE/x_y_dy_=_x^2_dy_+_y^2_dx/Proof_2
[ "Examples of First Order ODEs", "Examples of Bernoulli's Equation", "Examples of Homogeneous Differential Equation", "First Order ODE/x y dy = x^2 dy + y^2 dx" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Homogeneous Function/Real Space", "Definition:Homogeneous Function/Real Space/Degree", "Definition:Homogeneous Differential Equation", "Solution to Homogeneous Differential Equation", "Primitive of Power" ]
proofwiki-11053
First Order ODE/(exp x - 3 x^2 y^2) y' + y exp x = 2 x y^3
The first order ordinary differential equation: :$(1): \quad \paren {e^x - 3 x^2 y^2} y' + y e^x = 2 x y^3$ is an exact differential equation with solution: :$y e^x - x^2 y^3 = C$
Let $(1)$ be expressed as: :$\paren {y e^x - 2 x y^3} \rd x + \paren {e^x - 3 x^2 y^2} \rd y = 0$ Let: :$\map M {x, y} = y e^x - 2 x y^3$ :$\map N {x, y} = e^x - 3 x^2 y^2$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial M} {\partial y} | r = e^x - 6 x y^2 | c = }} {{eqn | l = \dfrac {\partial N} {\partia...
The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]: :$(1): \quad \paren {e^x - 3 x^2 y^2} y' + y e^x = 2 x y^3$ is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution to Differential Equation|solution]]: :$...
Let $(1)$ be expressed as: :$\paren {y e^x - 2 x y^3} \rd x + \paren {e^x - 3 x^2 y^2} \rd y = 0$ Let: :$\map M {x, y} = y e^x - 2 x y^3$ :$\map N {x, y} = e^x - 3 x^2 y^2$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial M} {\partial y} | r = e^x - 6 x y^2 | c = }} {{eqn | l = \dfrac {\partial N} {\pa...
First Order ODE/(exp x - 3 x^2 y^2) y' + y exp x = 2 x y^3
https://proofwiki.org/wiki/First_Order_ODE/(exp_x_-_3_x^2_y^2)_y'_+_y_exp_x_=_2_x_y^3
https://proofwiki.org/wiki/First_Order_ODE/(exp_x_-_3_x^2_y^2)_y'_+_y_exp_x_=_2_x_y^3
[ "Examples of Exact Differential Equation", "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Exact Differential Equation", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
proofwiki-11054
Second Order ODE/y'' + 2 x (y')^2 = 0
The second order ODE: :$(1): \quad y' ' + 2 x \paren {y'}^2 = 0$ has the general solution: :$C_1 \map \arctan {C_1 x} = y + C_2$
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable. Substitute $p$ for $y'$ in $(1)$ and rearranging: {{begin-eqn}} {{eqn | l = \dfrac {\d p} {\d x} | r = -2 x p^2 | c = }} {{eqn | ll= \leadsto | l = \int \frac {\d p} {p^2} | r = -2 \int x...
The [[Definition:Second Order ODE|second order ODE]]: :$(1): \quad y' ' + 2 x \paren {y'}^2 = 0$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$C_1 \map \arctan {C_1 x} = y + C_2$
The proof proceeds by using [[Solution of Second Order Differential Equation with Missing Dependent Variable]]. Substitute $p$ for $y'$ in $(1)$ and rearranging: {{begin-eqn}} {{eqn | l = \dfrac {\d p} {\d x} | r = -2 x p^2 | c = }} {{eqn | ll= \leadsto | l = \int \frac {\d p} {p^2} | r = -2 ...
Second Order ODE/y'' + 2 x (y')^2 = 0
https://proofwiki.org/wiki/Second_Order_ODE/y''_+_2_x_(y')^2_=_0
https://proofwiki.org/wiki/Second_Order_ODE/y''_+_2_x_(y')^2_=_0
[ "Examples of Second Order ODEs" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution of Second Order Differential Equation with Missing Dependent Variable", "Solution to Separable Differential Equation", "Solution to Separable Differential Equation", "Primitive of Reciprocal of x squared plus a squared/Arctangent Form" ]
proofwiki-11055
First Order ODE/(y over x^2) dx + (y - 1 over x) dy = 0
The first order ordinary differential equation: :$(1): \quad \dfrac y {x^2} \rd x + \paren {y - \dfrac 1 x} \rd y = 0$ is an exact differential equation with solution: :$\dfrac {y^2} 2 - \dfrac y x = C$
Let $M$ and $N$ be defined as: :$\map M {x, y} = \dfrac y {x^2}$ :$\map N {x, y} = y - \dfrac 1 x$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y} | r = \frac 1 {x^2} | c = }} {{eqn | l = \dfrac {\partial N} {\partial x} | r = \dfrac 1 {x^2} | c = }} {{end-eqn}} Thus $\dfrac {\...
The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]: :$(1): \quad \dfrac y {x^2} \rd x + \paren {y - \dfrac 1 x} \rd y = 0$ is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution to Differential Equation|solu...
Let $M$ and $N$ be defined as: :$\map M {x, y} = \dfrac y {x^2}$ :$\map N {x, y} = y - \dfrac 1 x$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y} | r = \frac 1 {x^2} | c = }} {{eqn | l = \dfrac {\partial N} {\partial x} | r = \dfrac 1 {x^2} | c = }} {{end-eqn}} Thus $\dfr...
First Order ODE/(y over x^2) dx + (y - 1 over x) dy = 0
https://proofwiki.org/wiki/First_Order_ODE/(y_over_x^2)_dx_+_(y_-_1_over_x)_dy_=_0
https://proofwiki.org/wiki/First_Order_ODE/(y_over_x^2)_dx_+_(y_-_1_over_x)_dy_=_0
[ "Examples of Exact Differential Equation", "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Exact Differential Equation", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Category:Examples of Exact Differential Equation", ...
proofwiki-11056
Focal Property of Parabola
Let $\KK$ be a parabola whose focus is at $F$ and whose directrix is $D$. Let $P$ be an arbitrary point on $\KK$. The tangent to $\KK$ at $P$ bisects the angle between the straight line $PF$ and the straight line through $P$ perpendicular to $D$. :500px
500px Let $BQ$ be the straight line constructed perpendicular to $D$ passing through $P$. Let $AT$ be the tangent to $\KK$ at $P$. Let $\angle APF$ be constructed equal to $\angle BPT$ where $F$ is on the axis of $\KK$. It is to be shown that $F$ is the focus of $\KK$. From Two Straight Lines make Equal Opposite Angles...
Let $\KK$ be a [[Definition:Parabola|parabola]] whose [[Definition:Focus of Parabola|focus]] is at $F$ and whose [[Definition:Directrix of Parabola|directrix]] is $D$. Let $P$ be an arbitrary [[Definition:Point|point]] on $\KK$. The [[Definition:Tangent Line|tangent]] to $\KK$ at $P$ [[Definition:Angle Bisector|bise...
[[File:Focal-property-of-parabola.png|500px]] Let $BQ$ be the [[Definition:Straight Line|straight line]] constructed [[Definition:Perpendicular|perpendicular]] to $D$ passing through $P$. Let $AT$ be the [[Definition:Tangent Line|tangent]] to $\KK$ at $P$. Let $\angle APF$ be constructed equal to $\angle BPT$ where...
Focal Property of Parabola
https://proofwiki.org/wiki/Focal_Property_of_Parabola
https://proofwiki.org/wiki/Focal_Property_of_Parabola
[ "Focal Property of Parabola", "Foci of Parabolas", "Parabolas" ]
[ "Definition:Parabola", "Definition:Parabola/Focus", "Definition:Parabola/Directrix", "Definition:Point", "Definition:Tangent Line", "Definition:Angle Bisector", "Definition:Angle", "Definition:Line/Straight Line", "Definition:Line/Straight Line", "Definition:Right Angle/Perpendicular", "File:Par...
[ "File:Focal-property-of-parabola.png", "Definition:Line/Straight Line", "Definition:Right Angle/Perpendicular", "Definition:Tangent Line", "Definition:Parabola/Axis", "Definition:Parabola/Focus", "Two Straight Lines make Equal Opposite Angles", "Definition:Triangle (Geometry)", "Triangle Side-Angle-...
proofwiki-11057
First Order ODE/(3 x^2 over y^4 - 1 over y^2) dy - 2 x over y^3 dx = 0
The first order ODE: :$(1): \quad \paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y - \dfrac {2 x} {y^3} \rd x = 0$ is an exact differential equation with general solution: :$\dfrac 1 y - \dfrac {x^2} {y^3} = C$
Let $M$ and $N$ be defined as: :$\map M {x, y} = -\dfrac {2 x} {y^3}$ :$\map N {x, y} = \dfrac {3 x^2} {y^4} - \dfrac 1 {y^2}$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y} | r = \frac {6 x} {2 y^4} | c = }} {{eqn | l = \dfrac {\partial N} {\partial x} | r = \dfrac {6 x} {y^4} ...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad \paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y - \dfrac {2 x} {y^3} \rd x = 0$ is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution to Differential Equation|general solution]]: :$\dfrac 1...
Let $M$ and $N$ be defined as: :$\map M {x, y} = -\dfrac {2 x} {y^3}$ :$\map N {x, y} = \dfrac {3 x^2} {y^4} - \dfrac 1 {y^2}$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y} | r = \frac {6 x} {2 y^4} | c = }} {{eqn | l = \dfrac {\partial N} {\partial x} | r = \dfrac {6 x} {y^4} ...
First Order ODE/(3 x^2 over y^4 - 1 over y^2) dy - 2 x over y^3 dx = 0
https://proofwiki.org/wiki/First_Order_ODE/(3_x^2_over_y^4_-_1_over_y^2)_dy_-_2_x_over_y^3_dx_=_0
https://proofwiki.org/wiki/First_Order_ODE/(3_x^2_over_y^4_-_1_over_y^2)_dy_-_2_x_over_y^3_dx_=_0
[ "Examples of Exact Differential Equation", "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Exact Differential Equation", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Category:Examples of Exact Differential Equation", ...
proofwiki-11058
First Order ODE/(y - 1 over x) dx + (x - y) dy = 0
The first order ODE: :$(1): \quad \paren {y - \dfrac 1 x} \rd x + \paren {x - y} \rd y = 0$ is an exact differential equation with solution: :$x y - \ln x - \dfrac {y^2} 2 + C$
Let $M$ and $N$ be defined as: :$\map M {x, y} = y - \dfrac 1 x$ :$\map N {x, y} = x - y$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y} | r = 1 | c = }} {{eqn | l = \dfrac {\partial N} {\partial x} | r = 1 | c = }} {{end-eqn}} Thus $\dfrac {\partial M} {\partial y} = \dfrac {...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad \paren {y - \dfrac 1 x} \rd x + \paren {x - y} \rd y = 0$ is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution to Differential Equation|solution]]: :$x y - \ln x - \dfrac {y^2} 2 + C$
Let $M$ and $N$ be defined as: :$\map M {x, y} = y - \dfrac 1 x$ :$\map N {x, y} = x - y$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y} | r = 1 | c = }} {{eqn | l = \dfrac {\partial N} {\partial x} | r = 1 | c = }} {{end-eqn}} Thus $\dfrac {\partial M} {\partial y} = \df...
