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Example 4. Using the method of isoclines, construct the integral curves of the equation $\frac{d y}{d x}=\frac{y-x}{y+x}$. | Solution. Assuming $y^{\prime}=k, k=$ const, we obtain the equation of the family of isoclines $\frac{y-x}{y+x}=k$. Thus, the isoclines are straight lines passing through the origin $O(0,0)$.
For $k=-1$, we get the isocline $y=0$; for $k=0$, the isocline $y=x$; for $k=1$, the isocline $x=0$.
Considering the "inverted... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,408 |
Example 2. Using the method of successive approximations, find an approximate solution to the equation $y^{\prime}=x^{2}+y^{2}$, satisfying the initial condition $\left.y\right|_{x=0}=0$ in the rectangle $-1 \leqslant x \leqslant 1,-1 \leqslant y \leqslant 1$. | Solution. We have $|f(x, y)|=x^{2}+y^{2} \leqslant 2$, i.e., $M=2$. For $h$, we take the smaller of the numbers $a=1, \frac{b}{M}=\frac{1}{2}$, i.e., $h=\frac{1}{2}$. According to (4), the successive approximations will converge in the interval $-\frac{1}{2}<x<\frac{1}{2}$. We construct them:
$$
\begin{aligned}
y_{0}(... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,409 |
Example 1. Solve the equation
$$
3 e^{x} \tan y d x+\left(2-e^{x}\right) \sec ^{2} y d y=0
$$ | Solution. Divide both sides of the equation by the product $\operatorname{tg} y \cdot\left(2-e^{x}\right)$:
$$
\frac{3 e^{x} d x}{2-e^{x}}+\frac{\sec ^{2} y d y}{\operatorname{tg} y}=0
$$
We obtained an equation with separated variables. Integrating it, we find
$$
-3 \ln \left|2-e^{x}\right|+\ln |\operatorname{tg} y... | \operatorname{tg}y-C(2-e^{x})^{3}=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,410 |
Example 2. Find the particular solution of the equation
$$
\left(1+e^{x}\right) y y^{\prime}=e^{x}
$$
satisfying the initial condition $\left.y\right|_{x=0}=1$. | Solution. We have
$$
\left(1+e^{x}\right) y \frac{d y}{d x}=e^{x}
$$
Separating variables, we get
$$
y d y=\frac{e^{x} d x}{1+e^{x}}
$$
Integrating, we find the general integral
$$
\frac{y^{2}}{2}=\ln \left(1+e^{x}\right)+C
$$
Assuming in (1) $x=0$ and $y=1$, we have
$$
\frac{1}{2}=\ln 2+C, \quad \text { from wh... | \sqrt{1+\ln(\frac{1+e^{x}}{2})^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,411 |
Example 3. Find the particular solutions of the equation
$$
y^{\prime} \sin x=y \ln y
$$
satisfying the initial conditions:
a) $\left.y\right|_{x=\pi / 2}=e ;$
b) $\left.y\right|_{x=\pi / 2}=1$. | ## Solution. We have
$$
\frac{d y}{d x} \sin x=y \ln y
$$
Separating variables
$$
\frac{d y}{y \ln y}=\frac{d x}{\sin x}
$$
Integrating, we find the general integral
$$
\ln |\ln y|=\ln \left|\operatorname{tg} \frac{x}{2}\right|+\ln C
$$
After exponentiation, we get
$$
\ln y=C \cdot \operatorname{tg} \frac{x}{2},... | e^{\operatorname{tg}(x/2)}for),\quady\equiv1forb) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,412 |
Example 4. Find a curve passing through the point $(0,-2)$ such that the tangent of the angle of inclination of the tangent at any point on the curve equals the ordinate of that point increased by three units. | Solution. Based on the geometric property of the first derivative, we obtain the differential equation of the family of curves satisfying the required property in the problem, namely
$$
\frac{d y}{d x}=y+3
$$
By separating variables and integrating, we obtain the general solution
$$
y=C e^{x}-3
$$
Since the desired... | e^{x}-3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,413 |
Example 5. Find the curve that has the property that the length of its arc, enclosed between any two points $P$ and $Q$, is proportional to the difference in the distances of points $P$ and $Q$ from a fixed point $O$. | Solution. If we fix point $P$, then the arc $Q P$ will change proportionally to the difference $O Q$ and the constant $O P$. Let's introduce polar coordinates,
Fig. 11
taking point $O$ as ... | Ce^{\varphi} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,414 |
Example 6. Suppose that at a constant temperature, the dissolution rate of a solid in a liquid is proportional to the amount of this substance that can still dissolve in the liquid until it becomes saturated (it is assumed that the substances entering the solution do not chemically interact with each other, and the sol... | Solution. Let $P$ be the amount of substance that gives a saturated solution, and $x$ be the amount of substance already dissolved. Then we obtain the differential equation
$$
\frac{d x}{d t}=k(P-x)
$$
where $k$ is a proportionality coefficient known from experience, and $t$ is time. Separating variables, we find
$$... | P(1-e^{-k}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,415 |
Example 7. In a cylindrical vessel with a volume of $V_{0}$, atmospheric air is adiabatically (without heat exchange with the environment) compressed to a volume of $V_{1}$. Calculate the work of compression. | Solution. It is known that the adiabatic process is characterized by the Poisson equation
$$
\frac{p}{p_{0}}=\left(\frac{V_{0}}{V}\right)^{k}
$$
where $V_{0}$ is the initial volume of the gas, $p_{0}$ is the initial pressure of the gas, and $k$ is a constant for the given gas. Let $V$ and $p$ denote the volume and pr... | W_{1}=\frac{p_{0}V_{0}}{k-1}[(\frac{V_{0}}{V_{1}})^{k-1}-1] | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,416 |
Example 8. Find the solution of the equation
$$
x^{3} \sin y \cdot y^{\prime}=2
$$
satisfying the condition
$$
y \rightarrow \frac{\pi}{2} \quad \text { as } \quad x \rightarrow \infty
$$ | Solution. By separating variables and integrating, we find the general integral of equation (6):
$$
\cos y=\frac{1}{x^{2}}+C
$$
$U_{\text {condition (7) gives }} \cos \frac{\pi}{2}=C$, i.e., $C=0$, so the particular integral will have the form $\cos y=\frac{1}{x^{2}}$. It corresponds to an infinite set of particular ... | \arccos\frac{1}{x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,417 |
Example 1. Solve the equation $x y^{\prime}=\sqrt{x^{2}-y^{2}}+y$. | Solution. Let's write the equation in the form
$$
y^{\prime}=\sqrt{1-\left(\frac{y}{x}\right)^{2}}+\frac{y}{x}
$$
so that the given equation turns out to be homogeneous with respect to $x$ and $y$. Let $u=\frac{y}{x}$, or $y=u x$. Then $y^{\prime}=x u^{\prime}+u$. Substituting the expressions for $y$ and $y^{\prime}$... | x\sin\lnCx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,418 |
Example 3. Solve the equation
$$
(x+y-2) d x+(x-y+4) d y=0
$$ | Solution. Consider the system of linear algebraic equations
$$
\left\{\begin{array}{l}
x+y-2=0 \\
x-y+4=0
\end{array}\right.
$$
The determinant of this system is
$$
\Delta=\left|\begin{array}{rr}
1 & 1 \\
1 & -1
\end{array}\right|=-2 \neq 0
$$
The system has a unique solution $x_{0}=-1, y_{0}=3$. We make the substi... | x^{2}+2xy-y^{2}-4x+8C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,420 |
Example 4. Solve the equation $(x+y+1) d x+(2 x+2 y-1) d y=0$.