First Order ODE/(y - 1 over x) dx + (x - y) dy = 0
https://proofwiki.org/wiki/First_Order_ODE/(y_-_1_over_x)_dx_+_(x_-_y)_dy_=_0
https://proofwiki.org/wiki/First_Order_ODE/(y_-_1_over_x)_dx_+_(x_-_y)_dy_=_0
[ "Examples of Exact Differential Equation", "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Exact Differential Equation", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Category:Examples of Exact Differential Equation", ...
proofwiki-11059
First Order ODE/1 over x^3 y^2 dx + (1 over x^2 y^3 + 3 y) dy = 0
The first order ODE: :$(1): \quad \dfrac 1 {x^3 y^2} \rd x + \paren {\dfrac 1 {x^2 y^3} + 3 y} \rd y = 0$ is an exact differential equation with solution: :$-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$
Let $M$ and $N$ be defined as: :$\map M {x, y} = \dfrac 1 {x^3 y^2}$ :$\map N {x, y} = \dfrac 1 {x^2 y^3} + 3 y$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y} | r = -\frac 1 {x^3 y^2} | c = }} {{eqn | l = \dfrac {\partial N} {\partial x} | r = -\frac 1 {x^3 y^2} | c = }} {{en...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad \dfrac 1 {x^3 y^2} \rd x + \paren {\dfrac 1 {x^2 y^3} + 3 y} \rd y = 0$ is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution to Differential Equation|solution]]: :$-\dfrac 1 {2 x^2 y^2} + \dfrac ...
Let $M$ and $N$ be defined as: :$\map M {x, y} = \dfrac 1 {x^3 y^2}$ :$\map N {x, y} = \dfrac 1 {x^2 y^3} + 3 y$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y} | r = -\frac 1 {x^3 y^2} | c = }} {{eqn | l = \dfrac {\partial N} {\partial x} | r = -\frac 1 {x^3 y^2} | c = }} ...
First Order ODE/1 over x^3 y^2 dx + (1 over x^2 y^3 + 3 y) dy = 0
https://proofwiki.org/wiki/First_Order_ODE/1_over_x^3_y^2_dx_+_(1_over_x^2_y^3_+_3_y)_dy_=_0
https://proofwiki.org/wiki/First_Order_ODE/1_over_x^3_y^2_dx_+_(1_over_x^2_y^3_+_3_y)_dy_=_0
[ "Examples of Exact Differential Equation", "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Category:Examples of Exact Differential Equation", "Category:Examples of First Order ODEs" ]
proofwiki-11060
Linear First Order ODE/y' - (y over x) = 3 x
The linear first order ODE: :$(1): \quad \dfrac {\d y} {\d x} - \dfrac y x = 3 x$ has the general solution: :$\dfrac y x = 3 x + C$ or: :$y = 3 x^2 + C x$
This is an instance of Linear First Order ODE: $y' - \dfrac y x = k x$. Its solution is: :$\dfrac y x = k x + C$ or: :$y = k x^2 + C x$ from which the solution to $(1)$ is found by substituting $3$ for $k$. {{qed}} Category:Examples of Linear First Order ODEs 5zl5xx9teqim55grzediaigsppbf6nk
The [[Definition:Linear First Order ODE|linear first order ODE]]: :$(1): \quad \dfrac {\d y} {\d x} - \dfrac y x = 3 x$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$\dfrac y x = 3 x + C$ or: :$y = 3 x^2 + C x$
This is an instance of [[Linear First Order ODE/y' - (y over x) = k x|Linear First Order ODE: $y' - \dfrac y x = k x$]]. Its [[Definition:General Solution of Differential Equation|solution]] is: :$\dfrac y x = k x + C$ or: :$y = k x^2 + C x$ from which the [[Definition:General Solution of Differential Equation|soluti...
Linear First Order ODE/y' - (y over x) = 3 x
https://proofwiki.org/wiki/Linear_First_Order_ODE/y'_-_(y_over_x)_=_3_x
https://proofwiki.org/wiki/Linear_First_Order_ODE/y'_-_(y_over_x)_=_3_x
[ "Examples of Linear First Order ODEs" ]
[ "Definition:Linear First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Linear First Order ODE/y' - (y over x) = k x", "Definition:Differential Equation/Solution/General Solution", "Definition:Differential Equation/Solution/General Solution", "Category:Examples of Linear First Order ODEs" ]
proofwiki-11061
Characterization of Closure by Basis
Let $T = \struct {S, \tau}$ be a topological space. Let $\BB \subseteq \tau$ be a basis. Let $A$ be a subset of $S$. Let $x$ be a point of $S$. Then $x \in A^-$ {{iff}}: :$\forall U \in \BB: x \in U \implies A \cap U \ne \O$ where: :$A^-$ denotes the closure of $A$
=== Sufficient Condition === Let $x \in A^-$. Let $U \in \BB$. By definition of basis, $U$ is an open set of $T$. Thus from Condition for Point being in Closure: :if $x \in U$ then $A \cap U \ne \O$. {{qed|lemma}}
Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]]. Let $\BB \subseteq \tau$ be a [[Definition:Analytic Basis|basis]]. Let $A$ be a [[Definition:Subset|subset]] of $S$. Let $x$ be a point of $S$. Then $x \in A^-$ {{iff}}: :$\forall U \in \BB: x \in U \implies A \cap U \ne \O$ wher...
=== Sufficient Condition === Let $x \in A^-$. Let $U \in \BB$. By definition of [[Definition:Analytic Basis|basis]], $U$ is an [[Definition:Open Set (Topology)|open set of $T$]]. Thus from [[Condition for Point being in Closure]]: :if $x \in U$ then $A \cap U \ne \O$. {{qed|lemma}}
Characterization of Closure by Basis
https://proofwiki.org/wiki/Characterization_of_Closure_by_Basis
https://proofwiki.org/wiki/Characterization_of_Closure_by_Basis
[ "Set Closures" ]
[ "Definition:Topological Space", "Definition:Basis (Topology)/Analytic Basis", "Definition:Subset", "Definition:Closure (Topology)" ]
[ "Definition:Basis (Topology)/Analytic Basis", "Definition:Open Set/Topology", "Condition for Point being in Closure", "Condition for Point being in Closure", "Definition:Open Set/Topology", "Definition:Open Set/Topology", "Definition:Basis (Topology)/Analytic Basis" ]
proofwiki-11062
Sorgenfrey Line is Separable
The Sorgenfrey line is separable.
{{Recall|Separable Space|separable space}} {{:Definition:Separable Space}} By Rationals are Everywhere Dense in Sorgenfrey Line: :$\Q$ is dense in the Sorgenfrey line. By Rational Numbers are Countably Infinite: :$\Q$ is countable. Thus by definition: :The Sorgenfrey line is separable. {{qed}}
The [[Definition:Sorgenfrey Line|Sorgenfrey line]] is [[Definition:Separable Space|separable]].
{{Recall|Separable Space|separable space}} {{:Definition:Separable Space}} By [[Rationals are Everywhere Dense in Sorgenfrey Line]]: :$\Q$ is [[Definition:Everywhere Dense|dense]] in the [[Definition:Sorgenfrey Line|Sorgenfrey line]]. By [[Rational Numbers are Countably Infinite]]: :$\Q$ is [[Definition:Countable Set...
Sorgenfrey Line is Separable
https://proofwiki.org/wiki/Sorgenfrey_Line_is_Separable
https://proofwiki.org/wiki/Sorgenfrey_Line_is_Separable
[ "Sorgenfrey Line", "Examples of Separable Spaces" ]
[ "Definition:Sorgenfrey Line", "Definition:Separable Space" ]
[ "Rationals are Everywhere Dense in Sorgenfrey Line", "Definition:Everywhere Dense", "Definition:Sorgenfrey Line", "Rational Numbers are Countably Infinite", "Definition:Countable Set", "Definition:Sorgenfrey Line", "Definition:Separable Space" ]
proofwiki-11063
First Order ODE/y dx + x dy = 0
The first order ODE: :$(1): \quad y \rd x + x \rd y = 0$ has the general solution: :$x y = C$
$(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = x \rd y | r = -y \rd x | c = }} {{eqn | ll= \leadsto | l = y | r = C x^{-1} | c = First Order ODE: $x \rd y = k y \rd x$ }} {{eqn | ll= \leadsto | l = x y | r = C | c = multiplying through by $x$ }} {{end-eqn}} {{qed}}...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad y \rd x + x \rd y = 0$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$x y = C$
$(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = x \rd y | r = -y \rd x | c = }} {{eqn | ll= \leadsto | l = y | r = C x^{-1} | c = [[First Order ODE/x dy = k y dx|First Order ODE: $x \rd y = k y \rd x$]] }} {{eqn | ll= \leadsto | l = x y | r = C | c = multiplying th...
First Order ODE/y dx + x dy = 0
https://proofwiki.org/wiki/First_Order_ODE/y_dx_+_x_dy_=_0
https://proofwiki.org/wiki/First_Order_ODE/y_dx_+_x_dy_=_0
[ "Examples of First Order ODEs", "Examples of Solutions to Separable Differential Equation" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "First Order ODE/x dy = k y dx", "Category:Examples of First Order ODEs", "Category:Examples of Solutions to Separable Differential Equation" ]
proofwiki-11064
First Order ODE/y dy = k dx
Let $k \in \R$ be a real number. The first order ODE: :$y \, \dfrac {\d y} {\d x} = k$ has the general solution: :$y^2 = 2 k x + C$
{{begin-eqn}} {{eqn | l = y \, \dfrac {\d y} {\d x} | r = k | c = }} {{eqn | ll= \leadsto | l = \int y \rd y | r = \int k \rd x | c = Solution to Separable Differential Equation }} {{eqn | ll= \leadsto | l = \frac {y^2} 2 | r = k x + \frac C 2 | c = }} {{eqn | ll= \lead...
Let $k \in \R$ be a [[Definition:Real Number|real number]]. The [[Definition:First Order ODE|first order ODE]]: :$y \, \dfrac {\d y} {\d x} = k$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y^2 = 2 k x + C$
{{begin-eqn}} {{eqn | l = y \, \dfrac {\d y} {\d x} | r = k | c = }} {{eqn | ll= \leadsto | l = \int y \rd y | r = \int k \rd x | c = [[Solution to Separable Differential Equation]] }} {{eqn | ll= \leadsto | l = \frac {y^2} 2 | r = k x + \frac C 2 | c = }} {{eqn | ll= \...
First Order ODE/y dy = k dx
https://proofwiki.org/wiki/First_Order_ODE/y_dy_=_k_dx
https://proofwiki.org/wiki/First_Order_ODE/y_dy_=_k_dx
[ "Examples of First Order ODEs", "Examples of Solutions to Separable Differential Equation" ]
[ "Definition:Real Number", "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution to Separable Differential Equation", "Category:Examples of First Order ODEs", "Category:Examples of Solutions to Separable Differential Equation" ]
proofwiki-11065
First Order ODE/x dy = (y + y^3) dx
The first order ODE: :$(1): \quad x \dfrac {\d y} {\d x} = y + y^3$ has the general solution: :$y = \dfrac x {\sqrt {C^2 - x^2} }$
{{begin-eqn}} {{eqn | l = x \dfrac {\d y} {\d x} | r = y + y^3 | c = }} {{eqn | ll= \leadsto | l = \int \frac {\d y} {y \paren {1 + y^2} } | r = \int \frac {\d x} x | c = Solution to Separable Differential Equation }} {{eqn | ll= \leadsto | l = \frac 1 2 \, \map \ln {\frac {y^2} {y^...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad x \dfrac {\d y} {\d x} = y + y^3$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = \dfrac x {\sqrt {C^2 - x^2} }$
{{begin-eqn}} {{eqn | l = x \dfrac {\d y} {\d x} | r = y + y^3 | c = }} {{eqn | ll= \leadsto | l = \int \frac {\d y} {y \paren {1 + y^2} } | r = \int \frac {\d x} x | c = [[Solution to Separable Differential Equation]] }} {{eqn | ll= \leadsto | l = \frac 1 2 \, \map \ln {\frac {y^2}...
First Order ODE/x dy = (y + y^3) dx
https://proofwiki.org/wiki/First_Order_ODE/x_dy_=_(y_+_y^3)_dx
https://proofwiki.org/wiki/First_Order_ODE/x_dy_=_(y_+_y^3)_dx
[ "Examples of First Order ODEs", "Examples of Solutions to Separable Differential Equation" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution to Separable Differential Equation", "Primitive of Reciprocal of x by x squared plus a squared", "Category:Examples of First Order ODEs", "Category:Examples of Solutions to Separable Differential Equation" ]
proofwiki-11066
First Order ODE/y dy = k x dx
Let $k \in \R$ be a real number. The first order ODE: :$y \rd y = k x \rd x$ has the general solution: :$y^2 = k x^2 + C$
{{begin-eqn}} {{eqn | l = y \rd y | r = k x \rd x | c = }} {{eqn | ll= \leadsto | l = \int y \rd y | r = \int k x \rd x | c = Solution to Separable Differential Equation }} {{eqn | ll= \leadsto | l = \frac {y^2} 2 | r = \frac {k x^2} 2 + C_1 | c = }} {{eqn | ll= \leadst...
Let $k \in \R$ be a [[Definition:Real Number|real number]]. The [[Definition:First Order ODE|first order ODE]]: :$y \rd y = k x \rd x$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y^2 = k x^2 + C$
{{begin-eqn}} {{eqn | l = y \rd y | r = k x \rd x | c = }} {{eqn | ll= \leadsto | l = \int y \rd y | r = \int k x \rd x | c = [[Solution to Separable Differential Equation]] }} {{eqn | ll= \leadsto | l = \frac {y^2} 2 | r = \frac {k x^2} 2 + C_1 | c = }} {{eqn | ll= \le...
First Order ODE/y dy = k x dx
https://proofwiki.org/wiki/First_Order_ODE/y_dy_=_k_x_dx
https://proofwiki.org/wiki/First_Order_ODE/y_dy_=_k_x_dx
[ "Examples of First Order ODEs", "Examples of Solutions to Separable Differential Equation" ]
[ "Definition:Real Number", "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution to Separable Differential Equation", "Definition:Primitive (Calculus)/Constant of Integration", "Category:Examples of First Order ODEs", "Category:Examples of Solutions to Separable Differential Equation" ]
proofwiki-11067
Primitive of Reciprocal of Root of x squared plus k
Let $k \in \R$. Then: :$\ds \int \frac {\d x} {\sqrt {\size {x^2 + k} } } = \map \ln {x + \sqrt {\size {x^2 + k} } } + C$
There are three cases: :$0 \le k$ :$-x^2 < k < 0$ :$k < -x^2$
Let $k \in \R$. Then: :$\ds \int \frac {\d x} {\sqrt {\size {x^2 + k} } } = \map \ln {x + \sqrt {\size {x^2 + k} } } + C$
There are three cases: :$0 \le k$ :$-x^2 < k < 0$ :$k < -x^2$
Primitive of Reciprocal of Root of x squared plus k
https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_Root_of_x_squared_plus_k
https://proofwiki.org/wiki/Primitive_of_Reciprocal_of_Root_of_x_squared_plus_k
[ "Primitives involving Root of x squared plus a squared", "Primitives involving Root of x squared minus a squared", "Primitives involving Root of a squared minus x squared", "Primitives involving Reciprocals" ]
[]
[]
proofwiki-11068
Bernoulli's Equation/x^2 dy = (2 x y + y^2) dx
The first order ODE: :$(1): \quad x^2 \rd y = \paren {2 x y + y^2} \rd x$ has the general solution: :$y = -\dfrac {x^2} {x + C}$
Rearranging $(1)$: :$(2): \quad \dfrac {\d y} {\d x} - \dfrac 2 x y = \dfrac {y^2} {x^2}$ It can be seen that $(2)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$ where: :$\map P x = -\dfrac 2 x$ :$\map Q x = \dfrac 1 {x^2}$ :$n = 2$ and so is an example of Bernoulli's equation. By Solution to Bern...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad x^2 \rd y = \paren {2 x y + y^2} \rd x$ has the [[Definition:General Solution to Differential Equation|general solution]]: :$y = -\dfrac {x^2} {x + C}$
Rearranging $(1)$: :$(2): \quad \dfrac {\d y} {\d x} - \dfrac 2 x y = \dfrac {y^2} {x^2}$ It can be seen that $(2)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$ where: :$\map P x = -\dfrac 2 x$ :$\map Q x = \dfrac 1 {x^2}$ :$n = 2$ and so is an example of [[Definition:Bernoulli's Equation|Bern...
Bernoulli's Equation/x^2 dy = (2 x y + y^2) dx
https://proofwiki.org/wiki/Bernoulli's_Equation/x^2_dy_=_(2_x_y_+_y^2)_dx
https://proofwiki.org/wiki/Bernoulli's_Equation/x^2_dy_=_(2_x_y_+_y^2)_dx
[ "Examples of Bernoulli's Equation" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Bernoulli's Equation", "Solution to Bernoulli's Equation", "Definition:Differential Equation/Solution/General Solution", "Category:Examples of Bernoulli's Equation" ]
proofwiki-11069
Bernoulli's Equation/2 x y dx + (x^2 + 2 y) dy = 0
The first order ODE: :$(1): \quad 2 x y \rd x + \paren {x^2 + 2 y} \rd y = 0$ has the solution: :$y \paren {x^2 + y} = C$
{{begin-eqn}} {{eqn | l = 2 x y \rd x + \paren {x^2 + 2 y} \rd y | r = 0 | c = }} {{eqn | n = 2 | ll= \leadsto | l = \dfrac {\d x} {\d y} + \frac 1 {2 y} x | r = -\frac 1 x | c = }} {{end-eqn}} It can be seen that $(2)$ is in the form: :$\dfrac {\d x} {\d y} + \map P y x = \map Q y...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad 2 x y \rd x + \paren {x^2 + 2 y} \rd y = 0$ has the [[Definition:General Solution to Differential Equation|solution]]: :$y \paren {x^2 + y} = C$
{{begin-eqn}} {{eqn | l = 2 x y \rd x + \paren {x^2 + 2 y} \rd y | r = 0 | c = }} {{eqn | n = 2 | ll= \leadsto | l = \dfrac {\d x} {\d y} + \frac 1 {2 y} x | r = -\frac 1 x | c = }} {{end-eqn}} It can be seen that $(2)$ is in the form: :$\dfrac {\d x} {\d y} + \map P y x = \map Q...
Bernoulli's Equation/2 x y dx + (x^2 + 2 y) dy = 0
https://proofwiki.org/wiki/Bernoulli's_Equation/2_x_y_dx_+_(x^2_+_2_y)_dy_=_0
https://proofwiki.org/wiki/Bernoulli's_Equation/2_x_y_dx_+_(x^2_+_2_y)_dy_=_0
[ "Examples of Bernoulli's Equation" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Bernoulli's Equation", "Solution to Bernoulli's Equation", "Definition:Differential Equation/Solution/General Solution", "Category:Examples of Bernoulli's Equation" ]
proofwiki-11070
Linear First Order ODE/y' - (y over x) = k x
Let $k \in \R$ be a real number. The linear first order ODE: :$(1): \quad \dfrac {\d y} {\d x} - \dfrac y x = k x$ has the general solution: :$\dfrac y x = k x + C$ or: :$y = k x^2 + C x$
$(1)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where $\map P x = -\dfrac 1 x$. Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x | r = \int -\frac 1 x \rd x | c = }} {{eqn | r = -\ln x | c = }} {{eqn | ll= \leadsto | l = e^{\int P \rd x} | r = e^{-\ln x} | ...
Let $k \in \R$ be a [[Definition:Real Number|real number]]. The [[Definition:Linear First Order ODE|linear first order ODE]]: :$(1): \quad \dfrac {\d y} {\d x} - \dfrac y x = k x$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$\dfrac y x = k x + C$ or: :$y = k x^2 + C x$
$(1)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where $\map P x = -\dfrac 1 x$. Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x | r = \int -\frac 1 x \rd x | c = }} {{eqn | r = -\ln x | c = }} {{eqn | ll= \leadsto | l = e^{\int P \rd x} | r = e^{-\ln x} |...
Linear First Order ODE/y' - (y over x) = k x
https://proofwiki.org/wiki/Linear_First_Order_ODE/y'_-_(y_over_x)_=_k_x
https://proofwiki.org/wiki/Linear_First_Order_ODE/y'_-_(y_over_x)_=_k_x
[ "Examples of Linear First Order ODEs" ]
[ "Definition:Real Number", "Definition:Linear First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution to Linear First Order Ordinary Differential Equation/Solution by Integrating Factor", "Definition:Differential Equation/Solution/General Solution", "Category:Examples of Linear First Order ODEs" ]
proofwiki-11071
First Order ODE/x dy = k y dx
The first order ODE: :$(1): \quad x \rd y = k y \rd x$ has the general solution: :$y = C x^k$
$(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = x \rd y | r = k y \rd x | c = }} {{eqn | ll= \leadsto | l = \int \dfrac {\d y} y | r = k \int \dfrac {\d x} x | c = Solutions to Separable Differential Equation }} {{eqn | ll= \leadsto | l = \ln y | r = k \ln x + \ln C ...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad x \rd y = k y \rd x$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = C x^k$
$(1)$ can be expressed as: {{begin-eqn}} {{eqn | l = x \rd y | r = k y \rd x | c = }} {{eqn | ll= \leadsto | l = \int \dfrac {\d y} y | r = k \int \dfrac {\d x} x | c = [[Solutions to Separable Differential Equation]] }} {{eqn | ll= \leadsto | l = \ln y | r = k \ln x + \ln C ...