The above text has been translated into English, preserving the original text's line breaks and format. | Solution. The system of linear algebraic equations
$$
\left\{\begin{aligned}
x+y+1 & =0 \\
2 x+2 y-1 & =0
\end{aligned}\right.
$$
is inconsistent. In this case, the method used in the previous example is not applicable. To integrate the equation, we apply the substitution \(x+y=z\), \(d y=d z-d x\). The equation beco... | x+2y+3\ln|x+y-2|=C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,421 |
Example 1. Solve the equation
$$
y^{\prime}+2 x y=2 x e^{-x^{2}}
$$ | Solution. We apply the method of variation of constants. Consider the homogeneous equation
$$
y^{\prime}+2 x y=0
$$
corresponding to the given non-homogeneous equation. This is a separable equation. Its general solution is
$$
y=C e^{-x^{2}}
$$
The general solution of the non-homogeneous equation is sought in the fo... | (x^{2}+C)e^{-x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,422 |
Example 2. Solve the equation $\frac{d y}{d x}=\frac{1}{x \cos y+\sin 2 y}$. | Solution. The given equation is linear if we consider $x$ as a function of $y$:
$$
\frac{d x}{d y}-x \cos y=\sin 2 y
$$
We apply the method of variation of arbitrary constants. First, we solve the corresponding homogeneous equation
$$
\frac{d x}{d y}-x \cos y=0
$$
which is a separable variable equation. Its general... | Ce^{\siny}-2(1+\siny) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,423 |
Example 3. Solve the Cauchy problem:
$$
\begin{aligned}
x(x-1) y^{\prime}+y & =x^{2}(2 x-1) \\
\left.y\right|_{x=2} & =4
\end{aligned}
$$ | Solution. We look for the general solution of equation (9) in the form
$$
y=u(x) v(x)
$$
we have $y^{\prime}=u^{\prime} v+u v^{\prime}$.
Substituting the expressions for $y$ and $y^{\prime}$ into (9), we get
$$
x(x-1)\left(u^{\prime} v+u v^{\prime}\right)+u v=x^{2}(2 x-1)
$$
or
$$
x(x-1) v u^{\prime}+\left[x(x-1)... | x^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,424 |
Example 5. Given a family $C_{\alpha}$ of integral curves of the linear inhomogeneous equation $y^{\prime}+p(x) y=q(x)$.
Show that the tangents at corresponding points to the curves $C_{\alpha}$, defined by the linear equation, intersect at one point (Fig. 13). | Solution. Consider the tangent to some curve $C_{\alpha}$ at the point $M(x, y)$. The equation of the tangent at the point $M(x, y)$ is given by
$$
\eta-q(x)(\xi-x)=y[1-p(x)(\xi-x)]
$$
where $\xi, \eta$ are the current coordinates of the point on the tangent.
},+\frac{q(x)}{p(x)}) | Calculus | proof | Yes | Yes | olympiads | false | 34,425 |
Example 6. Find the solution of the equation $y^{\prime}-y=\cos x-\sin x$, satisfying the condition: $y$ is bounded as $x \rightarrow+\infty$. | Solution. The general solution of the given equation is
$$
y=C e^{x}+\sin x
$$
Any solution of the equation obtained from the general solution for $C \neq 0$ will be unbounded, as when $x \rightarrow+\infty$ the function $\sin x$ is bounded, while $e^{x} \rightarrow+\infty$. Therefore, the equation has a unique bound... | \sinx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,426 |
Example 7. Solve the Bernoulli equation $y^{\prime}-x y=-x y^{3}$. | Solution. Divide both sides of the equation by $y^{3}$:
$$
\frac{y^{\prime}}{y^{3}}-x \frac{1}{y^{2}}=-x
$$
Make the substitution $\frac{1}{y^{2}}=z,-\frac{2 y^{\prime}}{y^{3}}=z^{\prime}$, from which $\frac{y^{\prime}}{y^{3}}=-\frac{1}{2} z^{\prime}$. After substitution, the last equation will turn into a linear equ... | y^{2}(1+Ce^{-x^{2}})=1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,427 |
Example 8. Solve the Bernoulli equation
$$
x y^{\prime}+y=y^{2} \ln x
$$ | Solution. We apply the method of variation of arbitrary constant. The general solution of the corresponding homogeneous equation $x y^{\prime}+y=0$ is $y=\frac{C}{x}$. We seek the general solution of equation (16) in the form
$$
y=\frac{C(x)}{x}
$$
where $C(x)$ is a new unknown function.