First Order ODE/x dy = k y dx
https://proofwiki.org/wiki/First_Order_ODE/x_dy_=_k_y_dx
https://proofwiki.org/wiki/First_Order_ODE/x_dy_=_k_y_dx
[ "Examples of First Order ODEs", "Examples of Solutions to Separable Differential Equation", "First Order ODE/x dy = k y dx" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solutions to Separable Differential Equation", "Category:Examples of First Order ODEs", "Category:Examples of Solutions to Separable Differential Equation", "Category:First Order ODE/x dy = k y dx" ]
proofwiki-11072
First Order ODE/dy = k y dx
Let $k \in \R$ be a real number. The first order ODE: :$\dfrac {\d y} {\d x} = k y$ has the general solution: :$y = C e^{k x}$
{{begin-eqn}} {{eqn | l = \dfrac {\d y} {\d x} | r = k y | c = }} {{eqn | ll= \leadsto | l = \int \dfrac {\d y} y | r = \int k \rd x | c = Solution to Separable Differential Equation }} {{eqn | ll= \leadsto | l = \ln y | r = k x + C' | c = Primitive of Reciprocal, Primit...
Let $k \in \R$ be a [[Definition:Real Number|real number]]. The [[Definition:First Order ODE|first order ODE]]: :$\dfrac {\d y} {\d x} = k y$ has the [[Definition:General Solution to Differential Equation|general solution]]: :$y = C e^{k x}$
{{begin-eqn}} {{eqn | l = \dfrac {\d y} {\d x} | r = k y | c = }} {{eqn | ll= \leadsto | l = \int \dfrac {\d y} y | r = \int k \rd x | c = [[Solution to Separable Differential Equation]] }} {{eqn | ll= \leadsto | l = \ln y | r = k x + C' | c = [[Primitive of Reciprocal]]...
First Order ODE/dy = k y dx/Proof 1
https://proofwiki.org/wiki/First_Order_ODE/dy_=_k_y_dx
https://proofwiki.org/wiki/First_Order_ODE/dy_=_k_y_dx/Proof_1
[ "First Order ODE/dy = k y dx", "Examples of Linear First Order ODEs", "Examples of Solutions to Separable Differential Equation" ]
[ "Definition:Real Number", "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution to Separable Differential Equation", "Primitive of Reciprocal", "Primitive of Constant" ]
proofwiki-11073
First Order ODE/dy = k y dx
Let $k \in \R$ be a real number. The first order ODE: :$\dfrac {\d y} {\d x} = k y$ has the general solution: :$y = C e^{k x}$
Write the differential equation as: :$y' - k y = 0$ Taking Laplace transforms: :$\laptrans {y' - k y} = \laptrans 0$ From Laplace Transform of Constant Mapping, we have: :$\laptrans 0 = 0$ We also have: {{begin-eqn}} {{eqn | l = \laptrans {y' - k y} | r = \laptrans {y'} - k \laptrans y | c = Linear Combination of ...
Let $k \in \R$ be a [[Definition:Real Number|real number]]. The [[Definition:First Order ODE|first order ODE]]: :$\dfrac {\d y} {\d x} = k y$ has the [[Definition:General Solution to Differential Equation|general solution]]: :$y = C e^{k x}$
Write the differential equation as: :$y' - k y = 0$ Taking [[Definition:Laplace Transform|Laplace transforms]]: :$\laptrans {y' - k y} = \laptrans 0$ From [[Laplace Transform of Constant Mapping]], we have: :$\laptrans 0 = 0$ We also have: {{begin-eqn}} {{eqn | l = \laptrans {y' - k y} | r = \laptrans {y'} -...
First Order ODE/dy = k y dx/Proof 2
https://proofwiki.org/wiki/First_Order_ODE/dy_=_k_y_dx
https://proofwiki.org/wiki/First_Order_ODE/dy_=_k_y_dx/Proof_2
[ "First Order ODE/dy = k y dx", "Examples of Linear First Order ODEs", "Examples of Solutions to Separable Differential Equation" ]
[ "Definition:Real Number", "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Laplace Transform", "Laplace Transform of Constant Mapping", "Linear Combination of Laplace Transforms", "Laplace Transform of Derivative", "Linear Combination of Laplace Transforms", "Laplace Transform of Exponential" ]
proofwiki-11074
First Order ODE/dy = k y dx
Let $k \in \R$ be a real number. The first order ODE: :$\dfrac {\d y} {\d x} = k y$ has the general solution: :$y = C e^{k x}$
{{begin-eqn}} {{eqn | l = \dfrac {\d y} {\d x} | r = k y | c = }} {{eqn | ll= \leadsto | l = \dfrac {\d y} {\d x} - k y | r = 0 | c = }} {{eqn | ll= \leadsto | l = e^{-k x} \paren {\dfrac {\d y} {\d x} - k y} | r = 0 | c = Solution to Linear First Order Ordinary Differe...
Let $k \in \R$ be a [[Definition:Real Number|real number]]. The [[Definition:First Order ODE|first order ODE]]: :$\dfrac {\d y} {\d x} = k y$ has the [[Definition:General Solution to Differential Equation|general solution]]: :$y = C e^{k x}$
{{begin-eqn}} {{eqn | l = \dfrac {\d y} {\d x} | r = k y | c = }} {{eqn | ll= \leadsto | l = \dfrac {\d y} {\d x} - k y | r = 0 | c = }} {{eqn | ll= \leadsto | l = e^{-k x} \paren {\dfrac {\d y} {\d x} - k y} | r = 0 | c = [[Solution to Linear First Order Ordinary Diffe...
First Order ODE/dy = k y dx/Proof 3
https://proofwiki.org/wiki/First_Order_ODE/dy_=_k_y_dx
https://proofwiki.org/wiki/First_Order_ODE/dy_=_k_y_dx/Proof_3
[ "First Order ODE/dy = k y dx", "Examples of Linear First Order ODEs", "Examples of Solutions to Separable Differential Equation" ]
[ "Definition:Real Number", "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution to Linear First Order Ordinary Differential Equation", "Solution to Linear First Order Ordinary Differential Equation" ]
proofwiki-11075
First Order ODE/(1 over x^3 y^2 + 1 over x) dx + (1 over x^2 y^3 - 1 over y) dy = 0
The first order ordinary differential equation: :$(1): \quad \paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y = 0$ is an exact differential equation with solution: :$-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$
Let: :$\map M {x, y} = \dfrac 1 {x^3 y^2} + \dfrac 1 x$ :$\map N {x, y} = \dfrac 1 {x^2 y^3} - \dfrac 1 y$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial M} {\partial y} | r = -\dfrac 2 {x^3 y^3} | c = }} {{eqn | l = \dfrac {\partial N} {\partial x} | r = -\dfrac 2 {x^3 y^3} | c = }} {{end-e...
The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]: :$(1): \quad \paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y = 0$ is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:Gen...
Let: :$\map M {x, y} = \dfrac 1 {x^3 y^2} + \dfrac 1 x$ :$\map N {x, y} = \dfrac 1 {x^2 y^3} - \dfrac 1 y$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial M} {\partial y} | r = -\dfrac 2 {x^3 y^3} | c = }} {{eqn | l = \dfrac {\partial N} {\partial x} | r = -\dfrac 2 {x^3 y^3} | c = }} {{end-...
First Order ODE/(1 over x^3 y^2 + 1 over x) dx + (1 over x^2 y^3 - 1 over y) dy = 0
https://proofwiki.org/wiki/First_Order_ODE/(1_over_x^3_y^2_+_1_over_x)_dx_+_(1_over_x^2_y^3_-_1_over_y)_dy_=_0
https://proofwiki.org/wiki/First_Order_ODE/(1_over_x^3_y^2_+_1_over_x)_dx_+_(1_over_x^2_y^3_-_1_over_y)_dy_=_0
[ "Examples of Exact Differential Equation", "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Exact Differential Equation", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Category:Examples of Exact Differential Equation", ...
proofwiki-11076
Linear First Order ODE/y' + (y over x) = k x^n
Let $k, n \in \R$ be real numbers. The linear first order ODE: :$(1): \quad \dfrac {\d y} {\d x} + \dfrac y x = k x^n$ has the general solution: :<nowiki>$y = \begin{cases} \dfrac {k x^{n + 1} } {n + 2} + \dfrac C x & : n \ne -2 \\ & \\ \dfrac {k \ln x} x + \dfrac C x & : n = -2 \end{cases}$</nowiki>
$(1)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = \dfrac 1 x$ :$\map Q x = k x^n$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x | r = \int \frac 1 x \rd x | c = }} {{eqn | r = \ln x | c = }} {{eqn | ll= \leadsto | l = e^{\int P \rd x} | r = e...
Let $k, n \in \R$ be [[Definition:Real Number|real numbers]]. The [[Definition:Linear First Order ODE|linear first order ODE]]: :$(1): \quad \dfrac {\d y} {\d x} + \dfrac y x = k x^n$ has the [[Definition:General Solution of Differential Equation|general solution]]: :<nowiki>$y = \begin{cases} \dfrac {k x^{n + 1} } {...
$(1)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = \dfrac 1 x$ :$\map Q x = k x^n$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x | r = \int \frac 1 x \rd x | c = }} {{eqn | r = \ln x | c = }} {{eqn | ll= \leadsto | l = e^{\int P \rd x} | r = ...
Linear First Order ODE/y' + (y over x) = k x^n
https://proofwiki.org/wiki/Linear_First_Order_ODE/y'_+_(y_over_x)_=_k_x^n
https://proofwiki.org/wiki/Linear_First_Order_ODE/y'_+_(y_over_x)_=_k_x^n
[ "Examples of Linear First Order ODEs" ]
[ "Definition:Real Number", "Definition:Linear First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution to Linear First Order Ordinary Differential Equation/Solution by Integrating Factor", "Primitive of Power", "Primitive of Reciprocal", "Category:Examples of Linear First Order ODEs" ]
proofwiki-11077
Linear First Order ODE/(x^2 + y) dx = x dy
The linear first order ODE: :$(1): \quad \paren {x^2 + y} \rd x = x \rd y$ has the general solution: :$y = x^2 + C x$
Rearranging $(1)$: :$(2): \quad \dfrac {\d y} {\d x} - \dfrac y x = x$ $(2)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = -\dfrac 1 x$ :$\map Q x = x$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x | r = \int -\dfrac 1 x \rd x | c = }} {{eqn | r = -\ln x | ...
The [[Definition:Linear First Order ODE|linear first order ODE]]: :$(1): \quad \paren {x^2 + y} \rd x = x \rd y$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = x^2 + C x$
Rearranging $(1)$: :$(2): \quad \dfrac {\d y} {\d x} - \dfrac y x = x$ $(2)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = -\dfrac 1 x$ :$\map Q x = x$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x | r = \int -\dfrac 1 x \rd x | c = }} {{eqn | r = -\ln x ...
Linear First Order ODE/(x^2 + y) dx = x dy
https://proofwiki.org/wiki/Linear_First_Order_ODE/(x^2_+_y)_dx_=_x_dy
https://proofwiki.org/wiki/Linear_First_Order_ODE/(x^2_+_y)_dx_=_x_dy
[ "Examples of Linear First Order ODEs" ]
[ "Definition:Linear First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution to Linear First Order Ordinary Differential Equation/Solution by Integrating Factor", "Definition:Differential Equation/Solution/General Solution" ]
proofwiki-11078
Linear First Order ODE/x y' + y = x^2 cosine x
The linear first order ODE: :$x \, \dfrac {\d y} {\d x} + y = x^2 \cos x$ has the general solution: :$y = 2 \cos x + x \sin x - \dfrac 2 x \sin x + \dfrac C x$
Rearranging: :$\dfrac {\d y} {\d x} + \dfrac y x = x \cos x$ This is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = \dfrac 1 x$ :$\map Q x = x \cos x$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x | r = \int \dfrac 1 x \rd x | c = }} {{eqn | r = \ln x | c = }}...
The [[Definition:Linear First Order ODE|linear first order ODE]]: :$x \, \dfrac {\d y} {\d x} + y = x^2 \cos x$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = 2 \cos x + x \sin x - \dfrac 2 x \sin x + \dfrac C x$
Rearranging: :$\dfrac {\d y} {\d x} + \dfrac y x = x \cos x$ This is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = \dfrac 1 x$ :$\map Q x = x \cos x$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x | r = \int \dfrac 1 x \rd x | c = }} {{eqn | r = \ln x | c = ...
Linear First Order ODE/x y' + y = x^2 cosine x/Proof 1
https://proofwiki.org/wiki/Linear_First_Order_ODE/x_y'_+_y_=_x^2_cosine_x
https://proofwiki.org/wiki/Linear_First_Order_ODE/x_y'_+_y_=_x^2_cosine_x/Proof_1
[ "Examples of Linear First Order ODEs", "Linear First Order ODE/x y' + y = x^2 cosine x" ]
[ "Definition:Linear First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution to Linear First Order Ordinary Differential Equation/Solution by Integrating Factor", "Primitive of x squared by Cosine of a x" ]
proofwiki-11079
Linear First Order ODE/x y' + y = x^2 cosine x
The linear first order ODE: :$x \, \dfrac {\d y} {\d x} + y = x^2 \cos x$ has the general solution: :$y = 2 \cos x + x \sin x - \dfrac 2 x \sin x + \dfrac C x$
{{begin-eqn}} {{eqn | l = x \dfrac {\d y} {\d x} + y | r = x^2 \cos x | c = }} {{eqn | ll= \leadsto | l = x \rd y + y \rd x | r = x^2 \cos x \rd x | c = }} {{eqn | ll= \leadsto | l = \map \d {x y} | r = x^2 \cos x \rd x | c = Product Rule for Differentials }} {{eqn | ll...
The [[Definition:Linear First Order ODE|linear first order ODE]]: :$x \, \dfrac {\d y} {\d x} + y = x^2 \cos x$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = 2 \cos x + x \sin x - \dfrac 2 x \sin x + \dfrac C x$
{{begin-eqn}} {{eqn | l = x \dfrac {\d y} {\d x} + y | r = x^2 \cos x | c = }} {{eqn | ll= \leadsto | l = x \rd y + y \rd x | r = x^2 \cos x \rd x | c = }} {{eqn | ll= \leadsto | l = \map \d {x y} | r = x^2 \cos x \rd x | c = [[Product Rule for Differentials]] }} {{eqn ...
Linear First Order ODE/x y' + y = x^2 cosine x/Proof 2
https://proofwiki.org/wiki/Linear_First_Order_ODE/x_y'_+_y_=_x^2_cosine_x
https://proofwiki.org/wiki/Linear_First_Order_ODE/x_y'_+_y_=_x^2_cosine_x/Proof_2
[ "Examples of Linear First Order ODEs", "Linear First Order ODE/x y' + y = x^2 cosine x" ]
[ "Definition:Linear First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Product Rule for Differentials", "Primitive of x squared by Cosine of a x" ]
proofwiki-11080
Linear First Order ODE/x y' + y = x^2 cosine x
The linear first order ODE: :$x \, \dfrac {\d y} {\d x} + y = x^2 \cos x$ has the general solution: :$y = 2 \cos x + x \sin x - \dfrac 2 x \sin x + \dfrac C x$
{{begin-eqn}} {{eqn | l = x \dfrac {\d y} {\d x} + y | r = x^2 \cos x | c = }} {{eqn | ll= \leadsto | l = x y | r = \int x^2 \cos x \d x + C | c = Linear First Order ODE: $x y' + y = \map f x$ }} {{eqn | r = 2 x \cos x + \paren {x^2 - 2} \sin x + C | c = Primitive of $x^2 \cos a x$ ...
The [[Definition:Linear First Order ODE|linear first order ODE]]: :$x \, \dfrac {\d y} {\d x} + y = x^2 \cos x$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = 2 \cos x + x \sin x - \dfrac 2 x \sin x + \dfrac C x$
{{begin-eqn}} {{eqn | l = x \dfrac {\d y} {\d x} + y | r = x^2 \cos x | c = }} {{eqn | ll= \leadsto | l = x y | r = \int x^2 \cos x \d x + C | c = [[Linear First Order ODE/x y' + y = f (x)|Linear First Order ODE: $x y' + y = \map f x$]] }} {{eqn | r = 2 x \cos x + \paren {x^2 - 2} \sin x ...
Linear First Order ODE/x y' + y = x^2 cosine x/Proof 3
https://proofwiki.org/wiki/Linear_First_Order_ODE/x_y'_+_y_=_x^2_cosine_x
https://proofwiki.org/wiki/Linear_First_Order_ODE/x_y'_+_y_=_x^2_cosine_x/Proof_3
[ "Examples of Linear First Order ODEs", "Linear First Order ODE/x y' + y = x^2 cosine x" ]
[ "Definition:Linear First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Linear First Order ODE/x y' + y = f (x)", "Primitive of x squared by Cosine of a x" ]
proofwiki-11081
First Order ODE/(6x + 4y + 3) dx + (3x + 2y + 2) dy = 0
The first order ODE: :$(1): \quad \paren {6 x + 4 y + 3} \rd x + \paren {3 x + 2 y + 2} \rd y = 0$ has the general solution: :$3 x + 2 y + \map \ln {\paren {3 x + 2 y}^2} + x = C$
We put $(1)$ in the form: :$\dfrac {\d y} {\d x} = -\dfrac {6 x + 4 y + 3} {3 x + 2 y + 2}$ Substitute $z = 3 x + 2 y$, which gives: :$\dfrac {\d z} {\d x} = 3 + 2 \dfrac {\d y} {\d x}$ and so: :$\dfrac {\d z} {\d x} = -2 \dfrac {2 z + 3} {z + 2} + 3$ which simplifies down to: :$\dfrac {\d z} {\d x} = \dfrac {-z} {z + ...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad \paren {6 x + 4 y + 3} \rd x + \paren {3 x + 2 y + 2} \rd y = 0$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$3 x + 2 y + \map \ln {\paren {3 x + 2 y}^2} + x = C$
We put $(1)$ in the form: :$\dfrac {\d y} {\d x} = -\dfrac {6 x + 4 y + 3} {3 x + 2 y + 2}$ Substitute $z = 3 x + 2 y$, which gives: :$\dfrac {\d z} {\d x} = 3 + 2 \dfrac {\d y} {\d x}$ and so: :$\dfrac {\d z} {\d x} = -2 \dfrac {2 z + 3} {z + 2} + 3$ which simplifies down to: :$\dfrac {\d z} {\d x} = \dfrac {-z} {...
First Order ODE/(6x + 4y + 3) dx + (3x + 2y + 2) dy = 0
https://proofwiki.org/wiki/First_Order_ODE/(6x_+_4y_+_3)_dx_+_(3x_+_2y_+_2)_dy_=_0
https://proofwiki.org/wiki/First_Order_ODE/(6x_+_4y_+_3)_dx_+_(3x_+_2y_+_2)_dy_=_0
[ "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[]
proofwiki-11082
First Order ODE/Cosine (x + y) dx = sine (x + y) dx + x sine (x + y) dy
The first order ordinary differential equation: :$\map \cos {x + y} \rd x = \map \sin {x + y} \rd x + x \map \sin {x + y} \rd y$ is an exact differential equation with solution: :$x \map \cos {x + y} = C$
Express it in the form: :$\paren {x \map \sin {x + y} - \map \cos {x + y} } \rd x + x \map \sin {x + y} \rd y$ Let: :$\map M {x, y} = x \map \sin {x + y} - \map \cos {x + y}$ :$\map N {x, y} = x \map \sin {x + y}$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y} | r = x \map \cos {x + y} + \map \si...
The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]: :$\map \cos {x + y} \rd x = \map \sin {x + y} \rd x + x \map \sin {x + y} \rd y$ is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution to Differential Equa...
Express it in the form: :$\paren {x \map \sin {x + y} - \map \cos {x + y} } \rd x + x \map \sin {x + y} \rd y$ Let: :$\map M {x, y} = x \map \sin {x + y} - \map \cos {x + y}$ :$\map N {x, y} = x \map \sin {x + y}$ Then: {{begin-eqn}} {{eqn | l = \frac {\partial M} {\partial y} | r = x \map \cos {x + y} + \map \...
First Order ODE/Cosine (x + y) dx = sine (x + y) dx + x sine (x + y) dy
https://proofwiki.org/wiki/First_Order_ODE/Cosine_(x_+_y)_dx_=_sine_(x_+_y)_dx_+_x_sine_(x_+_y)_dy
https://proofwiki.org/wiki/First_Order_ODE/Cosine_(x_+_y)_dx_=_sine_(x_+_y)_dx_+_x_sine_(x_+_y)_dy
[ "Examples of Exact Differential Equation", "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Exact Differential Equation", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
proofwiki-11083
Linear First Order ODE/x y' + y = f (x)
The linear first order ODE: :$(1): \quad x \, \dfrac {\d y} {\d x} + y = \map f x$ has the general solution: :$\ds x y = \int \map f x \rd x + C$
{{begin-eqn}} {{eqn | l = x \dfrac {\d y} {\d x} + y | r = \map f x | c = }} {{eqn | ll= \leadsto | l = x \rd y + y \rd x | r = \map f x \rd x | c = }} {{eqn | ll= \leadsto | l = \map \d {x y} | r = \map f x \rd x | c = Product Rule for Differentials }} {{eqn | ll= \lea...