Substituting (17) into (16),... | \frac{1}{1+Cx+\lnx} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,428 |
Example 10. Solve the equation
$$
x \int_{0}^{x} y(t) d t=(x+1) \int_{0}^{x} t y(t) d t, \quad x>0
$$ | Solution. Differentiating both sides of this equation with respect to $x$, we obtain
$$
\int_{0}^{x} y(t) d t+x y(x)=\int_{0}^{x} t y(t) d t+(x+1) x y(x)
$$
or
$$
\int_{0}^{x} y(t) d t=\int_{0}^{x} t y(t) d t+x^{2} y(x)
$$
Differentiating again with respect to $x$, we will have a linear homogeneous equation with re... | C\frac{1}{x^{3}}e^{-1/x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,429 |
Example 1. Solve the differential equation
$$
(\sin x y+x y \cos x y) d x+x^{2} \cos x y d y=0
$$ | Solution. Let's check that the given equation is an equation in total differentials:
$$
\begin{aligned}
\frac{\partial M}{\partial y} & =\frac{\partial}{\partial y}(\sin x y+x y \cos x y)= \\
& =x \cos x y+x \cos x y-x^{2} y \sin x y=2 x \cos x y-x^{2} y \sin x y \\
\frac{\partial N}{\partial x} & =\frac{\partial}{\pa... | x\sinxy=C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,430 |
Example 2. Solve the differential equation
$$
\left(x^{3}+x y^{2}\right) d x+\left(x^{2} y+y^{3}\right) d y=0
$$ | Solution. Here $\frac{\partial M}{\partial y}=2 x y, \frac{\partial N}{\partial x}=2 x y$, so condition (2) is satisfied and, consequently, the given equation is an equation in total differentials. This equation can easily be brought to the form $d u=0$ by direct grouping of its terms. For this purpose, rewrite it as f... | x^{4}+2(xy)^{2}+y^{4}=C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,431 |
Example 3. Solve the equation $\left(x+y^{2}\right) d x-2 x y d y=0$. | Solution. Here $M=x+y^{2}, N=-2 x y$. We have
$$
\frac{(\partial M / \partial y)-(\partial N / \partial x)}{N}=\frac{2 y+2 y}{-2 x y}=-\frac{2}{x}
$$
therefore,
$$
\frac{d \ln \mu}{d x}=-\frac{2}{x}, \quad \ln \mu=-2 \ln |x|, \quad \mu=\frac{1}{x^{2}}
$$
## Equation
$$
\frac{x+y^{2}}{x^{2}} d x-2 \frac{x y}{x^{2}}... | C\cdote^{y^{2}/x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,432 |
Example 4. Solve the equation $2 x y \ln y d x+\left(x^{2}+y^{2} \sqrt{y^{2}+1}\right) d y=0$. | Solution. Here $M=2 x y \ln y, N=x^{2}+y^{2} \sqrt{y^{2}+1}$. We have
$$
\frac{(\partial N / \partial x)-(\partial M / \partial y)}{M}=\frac{2 x-2 x(\ln y+1)}{2 x y \ln y}=-\frac{1}{y},
$$
therefore,
$$
\frac{d \ln \mu}{d y}=-\frac{1}{y}, \quad \mu=\frac{1}{y}
$$
The equation
$$
\frac{2 x y \ln y d x}{y}+\frac{x^{... | x^{2}\lny+\frac{1}{3}(y^{2}+1)^{3/2}=C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,433 |
Example 5. Solve the equation $\left(3 x+2 y+y^{2}\right) d x+\left(x+4 x y+5 y^{2}\right) d y=0$, if its integrating factor has the form $\mu=\varphi\left(x+y^{2}\right)$. | Solution. Let $z=x+y^{2}$, then $\mu=\varphi(z)$, and, consequently,
$$
\frac{\partial \ln \mu}{\partial x}=\frac{d \ln \mu}{d z} \cdot \frac{\partial z}{\partial x}=\frac{\partial \ln \mu}{d z}, \quad \frac{\partial \ln \mu}{\partial y}=\frac{d \ln \mu}{d z} \cdot \frac{\partial z}{\partial y}=\frac{\partial \ln \mu}... | (x+y)(x+y^{2})^{2}=\widehat{C} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,434 |
Example 1. Solve the equation $y\left(y^{\prime}\right)^{2}+(x-y) y^{\prime}-x=0$. | Solution. We will solve this equation with respect to $y^{\prime}$:
$$
y^{\prime}=\frac{y-x \pm \sqrt{(x-y)^{2}+4 x y}}{2 y} ; \quad y^{\prime}=1, \quad y^{\prime}=-\frac{x}{y},
$$
from which
$$
y=x+C, \quad y^{2}+x^{2}=C^{2}
$$ | x+C,\quady^{2}+x^{2}=C^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,435 |
Example 2. Solve the equation $2\left(y^{\prime}\right)^{2}-2 x y^{\prime}-2 y+x^{2}=0$. | Solution. Let's solve the equation with respect to $y$:
$$
y=\left(y^{\prime}\right)^{2}-x y^{\prime}+\frac{x^{2}}{2}
$$
Let $y^{\prime}=p$, where $p$ is a parameter; then we get
$$
y=p^{2}-x p+\frac{x^{2}}{2}
$$
Differentiating (2), we find
$$
d y=2 p d p-p d x-x d p+x d x
$$
But since $d y=p d x$, we have
$$
p... | Cx+C^{2}+\frac{x^{2}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,436 |
Example 4. Solve the equation $y^{2 / 3}+\left(y^{\prime}\right)^{2 / 3}=1$. | Solution. Let $y=\cos ^{3} t, p=\sin ^{3} t$,
$$
d x=\frac{d y}{p}=\frac{-3 \cos ^{2} t \sin t d t}{\sin ^{3} t}=-3 \frac{\cos ^{2} t}{\sin ^{2} t} d t
$$
From this
$$
x=\int\left(3-\frac{3}{\sin ^{2} t}\right) d t=3 t+3 \operatorname{ctg} t+C
$$
the general solution is
$$
x=3 t+3 \operatorname{ctg} t+C, \quad y=\... | 3+3\operatorname{ctg}+C,\quad\cos^{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,437 |
Example 5. Solve the equation $a \frac{d y}{d x}+b\left(\frac{d y}{d x}\right)^{2}=x$. | Solution. Let $\frac{d y}{d x}=p$, then
$$
\begin{aligned}
x & =a p+b p^{2}, \quad d x=a d p+2 b p d p \\
d y & =p d x=a p d p+2 b p^{2} d p, \quad y=\frac{a}{2} p^{2}+\frac{2}{3} b p^{3}+C
\end{aligned}
$$
Thus,
$$
x=a p+b p^{2}, \quad y=\frac{a}{2} p^{2}+\frac{2}{3} b p^{3}+C-\text { general solution. }
$$
Simila... | +^{2},\quad\frac{}{2}p^{2}+\frac{2}{3}^{3}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,438 |
Example 6. Integrate the equation $y=2 x y^{\prime}+\ln y^{\prime}$. | Solution. Let $y^{\prime}=p$, then $y=2 x p+\ln p$. Differentiating, we find
$$
p d x=2 p d x+2 x d p+\frac{d p}{p}
$$
from which $p \frac{d x}{d p}=-2 x-\frac{1}{p}$ or $\frac{d x}{d p}=-\frac{2}{p} x-\frac{1}{p^{2}}$.
We have obtained a first-order equation, linear in $x$; solving it, we find
$$
x=\frac{C}{p^{2}}... | \frac{C}{p^{2}}-\frac{1}{p},\quad\lnp+\frac{2C}{p}-2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,439 |
Example 7. Integrate the equation $y=x y^{\prime}+\frac{a}{2 y^{\prime}}$ ( $a=$ const). | Solution. Setting $y^{\prime}=p$, we get
$$
y=x p+\frac{a}{2 p}
$$
Differentiating the last equation and replacing $d y$ with $p d x$, we find
$$
p d x=p d x+x d p-\frac{a}{2 p^{2}} d p
$$
from which
$$
d p\left(x-\frac{a}{2 p^{2}}\right)=0
$$
Setting the first factor to zero, we get $d p=0$, hence $p=C$ and the ... | Cx+\frac{}{2C},\quady^{2}=2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,440 |
Example 1. Solve the Riccati equation
$$
y^{\prime}-y^{2}+2 e^{x} y=e^{2 x}+e^{x}
$$
knowing its particular solution $y_{1}=e^{x}$. | Solution. Let $y=e^{x}+z(x)$ and substitute into equation (4); we get
$$
\frac{d z}{d x}=z^{2}
$$
from which
$$
-\frac{1}{z}=x-C, \quad \text { or } \quad z=\frac{1}{C-x} .