The [[Definition:Linear First Order ODE|linear first order ODE]]: :$(1): \quad x \, \dfrac {\d y} {\d x} + y = \map f x$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$\ds x y = \int \map f x \rd x + C$
{{begin-eqn}} {{eqn | l = x \dfrac {\d y} {\d x} + y | r = \map f x | c = }} {{eqn | ll= \leadsto | l = x \rd y + y \rd x | r = \map f x \rd x | c = }} {{eqn | ll= \leadsto | l = \map \d {x y} | r = \map f x \rd x | c = [[Product Rule for Differentials]] }} {{eqn | ll= ...
Linear First Order ODE/x y' + y = f (x)
https://proofwiki.org/wiki/Linear_First_Order_ODE/x_y'_+_y_=_f_(x)
https://proofwiki.org/wiki/Linear_First_Order_ODE/x_y'_+_y_=_f_(x)
[ "Examples of Linear First Order ODEs" ]
[ "Definition:Linear First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Product Rule for Differentials" ]
proofwiki-11084
Second Order ODE/x^2 y'' + x y' = 1
The second order ODE: :$x^2 y' ' + x y' = 1$ has the general solution: :$y = \dfrac {\paren {\ln x}^2} 2 + C_1 \ln x + C_2$
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable. Substitute $p$ for $y'$: {{begin-eqn}} {{eqn | l = x^2 \dfrac {\d p} {\d x} + x p | r = 1 | c = }} {{eqn | ll= \leadsto | l = x \dfrac {\d p} {\d x} + p | r = \frac 1 x | c = }} {{...
The [[Definition:Second Order ODE|second order ODE]]: :$x^2 y' ' + x y' = 1$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = \dfrac {\paren {\ln x}^2} 2 + C_1 \ln x + C_2$
The proof proceeds by using [[Solution of Second Order Differential Equation with Missing Dependent Variable]]. Substitute $p$ for $y'$: {{begin-eqn}} {{eqn | l = x^2 \dfrac {\d p} {\d x} + x p | r = 1 | c = }} {{eqn | ll= \leadsto | l = x \dfrac {\d p} {\d x} + p | r = \frac 1 x | c = ...
Second Order ODE/x^2 y'' + x y' = 1
https://proofwiki.org/wiki/Second_Order_ODE/x^2_y''_+_x_y'_=_1
https://proofwiki.org/wiki/Second_Order_ODE/x^2_y''_+_x_y'_=_1
[ "Examples of Homogeneous LSOODEs" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution of Second Order Differential Equation with Missing Dependent Variable", "Linear First Order ODE/x y' + y = f (x)", "Primitive of Reciprocal", "Primitive of Logarithm of x over x" ]
proofwiki-11085
Sorgenfrey Line is Lindelöf
The Sorgenfrey line is a Lindelöf space.
Let $T = \struct {\R, \tau}$ be the Sorgenfrey line. Let $\CC$ be an open cover for $\R$. Define $\VV = \set {U^\circ_\R: U \in \CC}$ where $U^\circ_\R$ denotes the interior of $U$ in the real number line $R = \struct {\R, \tau_d}$ with the usual (Euclidean) topology. By definition of interior: :$\forall U \in \CC: U^\...
The [[Definition:Sorgenfrey Line|Sorgenfrey line]] is a [[Definition:Lindelöf Space|Lindelöf space]].
Let $T = \struct {\R, \tau}$ be the [[Definition:Sorgenfrey Line|Sorgenfrey line]]. Let $\CC$ be an [[Definition:Open Cover|open cover]] for $\R$. Define $\VV = \set {U^\circ_\R: U \in \CC}$ where $U^\circ_\R$ denotes the [[Definition:Interior (Topology)|interior]] of $U$ in the [[Definition:Real Number Line with Eu...
Sorgenfrey Line is Lindelöf
https://proofwiki.org/wiki/Sorgenfrey_Line_is_Lindelöf
https://proofwiki.org/wiki/Sorgenfrey_Line_is_Lindelöf
[ "Sorgenfrey Line", "Examples of Lindelöf Spaces" ]
[ "Definition:Sorgenfrey Line", "Definition:Lindelöf Space" ]
[ "Definition:Sorgenfrey Line", "Definition:Open Cover", "Definition:Interior (Topology)", "Definition:Euclidean Space/Euclidean Topology/Real Number Line", "Definition:Interior (Topology)", "Union is Smallest Superset/Set of Sets", "Topological Subspace of Real Number Line is Lindelöf", "Definition:Lin...
proofwiki-11086
First Order ODE/(y^2 exp x y + cosine x) dx + (exp x y + x y exp x y) dy = 0
The first order ordinary differential equation: :$(1): \quad \paren {y^2 e^{x y} + \cos x} \rd x + \paren {e^{x y} + x y e^{x y} } \rd y = 0$ is an exact differential equation with solution: :$y e^{x y} + \sin x = C$
Let: :$\map M {x, y} = y^2 e^{x y} + \cos x$ :$\map N {x, y} = e^{x y} + x y e^{x y}$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial M} {\partial y} | r = 2 y e^{x y} + x y^2 e^{x y} | c = }} {{eqn | l = \dfrac {\partial N} {\partial x} | r = y e^{x y} + x y^2 e^{x y} + y e^{x y} | c = }} {{...
The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]: :$(1): \quad \paren {y^2 e^{x y} + \cos x} \rd x + \paren {e^{x y} + x y e^{x y} } \rd y = 0$ is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution to Dif...
Let: :$\map M {x, y} = y^2 e^{x y} + \cos x$ :$\map N {x, y} = e^{x y} + x y e^{x y}$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial M} {\partial y} | r = 2 y e^{x y} + x y^2 e^{x y} | c = }} {{eqn | l = \dfrac {\partial N} {\partial x} | r = y e^{x y} + x y^2 e^{x y} + y e^{x y} | c = }} {...
First Order ODE/(y^2 exp x y + cosine x) dx + (exp x y + x y exp x y) dy = 0
https://proofwiki.org/wiki/First_Order_ODE/(y^2_exp_x_y_+_cosine_x)_dx_+_(exp_x_y_+_x_y_exp_x_y)_dy_=_0
https://proofwiki.org/wiki/First_Order_ODE/(y^2_exp_x_y_+_cosine_x)_dx_+_(exp_x_y_+_x_y_exp_x_y)_dy_=_0
[ "Examples of Exact Differential Equation", "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Exact Differential Equation", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Primitive of x by Exponential of a x", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
proofwiki-11087
First Order ODE/y' ln (x - y) = 1 + ln (x - y)
The first order ordinary differential equation: :$(1): \quad \dfrac {\d y} {\d x} \map \ln {x - y} \d x = 1 + \map \ln {x - y}$ is an exact differential equation with solution: :$\paren {x - y} \map \ln {x - y} = C - y$
Let $(1)$ be expressed as: :$\paren {1 + \map \ln {x - y} } \rd x - \map \ln {x - y} \rd y = 0$ Let: :$\map M {x, y} = 1 + \map \ln {x - y}$ :$\map N {x, y} = -\map \ln {x - y}$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial M} {\partial y} | r = -\frac 1 {x - y} | c = }} {{eqn | l = \dfrac {\partial N} ...
The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]: :$(1): \quad \dfrac {\d y} {\d x} \map \ln {x - y} \d x = 1 + \map \ln {x - y}$ is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution to Differential Equa...
Let $(1)$ be expressed as: :$\paren {1 + \map \ln {x - y} } \rd x - \map \ln {x - y} \rd y = 0$ Let: :$\map M {x, y} = 1 + \map \ln {x - y}$ :$\map N {x, y} = -\map \ln {x - y}$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial M} {\partial y} | r = -\frac 1 {x - y} | c = }} {{eqn | l = \dfrac {\partial ...
First Order ODE/y' ln (x - y) = 1 + ln (x - y)
https://proofwiki.org/wiki/First_Order_ODE/y'_ln_(x_-_y)_=_1_+_ln_(x_-_y)
https://proofwiki.org/wiki/First_Order_ODE/y'_ln_(x_-_y)_=_1_+_ln_(x_-_y)
[ "Examples of Exact Differential Equation", "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Exact Differential Equation", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Primitive of Logarithm of x", "Primitive of Logarithm of x", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General S...
proofwiki-11088
Linear First Order ODE/y' + 2 x y = exp -x^2
The linear first order ODE: :$\dfrac {\d y} {\d x} + 2 x y = \map \exp {-x^2}$ has the general solution: :$y = \paren {x + C} \map \exp {-x^2}$
$(1)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = 2 x$ :$\map Q x = \map \exp {-x^2}$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x | r = \int 2 x \rd x | c = }} {{eqn | r = x^2 | c = }} {{eqn | ll= \leadsto | l = e^{\int P \rd x} | r = \map ...
The [[Definition:Linear First Order ODE|linear first order ODE]]: :$\dfrac {\d y} {\d x} + 2 x y = \map \exp {-x^2}$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = \paren {x + C} \map \exp {-x^2}$
$(1)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x$ where: :$\map P x = 2 x$ :$\map Q x = \map \exp {-x^2}$ Thus: {{begin-eqn}} {{eqn | l = \int \map P x \rd x | r = \int 2 x \rd x | c = }} {{eqn | r = x^2 | c = }} {{eqn | ll= \leadsto | l = e^{\int P \rd x} | r = \map...
Linear First Order ODE/y' + 2 x y = exp -x^2
https://proofwiki.org/wiki/Linear_First_Order_ODE/y'_+_2_x_y_=_exp_-x^2
https://proofwiki.org/wiki/Linear_First_Order_ODE/y'_+_2_x_y_=_exp_-x^2
[ "Examples of Linear First Order ODEs" ]
[ "Definition:Linear First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution to Linear First Order Ordinary Differential Equation/Solution by Integrating Factor" ]
proofwiki-11089
First Order ODE/(y^2 - 3 x y - 2 x^2) dx = (x^2 - x y) dy
The first order ordinary differential equation: :$(1): \quad \paren {y^2 - 3 x y - 2 x^2} \rd x = \paren {x^2 - x y} \rd y$ is a homogeneous differential equation with solution: :$y^2 x^2 - 2 y x^3 + x^4 = C$
$(1)$ can also be rendered: :$\dfrac {\rd y} {\rd x} = \dfrac {y^2 - 3 x y - 2 x^2} {x^2 - x y}$ Let: :$\map M {x, y} = y^2 - 3 x y - 2 x^2$ :$\map N {x, y} = x^2 - x y$ Put $t x, t y$ for $x, y$: {{begin-eqn}} {{eqn | l = \map M {t x, t y} | r = \paren {t y}^2 - 3 t x t y - 2 \paren {t x}^2 | c = }} {{eqn...