$$
Thus, the general solution of equation (4) is
$$
y=e^{x}+\frac{1}{C-x}
$$
Remark. Instead of substitution (2), it is often more practical ... | e^{x}+\frac{1}{C-x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,441 |
Example 2. Equation
$$
\frac{d y}{d x}=\frac{m^{2}}{x^{4}}-y^{2}, \quad m=\text { const }
$$
has particular solutions
$$
y_{1}=\frac{1}{x}+\frac{m}{x^{2}}, \quad y_{2}=\frac{1}{x}-\frac{m}{x^{2}}
$$
Find its general integral. | Solution. Using formula (7), we obtain the general integral of the original equation
$$
\frac{y-y_{1}}{y-y_{2}}=C e^{-\int\left(2 m / x^{2}\right) d x}, \quad \text { from which } \quad \frac{x^{2} y-x-m}{x^{2} y-x+m}=C e^{2 m / x}
$$
## Problems for Independent Solution
Integrate the following Riccati equations, kn... | \frac{x^{2}y-x-}{x^{2}y-x+}=Ce^{2/x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,442 |
Example 1. Find the differential equation of the family of hyperbolas $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{1}=1$. | Solution. Differentiating this equation with respect to $x$, we get
$$
\frac{2 x}{a^{2}}-2 y y^{\prime}=0, \quad \text { or } \quad \frac{x}{a^{2}}=y y^{\prime}
$$
Multiplying both sides by $x$, then $\frac{x^{2}}{a^{2}}=x y y^{\prime}$. Substituting into the equation of the family, we find $x y y^{\prime}-y^{2}=1$.
... | xyy^{\}-y^{2}=1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,443 |
Example 3. Form a differential equation for the family of lines that are at a distance of one unit from the origin.
The text provided is a task to form a differential equation for a family of lines that are equidistant from the origin, with the distance being one unit. | Solution. We will start from the normal equation of a straight line
$$
x \cos \alpha + y \sin \alpha - 1 = 0
$$
where $\alpha$ is a parameter.
Differentiating (6) with respect to $x$, we get $\cos \alpha + y' \sin \alpha = 0$, from which $y' = -\operatorname{ctg} \alpha$, hence,
$$
\sin \alpha = \frac{1}{\sqrt{1 + ... | xy'+\sqrt{1+(y')^2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,444 |
Example 4. Find the orthogonal trajectories of the family of lines $y=k x$. | Solution. The family of lines $y=k x$ consists of straight lines passing through the origin. To find the differential equation of this family, we differentiate both sides of the equation $y=k x$ with respect to $x$. We have $y^{\prime}=k$. Eliminating the parameter $k$ from the system of equations
$$
\left\{\begin{arr... | x^{2}+y^{2}=C(C\geqslant0) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,445 |
Example 5. Find the equation of the family of lines orthogonal to the family $x^{2}+y^{2}=2 a x$. | Solution. The given family of lines represents a family of circles, the centers of which lie on the $O x$ axis and which are tangent to the $O y$ axis.
Differentiating both sides of the equation of this family with respect to $x$, we find $x + y y' = a$. By eliminating the parameter $a$ from the equations $x^2 + y^2 =... | x^2+y^2=Cy | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,446 |
Example 6. Find the orthogonal trajectories of the family of parabolas $y=a x^{2}$. | Solution. We form the differential equation of the family of parabolas. For this, we differentiate both parts of the given equation with respect to $x: y^{\prime}=2 a x$. Eliminating the parameter $a$, we find $\frac{y^{\prime}}{y}=\frac{2}{x}$, or $y^{\prime}=\frac{2 y}{x}$ - the differential equation of the given fam... | \frac{x^{2}}{2}+y^{2}=C,C>0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,447 |
Example 7. $\rho^{2}=a \cos 2 \varphi$. Find the orthogonal trajectories of the family of lemniscates | Solution. We have
$\rho^{2}=a \cos 2 \varphi, \quad \rho \rho^{\prime}=-a \sin 2 \varphi$.
By eliminating the parameter $a$, we obtain the differential equation of this family of curves
$$
\rho^{\prime}=-\rho \operatorname{tg} 2 \varphi
$$
Replacing $\rho^{\prime}$ with $-\frac{\rho^{2}}{\rho^{\prime}}$, we find th... | \rho^{2}=C\sin2\varphi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,448 |
Example 1. Find the singular solutions of the differential equation
$$
x y^{\prime}+\left(y^{\prime}\right)^{2}-y=0
$$ | ## Solution.
a) We find the $p$-discriminant curve. In this case,
$$
F\left(x, y, y^{\prime}\right) \equiv x y^{\prime}+\left(y^{\prime}\right)^{2}-y
$$
and condition (2) takes the form
$$
\frac{\partial F}{\partial y^{\prime}} \equiv x+2 y^{\prime}=0
$$
from which $y^{\prime}=-\frac{x}{2}$. Substituting this expr... | -\frac{x^{2}}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,449 |
Example 2. Find the singular solutions of the differential equation
$$
x\left(y^{\prime}\right)^{2}-2 y y^{\prime}+4 x=0, \quad x>0
$$
knowing its general integral
$$
x^{2}=C(y-C) \text {. }
$$ | ## Solution.
a) We find the $C$-discriminant curve. We have
$$
\Phi(x, y, C) \equiv C(y-C)-x^{2}
$$
so
$$
\frac{\partial \Phi}{\partial C} \equiv y-2 C
$$
from which $C=\frac{y}{2}$. Substituting this value of $C$ into (14), we get
$$
x^{2}=\frac{y}{2}\left(y-\frac{y}{2}\right)
$$
from which
$$
(y-2 x)(y+2 x)=0... | \2x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,450 |
Example 3. Find the particular solution of the differential equation
$$
2 y\left(y^{\prime}+2\right)-x\left(y^{\prime}\right)^{2}=0
$$ | Solution. A particular solution, if it exists, is determined by the system
$$
\left\{\begin{array}{r}
2 y\left(y^{\prime}+2\right)-x\left(y^{\prime}\right)^{2}=0 \\
2 y-2 x y^{\prime}=0
\end{array}\right.
$$
where the second equation (19) is obtained by differentiating (18) with respect to $y^{\prime}$. By eliminatin... | 0-4x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,451 |
Example 4. Find the singular solutions of the differential equation
$$
\left(y^{\prime}\right)^{2}=4 x^{2}
$$ | Solution. Differentiating (23) with respect to $y_{1}$:
$$
2 y^{\prime}=0
$$
Excluding $y^{\prime}$ from (23) and (24), we get $x^{2}=0$. The discriminant curve is the y-axis. It is not an integral curve of equation (23), but according to scheme (16) it can be the geometric locus of points of tangency of integral cur... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,452 |
Example 5. Find the singular solutions of the differential equation
$$
\left(y^{\prime}\right)^{2}(2-3 y)^{2}=4(1-y)
$$ | Solution. Let's find the PDK. Excluding $y^{\prime}$ from the system of equations
$$
\left\{\begin{aligned}
\left(y^{\prime}\right)^{2}(2-3 y)^{2}-4(1-y) & =0 \\
y^{\prime}(2-3 y)^{2} & =0
\end{aligned}\right.
$$
we obtain
$$
(2-3 y)^{2}(1-y)=0 .
$$
Transforming equation (25) to the form
$$
\frac{d x}{d y}= \pm \f... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,453 |
Example 6. Find the singular solutions of the differential equation
$$
3 y=2 x y^{\prime}-\frac{2}{x}\left(y^{\prime}\right)^{2}
$$ | ## Solution.
a) We find the $p$-discriminant curve. Differentiating (28) with respect to $y'$, we get
$$
0=2 x-\frac{4}{x} y'
$$
from which
$$
y'=\frac{x^{2}}{2}
$$
Substituting (29) into (28), we find the equation of the PD curve:
$$
\text { PD } \equiv 6 y-x^{3}=0 .
$$
^{2}}$. | Solution. The given equation does not contain the unknown function $y$ and its derivative, so we assume $y^{\prime \prime}=p$. After this, the equation becomes
$$
\frac{d p}{d x}=\sqrt{1+p^{2}}
$$
Separating variables and integrating, we find
$$
p=\frac{e^{x+C_{1}}-e^{-\left(x+C_{1}\right)}}{2}
$$
Replacing $p$ wit... | \operatorname{sh}(x+C_{1})+C_{2}x+C_{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,458 |
Example 5. Solve the equation $y^{\prime \prime}+\left(y^{\prime}\right)^{2}=2 e^{-y}$. | Solution. The equation does not contain the independent variable $x$. Setting $y^{\prime}=p, y^{\prime \prime}=p \frac{d p}{d y}$, we obtain the Bernoulli equation
$$
p \frac{d p}{d y}+p^{2}=2 e^{-y}
$$
By the substitution $p^{2}=z$ it reduces to the linear equation
$$
\frac{d z}{d y}+2 z=4 e^{-y}
$$
the general so... | e^{y}+\tilde{C}_{1}=(x+C_{2})^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,460 |
Example 6. Solve the equation $x^{2} y y^{\prime \prime}=\left(y-x y^{\prime}\right)^{2}$. | Solution. The given equation is homogeneous with respect to $y, y^{\prime}, y^{\prime \prime}$. The order of this equation can be reduced by one using the substitution $y=e^{\int z d x}$, where $z$ is a new unknown function of $x$. We have
$$
y^{\prime}=z e^{\int z d x}, \quad y^{\prime \prime}=\left(z^{\prime}+z^{2}\... | C_{2}xe^{-C_{1}/x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,461 |
Example 7. Solve the equation $x^{3} y^{\prime \prime}=\left(y-x y^{\prime}\right)^{2}$. | Solution. We will show that this equation is a generalized homogeneous equation. Considering $x, y, y^{\prime}, y^{\prime \prime}$ as quantities of the 1st, $m$-th, $(m-1)$-th, and $(m-2)$-th dimensions respectively, and equating the dimensions of all terms, we get
$$
3+(m-2)=2 m
$$
from which $m=1$. The solvability ... | x\ln\frac{x}{C_{1}x+C_{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,462 |
Example 8. Solve the Cauchy problem $y^{\prime \prime}=2 y^{3} ;\left.y\right|_{x=0}=1,\left.y^{\prime}\right|_{x=0}=1$. | Solution. Setting $y^{\prime}=p$, we obtain
$$
p \frac{d p}{d y}=2 y^{3}
$$
from which
$$
p^{2}=y^{4}+C_{1}, \quad \text { or } \quad \frac{d y}{d x}=\sqrt{y^{4}+C_{1}} .
$$
Separating variables, we find
$$
x+C_{2}=\int\left(y^{4}+C_{1}\right)^{-1 / 2} d y
$$
In the right-hand side of the last equation, we have a... | \frac{1}{1-x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,463 |
Example 9. Find the plane curves for which the radius of curvature is proportional to the length of the normal. | Solution. Let $y=y(x)$ be the equation of the desired curve. Its radius of curvature $R=$ $\frac{\left(1+\left(y^{\prime}\right)^{2}\right)^{3 / 2}}{\left|y^{\prime \prime}\right|}$. The length of the normal $M N$ of the curve is (Fig. 24): $M N=|y| \sqrt{1+\left(y^{\prime}\right)^{2}}$.
The defining property of the c... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,464 |
Example 1. Show that the system of functions $1, x, x^{2}, x^{3}$ is linearly independent on the interval $(-\infty,+\infty)$. | Solution. Indeed, the equality $\alpha_{1} \cdot 1+\alpha_{2} x+\alpha_{3} x^{2}+\alpha_{4} x^{3}=0$ can only hold for all $x \in(-\infty,+\infty)$ if $\alpha_{1}=\alpha_{2}=\alpha_{3}=\alpha_{4}=0$. If at least one of these numbers is not zero, then in the left-hand side of the equality we will have a polynomial of de... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,465 |
Example 2. Show that the system of functions $e^{k_{1} x}, e^{k_{2} x}, e^{k_{3} x}$, where $k_{1}, k_{2}, k_{3}$ are pairwise distinct, is linearly independent on the interval $-\infty<x<+\infty$. | Solution. Suppose the opposite, i.e., that the given system of functions is linearly dependent on this interval. Then
$$
\alpha_{1} e^{k_{1} x}+\alpha_{2} e^{k_{2} x}+\alpha_{3} e^{k_{3} x} \equiv 0
$$
on the interval $(-\infty,+\infty)$, and at least one of the numbers $\alpha_{1}, \alpha_{2}, \alpha_{3}$ is not zer... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,466 |
Example 3. Show that the system of functions $e^{\alpha x} \sin \beta x, e^{\alpha x} \cos \beta x$, where $\beta \neq 0$, is linearly independent on the interval $-\infty < x < +\infty$. | Solution. Let us determine the values of $\alpha_{1}$ and $\alpha_{2}$ for which the identity
$$
\alpha_{1} e^{\alpha x} \sin \beta x+\alpha_{2} e^{\alpha x} \cos \beta x \equiv 0
$$
holds. Dividing both sides of the identity by $e^{\alpha x} \neq 0$:
$$
\alpha_{1} \sin \beta x+\alpha_{2} \cos \beta x \equiv 0
$$
S... | proof | Calculus | proof | Yes | Yes | olympiads | false | 34,467 |
Example 4. Prove that the functions
$$
\sin x, \quad \sin \left(x+\frac{\pi}{8}\right), \quad \sin \left(x-\frac{\pi}{8}\right)
$$
are linearly dependent in the interval $(-\infty,+\infty)$. | Solution. We will show that there exist numbers $\alpha_{1}, \alpha_{2}, \alpha_{3}$, not all zero, such that the identity
$$
\alpha_{1} \sin x+\alpha_{2} \sin \left(x+\frac{\pi}{8}\right)+\alpha_{3} \sin \left(x-\frac{\pi}{8}\right) \equiv 0
$$
holds for $-\infty<x<+\infty$.