The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]: :$(1): \quad \paren {y^2 - 3 x y - 2 x^2} \rd x = \paren {x^2 - x y} \rd y$ is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]] with [[Definition:General Solution to Differenti...
$(1)$ can also be rendered: :$\dfrac {\rd y} {\rd x} = \dfrac {y^2 - 3 x y - 2 x^2} {x^2 - x y}$ Let: :$\map M {x, y} = y^2 - 3 x y - 2 x^2$ :$\map N {x, y} = x^2 - x y$ Put $t x, t y$ for $x, y$: {{begin-eqn}} {{eqn | l = \map M {t x, t y} | r = \paren {t y}^2 - 3 t x t y - 2 \paren {t x}^2 | c = }} ...
First Order ODE/(y^2 - 3 x y - 2 x^2) dx = (x^2 - x y) dy
https://proofwiki.org/wiki/First_Order_ODE/(y^2_-_3_x_y_-_2_x^2)_dx_=_(x^2_-_x_y)_dy
https://proofwiki.org/wiki/First_Order_ODE/(y^2_-_3_x_y_-_2_x^2)_dx_=_(x^2_-_x_y)_dy
[ "Examples of Homogeneous Differential Equation", "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Homogeneous Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Homogeneous Function/Real Space", "Definition:Homogeneous Function/Real Space/Degree", "Definition:Homogeneous Differential Equation", "Solution to Homogeneous Differential Equation", "Primitive of Function under its Derivative" ]
proofwiki-11090
Linear First Order ODE/(1 + x^2) y' + 2 x y = 4 x^3
The linear first order ODE: :$\paren {1 + x^2} \dfrac {\d y} {\d x} + 2 x y = 4 x^3$ has the general solution: :$y = \dfrac {x^4} {1 + x^2} + \dfrac C {1 + x^2}$
It is noticed that: :$\dfrac {\d} {\d x} \paren {1 + x^2} = 2 x$ and so $(1)$ can be rendered in the form: {{begin-eqn}} {{eqn | l = \map \d {\paren {1 + x^2} y} | r = 4 x^3 | c = }} {{eqn | ll= \leadsto | l = \paren {1 + x^2} y | r = \int 4 x^3 \rd x | c = }} {{eqn | r = x^4 + C |...
The [[Definition:Linear First Order ODE|linear first order ODE]]: :$\paren {1 + x^2} \dfrac {\d y} {\d x} + 2 x y = 4 x^3$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = \dfrac {x^4} {1 + x^2} + \dfrac C {1 + x^2}$
It is noticed that: :$\dfrac {\d} {\d x} \paren {1 + x^2} = 2 x$ and so $(1)$ can be rendered in the form: {{begin-eqn}} {{eqn | l = \map \d {\paren {1 + x^2} y} | r = 4 x^3 | c = }} {{eqn | ll= \leadsto | l = \paren {1 + x^2} y | r = \int 4 x^3 \rd x | c = }} {{eqn | r = x^4 + C ...
Linear First Order ODE/(1 + x^2) y' + 2 x y = 4 x^3
https://proofwiki.org/wiki/Linear_First_Order_ODE/(1_+_x^2)_y'_+_2_x_y_=_4_x^3
https://proofwiki.org/wiki/Linear_First_Order_ODE/(1_+_x^2)_y'_+_2_x_y_=_4_x^3
[ "Examples of Linear First Order ODEs" ]
[ "Definition:Linear First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[]
proofwiki-11091
First Order ODE/exp x sine y dx + exp x cos y dy = y sine x y dx + x sine x y dy
The first order ordinary differential equation: :$(1): \quad e^x \sin y \rd x + e^x \cos y \rd y = y \sin x y \rd x + x \sin x y \rd y$ is an exact differential equation with solution: :$e^x \sin y + \cos x y = C$
Let $(1)$ be expressed as: :$\paren {e^x \sin y - y \sin x y} \rd x + \paren {e^x \cos y - x \sin x y} \rd y = 0$ Let: :$\map M {x, y} = e^x \sin y - y \sin x y$ :$\map N {x, y} = e^x \cos y - x \sin x y$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial M} {\partial y} | r = e^x \cos y - x y \cos x y - y \sin x y...
The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]: :$(1): \quad e^x \sin y \rd x + e^x \cos y \rd y = y \sin x y \rd x + x \sin x y \rd y$ is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution to Different...
Let $(1)$ be expressed as: :$\paren {e^x \sin y - y \sin x y} \rd x + \paren {e^x \cos y - x \sin x y} \rd y = 0$ Let: :$\map M {x, y} = e^x \sin y - y \sin x y$ :$\map N {x, y} = e^x \cos y - x \sin x y$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial M} {\partial y} | r = e^x \cos y - x y \cos x y - y \sin x...
First Order ODE/exp x sine y dx + exp x cos y dy = y sine x y dx + x sine x y dy
https://proofwiki.org/wiki/First_Order_ODE/exp_x_sine_y_dx_+_exp_x_cos_y_dy_=_y_sine_x_y_dx_+_x_sine_x_y_dy
https://proofwiki.org/wiki/First_Order_ODE/exp_x_sine_y_dx_+_exp_x_cos_y_dy_=_y_sine_x_y_dx_+_x_sine_x_y_dy
[ "Examples of Exact Differential Equation", "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Exact Differential Equation", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
proofwiki-11092
First Order ODE/(1 + x^2) y' + x y = 0
The first order ODE: :$\paren {1 + x^2} y' + x y = 0$ has the general solution: :$y = \dfrac C {\sqrt {1 + x^2} }$
{{begin-eqn}} {{eqn | l = \paren {1 + x^2} \dfrac {\d y} {\d x} | r = -x y | c = }} {{eqn | ll= \leadsto | l = \int \frac {\d y} y | r = -\int \frac {x \rd x} {1 + x^2} | c = Solution to Separable Differential Equation }} {{eqn | ll= \leadsto | l = \ln y | r = -\frac 1 2 \, \m...
The [[Definition:First Order ODE|first order ODE]]: :$\paren {1 + x^2} y' + x y = 0$ has the [[Definition:General Solution to Differential Equation|general solution]]: :$y = \dfrac C {\sqrt {1 + x^2} }$
{{begin-eqn}} {{eqn | l = \paren {1 + x^2} \dfrac {\d y} {\d x} | r = -x y | c = }} {{eqn | ll= \leadsto | l = \int \frac {\d y} y | r = -\int \frac {x \rd x} {1 + x^2} | c = [[Solution to Separable Differential Equation]] }} {{eqn | ll= \leadsto | l = \ln y | r = -\frac 1 2 \...
First Order ODE/(1 + x^2) y' + x y = 0
https://proofwiki.org/wiki/First_Order_ODE/(1_+_x^2)_y'_+_x_y_=_0
https://proofwiki.org/wiki/First_Order_ODE/(1_+_x^2)_y'_+_x_y_=_0
[ "Examples of Linear First Order ODEs", "Examples of Solutions to Separable Differential Equation" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution to Separable Differential Equation", "Primitive of x over x squared plus a squared", "Category:Examples of Linear First Order ODEs", "Category:Examples of Solutions to Separable Differential Equation" ]
proofwiki-11093
Linear Second Order ODE/(1 + x^2) y'' + x y' = 0
The second order ODE: :$\paren {1 + x^2} y' ' + x y' = 0$ has the general solution: :$y = C_1 \map \ln {x + \sqrt {x^2 + 1} } + C_2$
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable. Substitute $p$ for $y'$: {{begin-eqn}} {{eqn | l = \paren {1 + x^2} \dfrac {\d p} {\d x} + x p | r = 0 | c = }} {{eqn | ll= \leadsto | l = p = \dfrac {\d y} {\d x} | r = \frac {C_1} {\sqr...
The [[Definition:Second Order ODE|second order ODE]]: :$\paren {1 + x^2} y' ' + x y' = 0$ has the [[Definition:General Solution of Differential Equation|general solution]]: :$y = C_1 \map \ln {x + \sqrt {x^2 + 1} } + C_2$
The proof proceeds by using [[Solution of Second Order Differential Equation with Missing Dependent Variable]]. Substitute $p$ for $y'$: {{begin-eqn}} {{eqn | l = \paren {1 + x^2} \dfrac {\d p} {\d x} + x p | r = 0 | c = }} {{eqn | ll= \leadsto | l = p = \dfrac {\d y} {\d x} | r = \frac {C_1}...
Linear Second Order ODE/(1 + x^2) y'' + x y' = 0
https://proofwiki.org/wiki/Linear_Second_Order_ODE/(1_+_x^2)_y''_+_x_y'_=_0
https://proofwiki.org/wiki/Linear_Second_Order_ODE/(1_+_x^2)_y''_+_x_y'_=_0
[ "Examples of Homogeneous LSOODEs" ]
[ "Definition:Second Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Solution of Second Order Differential Equation with Missing Dependent Variable", "First Order ODE/(1 + x^2) y' + x y = 0", "Primitive of Reciprocal of Root of x squared plus a squared" ]
proofwiki-11094
First Order ODE/(x exp y + y - x^2) dy = (2 x y - exp y - x) dx
The first order ordinary differential equation: :$(1): \quad \paren {x e^y + y - x^2} \rd y = \paren {2 x y - e^y - x} \rd x$ is an exact differential equation with solution: :$2 x e^y + x^2 + y^2 - 2 x^2 y = C$
Let $(1)$ be expressed in the form: :$\paren {e^y + x - 2 x y} \rd x + \paren {x e^y + y - x^2} rd y = 0$ Let: :$\map M {x, y} = e^y + x - 2 x y$ :$\map N {x, y} = x e^y + y - x^2$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial M} {\partial y} | r = e^y - 2 x | c = }} {{eqn | l = \dfrac {\partial N} {\pa...
The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]: :$(1): \quad \paren {x e^y + y - x^2} \rd y = \paren {2 x y - e^y - x} \rd x$ is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution to Differential Equati...
Let $(1)$ be expressed in the form: :$\paren {e^y + x - 2 x y} \rd x + \paren {x e^y + y - x^2} rd y = 0$ Let: :$\map M {x, y} = e^y + x - 2 x y$ :$\map N {x, y} = x e^y + y - x^2$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\partial M} {\partial y} | r = e^y - 2 x | c = }} {{eqn | l = \dfrac {\partial N} ...
First Order ODE/(x exp y + y - x^2) dy = (2 x y - exp y - x) dx
https://proofwiki.org/wiki/First_Order_ODE/(x_exp_y_+_y_-_x^2)_dy_=_(2_x_y_-_exp_y_-_x)_dx
https://proofwiki.org/wiki/First_Order_ODE/(x_exp_y_+_y_-_x^2)_dy_=_(2_x_y_-_exp_y_-_x)_dx
[ "Examples of Exact Differential Equation", "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Exact Differential Equation", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution", "Solution to Exact Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
proofwiki-11095
First Order ODE/exp x (1 + x) dx = (x exp x - y exp y) dy
The first order ordinary differential equation: :$(1): \quad e^x \paren {1 + x} \rd x = \paren {x e^x - y e^y} \rd y$ has the solution: :$2 x e^x e^{-y} + y^2 + C$
Dividing $(1)$ through by $e^y$: :$(2): \quad \paren {1 + x} e^x e^{-y} \rd x = \paren {x e^x e^{-y} - y} \rd y$ Let $z = x e^x e^{-y}$. Then: :$\dfrac {\d z} {\d x} = \paren {1 + x} e^x e^{-y} - x e^x e^{-y} \dfrac {\d y} {\d x}$ or: :$\d z = \paren {1 + x} e^x e^{-y} \rd x - x e^x e^{-y} \rd y$ Substituting $z$ and $...
The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]: :$(1): \quad e^x \paren {1 + x} \rd x = \paren {x e^x - y e^y} \rd y$ has the [[Definition:General Solution to Differential Equation|solution]]: :$2 x e^x e^{-y} + y^2 + C$
Dividing $(1)$ through by $e^y$: :$(2): \quad \paren {1 + x} e^x e^{-y} \rd x = \paren {x e^x e^{-y} - y} \rd y$ Let $z = x e^x e^{-y}$. Then: :$\dfrac {\d z} {\d x} = \paren {1 + x} e^x e^{-y} - x e^x e^{-y} \dfrac {\d y} {\d x}$ or: :$\d z = \paren {1 + x} e^x e^{-y} \rd x - x e^x e^{-y} \rd y$ Substituting $z$ a...
First Order ODE/exp x (1 + x) dx = (x exp x - y exp y) dy
https://proofwiki.org/wiki/First_Order_ODE/exp_x_(1_+_x)_dx_=_(x_exp_x_-_y_exp_y)_dy
https://proofwiki.org/wiki/First_Order_ODE/exp_x_(1_+_x)_dx_=_(x_exp_x_-_y_exp_y)_dy
[ "Examples of First Order ODEs" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[]
proofwiki-11096
General Solution of Riccati Equation from Particular Solution
Consider the Riccati equation: :$(1): \quad y' = \map p x + \map q x y + \map r x y^2$ Let $\map {y_1} x$ be a particular solution to $(1)$. Then the general solution to $(1)$ has the form: :$\map y x = \map {y_1} x + \map z x$ where $\map z x$ is the general solution to the Bernoulli equation: :$z' - \paren {q - 2 r y...
Let $\map y x = \map {y_1} x + \map z x$ be a particular solution to $(1)$. Then: {{begin-eqn}} {{eqn | l = \dfrac {\d y} {\d x} | r = y_1' + z' | c = }} {{eqn | r = \map p x + \map q x \paren {y_1 + z} + \map r x \paren {y_1 + z}^2 | c = }} {{eqn | r = \map p x + \map q x \paren {y_1 + z} + \map r ...
Consider the [[Definition:Riccati Equation|Riccati equation]]: :$(1): \quad y' = \map p x + \map q x y + \map r x y^2$ Let $\map {y_1} x$ be a [[Definition:Particular Solution|particular solution]] to $(1)$. Then the [[Definition:General Solution to Differential Equation|general solution]] to $(1)$ has the form: :$\...
Let $\map y x = \map {y_1} x + \map z x$ be a [[Definition:Particular Solution|particular solution]] to $(1)$. Then: {{begin-eqn}} {{eqn | l = \dfrac {\d y} {\d x} | r = y_1' + z' | c = }} {{eqn | r = \map p x + \map q x \paren {y_1 + z} + \map r x \paren {y_1 + z}^2 | c = }} {{eqn | r = \map p x ...
General Solution of Riccati Equation from Particular Solution
https://proofwiki.org/wiki/General_Solution_of_Riccati_Equation_from_Particular_Solution
https://proofwiki.org/wiki/General_Solution_of_Riccati_Equation_from_Particular_Solution
[ "Riccati Equation" ]
[ "Definition:Riccati Equation", "Definition:Differential Equation/Solution/Particular Solution", "Definition:Differential Equation/Solution/General Solution", "Definition:Differential Equation/Solution/General Solution", "Definition:Bernoulli's Equation" ]
[ "Definition:Differential Equation/Solution/Particular Solution" ]
proofwiki-11097
Riccati Equation/y' = (y over x) + x^3 y^2 - x^5
The Riccati equation: :$(1): \quad y' = \dfrac y x + x^3 y^2 - x^5$ has the general solution: :$C \exp \dfrac {2 x^5} 5 = \dfrac {y - x} {y + x}$
It can be seen by inspection that: :$\map {y_1} x = x$ is a particular solution to $(1)$. Thus from General Solution of Riccati Equation from Particular Solution: :$y = x + \map z x$ where: :$z' - \paren {\dfrac 1 x + 2 x^4} z = x^3 z^2$ From Bernoulli's Equation: $y' - \paren {\dfrac 1 x + 2 x^4} y = x^3 y^2$: :$z = \...
The [[Definition:Riccati Equation|Riccati equation]]: :$(1): \quad y' = \dfrac y x + x^3 y^2 - x^5$ has the [[Definition:General Solution to Differential Equation|general solution]]: :$C \exp \dfrac {2 x^5} 5 = \dfrac {y - x} {y + x}$
It can be seen by inspection that: :$\map {y_1} x = x$ is a [[Definition:Particular Solution|particular solution]] to $(1)$. Thus from [[General Solution of Riccati Equation from Particular Solution]]: :$y = x + \map z x$ where: :$z' - \paren {\dfrac 1 x + 2 x^4} z = x^3 z^2$ From [[Bernoulli's Equation/y' - (1 over ...
Riccati Equation/y' = (y over x) + x^3 y^2 - x^5
https://proofwiki.org/wiki/Riccati_Equation/y'_=_(y_over_x)_+_x^3_y^2_-_x^5
https://proofwiki.org/wiki/Riccati_Equation/y'_=_(y_over_x)_+_x^3_y^2_-_x^5
[ "Examples of Riccati Equation" ]
[ "Definition:Riccati Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Differential Equation/Solution/Particular Solution", "General Solution of Riccati Equation from Particular Solution", "Bernoulli's Equation/y' - (1 over x + 2 x^4) y = x^3 y^2" ]
proofwiki-11098
Bernoulli's Equation/y' - (1 over x + 2 x^4) y = x^3 y^2
The first order ODE: :$(1): \quad y' - \paren {\dfrac 1 x + 2 x^4} y = x^3 y^2$ has the general solution: :$y = \dfrac {2 x} {C \map \exp {-\dfrac {2 x^5} 5} - 1}$
It can be seen that $(1)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$ where: :$\map P x = -\paren {\dfrac 1 x + 2 x^4}$ :$\map Q x = x^3$ :$n = 2$ and so is an example of Bernoulli's equation. By Solution to Bernoulli's Equation it has the general solution: :$(3): \quad \ds \frac {\map \mu x} {y...
The [[Definition:First Order ODE|first order ODE]]: :$(1): \quad y' - \paren {\dfrac 1 x + 2 x^4} y = x^3 y^2$ has the [[Definition:General Solution to Differential Equation|general solution]]: :$y = \dfrac {2 x} {C \map \exp {-\dfrac {2 x^5} 5} - 1}$
It can be seen that $(1)$ is in the form: :$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$ where: :$\map P x = -\paren {\dfrac 1 x + 2 x^4}$ :$\map Q x = x^3$ :$n = 2$ and so is an example of [[Definition:Bernoulli's Equation|Bernoulli's equation]]. By [[Solution to Bernoulli's Equation]] it has the [[Definition:...
Bernoulli's Equation/y' - (1 over x + 2 x^4) y = x^3 y^2
https://proofwiki.org/wiki/Bernoulli's_Equation/y'_-_(1_over_x_+_2_x^4)_y_=_x^3_y^2
https://proofwiki.org/wiki/Bernoulli's_Equation/y'_-_(1_over_x_+_2_x^4)_y_=_x^3_y^2
[ "Examples of Bernoulli's Equation" ]
[ "Definition:First Order Ordinary Differential Equation", "Definition:Differential Equation/Solution/General Solution" ]
[ "Definition:Bernoulli's Equation", "Solution to Bernoulli's Equation", "Definition:Differential Equation/Solution/General Solution", "Definition:Differential Equation/Solution/General Solution", "Category:Examples of Bernoulli's Equation" ]
proofwiki-11099
Motion of Body with Variable Mass
Let $B$ be a body undergoing a force $\mathbf F$. Let $B$ be travelling at a velocity $\mathbf v$ at time $t$. Let mass travelling at a velocity $\mathbf v + \mathbf w$ be added to $B$ at a rate of $\dfrac {\d m} {\d t}$. Let $m$ be the mass of $B$ at time $t$. Then the equation of motion of $B$ is given by: :$\mathbf ...
From Newton's Second Law of Motion: :$\mathbf F = \map {\dfrac \d {\d t} } {m \mathbf v}$ Then the added momentum being added to $B$ by the mass being added to it is given by: :$\paren {\mathbf v + \mathbf w} \dfrac {\d m} {\d t}$ Hence: {{begin-eqn}} {{eqn | l = \paren {\mathbf v + \mathbf w} \dfrac {\d m} {\d t} + \m...
Let $B$ be a [[Definition:Body|body]] undergoing a [[Definition:Force|force]] $\mathbf F$. Let $B$ be travelling at a [[Definition:Velocity|velocity]] $\mathbf v$ at [[Definition:Time|time]] $t$. Let [[Definition:Mass|mass]] travelling at a [[Definition:Velocity|velocity]] $\mathbf v + \mathbf w$ be added to $B$ at a...
From [[Newton's Second Law of Motion]]: :$\mathbf F = \map {\dfrac \d {\d t} } {m \mathbf v}$ Then the added [[Definition:Momentum|momentum]] being added to $B$ by the [[Definition:Mass|mass]] being added to it is given by: :$\paren {\mathbf v + \mathbf w} \dfrac {\d m} {\d t}$ Hence: {{begin-eqn}} {{eqn | l = \pare...
Motion of Body with Variable Mass
https://proofwiki.org/wiki/Motion_of_Body_with_Variable_Mass
https://proofwiki.org/wiki/Motion_of_Body_with_Variable_Mass
[ "Dynamics" ]
[ "Definition:Body", "Definition:Force", "Definition:Velocity", "Definition:Time", "Definition:Mass", "Definition:Velocity", "Definition:Rate", "Definition:Mass", "Definition:Time" ]
[ "Newton's Laws of Motion/Second Law", "Definition:Momentum", "Definition:Mass" ]