Assume that identity (7) is satisfied; f... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,468 |
Example 7. Find the Wronskian determinant for the functions $y_{1}(x)=e^{k_{1} x}$, $y_{2}(x)=e^{k_{2} x}, y_{3}(x)=e^{k_{3} x}$.
1) The linear dependence of the functions $\sin x, \sin \left(x+\frac{\pi}{8}\right), \sin \left(x-\frac{\pi}{8}\right)$ can be established by noting that $\sin \left(x+\frac{\pi}{8}\right)... | ## Solution. We have
$W\left[y_{1}, y_{2}, y_{3}\right]=\left|\begin{array}{lll}e^{k_{1} x} & e^{k_{2} x} & e^{k_{3} x} \\ k_{1} e^{k_{1} x} & k_{2} e^{k_{2} x} & k_{3} k_{1} k_{3} \\ k_{1}^{2} e^{k_{1} x} & k_{2}^{2} e^{k_{2} z} & k_{3}^{2} e^{k_{3} x}\end{array}\right|=e^{\left(k_{1}+k_{2}+k_{3}\right) x}\left(k_{2}... | e^{(k_{1}+k_{2}+k_{3})x}(k_{2}-k_{1})(k_{3}-k_{1})(k_{3}-k_{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,469 |
Example 8. Find the Wronskian determinant for the functions: $y_{1}(x)=\sin x$,
$$
y_{2}(x)=\sin \left(x+\frac{\pi}{8}\right), y_{3}(x)=\sin \left(x-\frac{\pi}{8}\right)
$$ | Solution. We have
$$
W\left[y_{1}, y_{2}, y_{3}\right]=\left|\begin{array}{rrr}
\sin x & \sin \left(x+\frac{\pi}{8}\right) & \sin \left(x-\frac{\pi}{8}\right) \\
\cos x & \cos \left(x+\frac{\pi}{8}\right) & \cos \left(x-\frac{\pi}{8}\right) \\
-\sin x & -\sin \left(x+\frac{\pi}{8}\right) & -\sin \left(x-\frac{\pi}{8}\... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,470 |
Example 10. Show that the functions $y_{1}=x, y_{2}=2 x$ are linearly dependent on the interval $[0,1]$. | ## Solution. We have
$$
\begin{gathered}
\left(y_{1}, y_{1}\right)=\int_{0}^{1} x^{2} d x=\frac{1}{3},\left(y_{1}, y_{2}\right)=\left(y_{2}, y_{1}\right)=\int_{0}^{1} 2 x^{2} d x=\frac{2}{3},\left(y_{2}, y_{2}\right)=\int_{0}^{1} 4 x^{2} d x=\frac{4}{3} \\
\Gamma\left(y_{1}, y_{2}\right)=\left|\begin{array}{ll}
\frac{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,471 |
Example 1. Find the general solution of the equation
$$
y^{\prime \prime \prime}-2 y^{\prime \prime}-3 y^{\prime}=0
$$ | Solution. We form the characteristic equation
$$
\lambda^{3}-2 \lambda^{2}-3 \lambda=0
$$
We find its roots: $\lambda_{1}=0, \lambda_{2}=-1, \lambda_{3}=3$. Since they are real and distinct, the general solution is
$$
y_{0.0}=C_{1}+C_{2} e^{-x}+C_{3} e^{3 x} .
$$ | y_{0}=C_{1}+C_{2}e^{-x}+C_{3}e^{3x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,472 |
Example 2. Find the general solution of the equation
$$
y^{\prime \prime \prime}+2 y^{\prime \prime}+y^{\prime}=0
$$ | Solution. The characteristic equation is
$$
\lambda^{3}+2 \lambda^{2}+\lambda=0 .
$$
From this, $\lambda_{1}=\lambda_{2}=-1, \lambda_{3}=0$. The roots are real, and one of them, namely $\lambda=-1$, is a double root, so the general solution is
$$
y_{0.0}=C_{1} e^{-x}+C_{2} x e^{-x}+C_{3}
$$ | y_{0}=C_{1}e^{-x}+C_{2}xe^{-x}+C_{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,473 |
Example 3. Find the general solution of the equation $y^{\prime \prime \prime}+4 y^{\prime \prime}+13 y^{\prime}=0$. | Solution. The characteristic equation
$$
\lambda^{3}+4 \lambda^{2}+13 \lambda=0
$$
has roots $\lambda_{1}=0, \lambda_{2}=-2-3 i, \lambda_{3}=-2+3 i$.
The general solution is
$$
y_{0.0}=C_{1}+C_{2} e^{-2 x} \cos 3 x+C_{3} e^{-2 x} \sin 3 x
$$ | y_{0}=C_{1}+C_{2}e^{-2x}\cos3x+C_{3}e^{-2x}\sin3x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,474 |
Example 4. Find the general solution of the equation
$$
y^{\mathrm{V}}-2 y^{\mathrm{IV}}+2 y^{\prime \prime \prime}-4 y^{\prime \prime}+y^{\prime}-2 y=0
$$ | Solution. The characteristic equation
$$
\lambda^{5}-2 \lambda^{4}+2 \lambda^{3}-4 \lambda^{2}+\lambda-2=0
$$
or
$$
(\lambda-2)\left(\lambda^{2}+1\right)^{2}=0
$$
has roots $\lambda=2$ - a single root and $\lambda= \pm i$ - a pair of double imaginary roots. The general solution is
$$
y_{0.0}=C_{1} e^{2 x}+\left(C_... | y_{0}=C_{1}e^{2x}+(C_{2}+C_{3}x)\cosx+(C_{4}+C_{5}x)\sinx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,475 |
Example 5. Solve the equation
$$
y^{\mathrm{IV}}+4 y^{\prime \prime \prime}+8 y^{\prime \prime}+8 y^{\prime}+4 y=0
$$ | Solution. We form the characteristic equation
$$
\lambda^{4}+4 \lambda^{3}+8 \lambda^{2}+8 \lambda+4=0
$$
or
$$
\left(\lambda^{2}+2 \lambda+2\right)^{2}=0
$$
It has double complex roots $\lambda_{1}=\lambda_{2}=-1-i, \lambda_{3}=\lambda_{4}=-1+i$ and, consequently, the general solution will have the form
$$
y_{0.0... | y_{0}=e^{-x}(C_{1}+C_{3}x)\cosx+e^{-x}(C_{2}+C_{4}x)\sinx | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,476 |
Example 3. Find the general solution of the equation $y^{\prime \prime}+y^{\prime}=4 x^{2} e^{x}$. | Solution. The characteristic equation $\lambda^{2}+\lambda=0$ has roots $\lambda_{1}=0$, $\lambda_{2}=-1$. Therefore, the general solution $y_{0.0}$ of the corresponding homogeneous equation is
$$
y_{0.0}=C_{1}+C_{2} e^{-x}
$$
Since $\alpha=1$ is not a root of the characteristic equation, the particular solution $y_{... | y(x)=C_{1}+C_{2}e^{-x}+(2x^{2}-6x+7)e^{x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,477 |
Example 4. Find the general solution of the equation
$$
y^{\prime \prime}+10 y^{\prime}+25 y=4 e^{-5 x}
$$ | Solution. The characteristic equation $\lambda^{2}+10 \lambda+25=0$ has a double root $\lambda_{1}=\lambda_{2}=-5$, therefore
$$
y_{0.0}=\left(C_{1}+C_{2} x\right) e^{-5 x}
$$
Since $\alpha=-5$ is a root of the characteristic equation of multiplicity $s=2$, we seek the particular solution $y_{\text {p }}$ of the non-... | y(x)=(C_{1}+C_{2}x)e^{-5x}+2x^{2}e^{-5x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,478 |
Example 5. Find the general solution of the equation
$$
y^{\prime \prime}+3 y^{\prime}+2 y=x \sin x
$$ | Solution. First method. The characteristic equation $\lambda^{2}+3 \lambda+2=0$ has roots $\lambda_{1}=-1, \lambda_{2}=-2$, therefore
$$
y_{0.0}=C_{1} e^{-x}+C_{2} e^{-2 x}
$$
Since the number $i$ is not a root of the characteristic equation, the particular solution $y_{\text {p.n }}$ of the non-homogeneous equation ... | y(x)=C_{1}e^{-x}+C_{2}e^{-2x}+(-\frac{3}{10}x+\frac{17}{50})\cosx+(\frac{1}{10}x+\frac{3}{25})\sinx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,479 |
Example 6. Find the general solution of the equation $y^{\prime \prime}+4 y=\sin 2 x$. | Solution. Consider the equation $z^{\prime \prime}+4 z=e^{2 i x}$.
We have
$$
\sin 2 x=\operatorname{Im} e^{2 i x}
$$
therefore
$$
y_{4, \mu}=\operatorname{Im} z_{4, H} .
$$
The characteristic equation $\lambda^{2}+4=0$ has simple roots $\lambda_{1,2}= \pm 2 i$. Therefore, we seek a particular solution in the form... | y_{4,11}=-\frac{1}{4}x\cos2x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,480 |
Example 7. Find the general solution of the equation $y^{\prime \prime}-6 y^{\prime}+9 y=25 e^{x} \sin x$. | Solution. The characteristic equation $\lambda^{2}-6 \lambda+9=0$ has roots $\lambda_{1}=\bar{\lambda}_{2}=3$; the general solution $y_{\text {g.s.}}$ of the homogeneous equation will be
$$
y_{0.0}=\left(C_{1}+C_{2} x\right) e^{3 x} .
$$
The numbers $1 \pm i$ are not roots of the characteristic equation, so the parti... | y(x)=(C_{1}+C_{2}x)e^{3x}+e^{x}(4\cosx+3\sinx) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,481 |
Example 8. Find the general solution of the equation
$$
y^{\prime \prime}+2 y^{\prime}+5 y=e^{-x} \cos 2 x
$$ | Solution. The characteristic equation $\lambda^{2}+2 \lambda+5=0$ has roots $\lambda_{1,2}=-1 \pm 2 i$, so
$$
y_{0.0}=\left(C_{1} \cos 2 x+C_{2} \sin 2 x\right) e^{-x}
$$
Since the number $\alpha+i \beta=-1+2 i$ is a simple root of the characteristic equation, we need to look for $y_{\text {ch.n }}$ in the form (see ... | y(x)=(C_{1}\cos2x+C_{2}\sin2x)e^{-x}+\frac{1}{4}xe^{-x}\sin2x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,482 |
Example 9. Solve the equation
$$
y^{\prime \prime}-6 y^{\prime}+9 y=4 e^{x}-16 e^{3 x}
$$ | Solution. The characteristic equation $\lambda^{2}-6 \lambda+9=0$ has roots $\lambda_{1}=\lambda_{2}=3$, and therefore the general solution of the corresponding homogeneous equation will be
$$
y_{0.0}=C_{1} e^{3 x}+C_{2} x e^{3 x}
$$
To find a particular solution of the non-homogeneous equation (14), we will find par... | (C_{1}+C_{2}x)e^{3x}+e^{x}-8x^{2}e^{3x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,483 |
Example 10. Solve the equation
$$
y^{\prime \prime \prime}-2 y^{\prime \prime}+2 y^{\prime}=4 \cos x \cos 3 x+6 \sin ^{2} x
$$ | Solution. Using known trigonometric identities, we transform the right-hand side of equation (17) to a "standard" form:
$$
4 \cos x \cos 3 x+6 \sin ^{2} x=2 \cos 4 x-\cos 2 x+3
$$
The original equation (17) can now be written as:
$$
y^{\prime \prime \prime}-2 y^{\prime \prime}+2 y^{\prime}=2 \cos 4 x-\cos 2 x+3
$$
... | C_{1}+(C_{2}\cosx+C_{3}\sinx)e^{x}+\frac{1}{65}(\cos4x-\frac{7}{4}\sin4x)+\frac{1}{10}(\frac{\sin2x}{2}-\cos2x)+\frac{3}{2}x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,484 |
Example 11. Find a particular solution of the equation
$$
y^{\prime \prime}-y=4 e^{x},
$$
satisfying the initial conditions
$$
y(0)=0, \quad y^{\prime}(0)=1
$$ | Solution. We find the particular solution of equation (22)
$$
y=C_{1} e^{x}+C_{2} e^{-x}+2 x e^{x}
$$
To solve the initial value problem (22)-(23) (Cauchy problem), it is necessary to determine the values of the constants $C_{1}$ and $C_{2}$ so that the solution (24) satisfies the initial conditions (23). Using the c... | 2xe^{x}-\operatorname{sh}x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,485 |
Example 12. Find a particular solution of the equation
$$
y^{\prime \prime}+4 y^{\prime}+5 y=8 \cos x
$$
bounded as $x \rightarrow-\infty$. | Solution. The general solution of the given equation is
$$
y=e^{-2 x}\left(C_{1} \cos x+C_{2} \sin x\right)+2(\cos x+\sin x)
$$
As $x \rightarrow-\infty$, the quantity $e^{-2 x} \rightarrow+\infty$ and for any $C_{1}$ and $C_{2}$, not both zero, the first term on the right-hand side (26) will be an unbounded function... | 2(\cosx+\sinx) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,486 |
Example 1. Find the general solution of the Euler equation $x^{2} y^{\prime \prime}+2 x y^{\prime}-$ $6 y=0$. | Solution. First method. In the equation, we make the substitution $x=e^{t}$, then
$$
\begin{aligned}
& y^{\prime}=\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=e^{-t} \frac{d y}{d t} \\
& y^{\prime \prime}=\frac{d y^{\prime}}{d x}=\frac{d y^{\prime} / d t}{d x / d t}=\frac{\left(d^{2} y / d t^{2}-d y / d t\right) e^{-t}... | \frac{C_{1}}{x^{3}}+C_{2}x^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,487 |
Example 2. Solve the Euler equation $x^{2} y^{\prime \prime}-x y^{\prime}+2 y=x \ln x$. | Solution. The characteristic equation $k(k-1)-k+2=0$, or $k^{2}-$ $2 k+2=0$ has roots, $k_{1}=1-i, k_{2}=1+i$. Therefore, the general solution of the corresponding homogeneous equation will be
$$
y_{0.0}=x\left(C_{1} \cos \ln x+C_{2} \sin \ln x\right)
$$
We seek a particular solution in the form $y_{4}=x(A \ln x+B)$;... | x(C_{1}\cos\lnx+C_{2}\sin\lnx)+x\lnx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,488 |
Example 1. Find the general solution of the equation $x y^{\prime \prime}+2 y^{\prime}+x y=0$, if $y_{1}=\frac{\sin x}{x}$ is its particular solution. then | Solution. Let $y=\frac{\sin x}{x} \cdot z$, where $z$ is a new unknown function of $x$;
$$
y^{\prime}=y_{1}^{\prime} z+y_{1} z^{\prime}, \quad y^{\prime \prime}=y_{1}^{\prime \prime} z+2 y_{1}^{\prime} z^{\prime}+y_{1} z^{\prime \prime}
$$
Substituting into the given equation, we get
$$
\left(x y_{1}^{\prime \prime}... | C_{1}\frac{\cosx}{x}+C_{2}\frac{\sinx}{x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,489 |
Example 2. Find the general solution of the equation $y^{\prime \prime}+\frac{2}{x} y^{\prime}+y=\frac{1}{x},(x \neq 0)$. | Solution. The general solution of the corresponding homogeneous equation has the form (see Example 1)
$$
y_{0.0}=C_{1} \frac{\sin x}{x}+C_{2} \frac{\cos x}{x}
$$
and therefore, its fundamental system of solutions will be
$$
y_{1}=\frac{\sin x}{x}, \quad y_{2}=\frac{\cos x}{x}
$$
We will seek the general solution of... | \widetilde{C}_{1}\frac{\sinx}{x}+\widetilde{C}_{2}\frac{\cosx}{x}+\frac{1}{x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,490 |
Example 3. Solve the equation $y^{\prime \prime}+y=\frac{1}{\cos x}$. | Solution. The corresponding homogeneous equation will be $y^{\prime \prime}+y=0$. Its characteristic equation $\lambda^{2}+1=0$ has imaginary roots $\lambda_{1}=-i, \lambda_{2}=i$, and the general solution of the homogeneous equation is
$$
y_{0.0}=C_{1} \cos x+C_{2} \sin x
$$
The general solution of the original equa... | \widetilde{C}_{1}\cosx+\widetilde{C}_{2}\sinx+\cosx\cdot\ln|\cosx|+x\sinx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,491 |
Example 4. Knowing the fundamental system of solutions $y_{1}=\ln x, y_{2}=x$ of the corresponding homogeneous equation, find a particular solution of the equation
$$
x^{2}(1-\ln x) y^{\prime \prime}+x y^{\prime}-y=\frac{(1-\ln x)^{2}}{x}
$$
satisfying the condition $\lim _{x \rightarrow+\infty} y=0$. | Solution. Applying the method of variation of constants, we find the general solution of equation (39):
$$
y=C_{1} \ln x+C_{2} x+\frac{1-2 \ln x}{4 x}
$$
As $x \rightarrow+\infty$, the first two terms on the right-hand side of (40) tend to infinity, and for any $C_{1}, C_{2}$, not both zero, the function $C_{1} \ln x... | \frac{1-2\lnx}{4x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,492 |
Example 1. Form a differential equation for which $y_{1}(x)=e^{x}, y_{2}(x)=e^{-x}$ form a fundamental system of solutions. | Solution. Applying formula (42), we get
$$
\left|\begin{array}{rrr}
e^{x} & e^{-x} & y \\
e^{x} & -e^{-x} & y^{\prime} \\
e^{x} & e^{-x} & y^{\prime \prime}
\end{array}\right|=0, \quad \text { or } \quad\left|\begin{array}{rrr}
1 & 1 & y^{\prime} \\
1 & -1 & y^{\prime} \\
1 & 1 & y^{\prime \prime}
\end{array}\right|=0... | y^{\\}-y=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,493 |
Example 2. Form a differential equation for which the functions $y_{1}(x)=e^{x^{2}}, y_{2}(x)=e^{-x^{2}}$ form a fundamental system of solutions. | Solution. Let us form an equation of the form (42):
$$
\left|\begin{array}{ccc}
e^{x^{2}} & e^{-x^{2}} & y \\
2 x e^{x^{2}} & -2 x e^{-x^{2}} & y^{\prime} \\
\left(2+4 x^{2}\right) e^{x^{2}} & \left(4 x^{2}-2\right) e^{-x^{2}} & y^{\prime \prime}
\end{array}\right|=0, \quad \text { or }\left|\begin{array}{ccc}
1 & 1 &... | xy^{\\}-y^{\}-4x^{3}0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,494 |
Example 1. Show that the linear differential equation $x y^{\prime \prime}-(x+2) y^{\prime}+y=0$ has a solution of the form $y_{1}=P(x)$, where $P(x)$ is some polynomial. Show that the second solution $y_{2}$ of this equation has the form $y_{2}=e^{x} Q(x)$, where $Q(x)$ is also a polynomial. | Solution. We will look for a solution $y_{1}(x)$ in the form of a polynomial, for example, of the first degree: $y_{1}=A x+B$. Substituting into the equation, we find that $-2 A+B=0$. Let $A=1$, then $B=2$; thus, the polynomial $y_{1}=x+2$ will be a solution to the given equation. Rewrite the given equation as
$$
y^{\... | y_{2}=A(x-2)e^{x} | Calculus | proof | Yes | Yes | olympiads | false | 34,495 |
Example 1. Construct the phase plane trajectories for the equation
$$
\frac{d^{2} x}{d t^{2}}+x=0
$$ | Solution. Let $\frac{d x}{d t}=v$. Equation (3) takes the form
$$
v \frac{d v}{d x}+x=0, \quad \text { or } \quad \frac{d v}{d x}=-\frac{x}{v} .
$$
The isocline equations for (4): $-\frac{x}{v}=k$. By constructing isoclines corresponding to different values of $k$, we find that the phase trajectories are circles cent... | x()=C_{1}\cos+C_{2}\sin | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,496 |
Example 1. Solve the boundary value problem
$$
y^{\prime \prime}-y=0, \quad y^{\prime}(0)=0, \quad y(1)=1
$$ | Solution. The general solution of the given equation is
$$
y(x)=C_{1} e^{x}+C_{2} e^{-x}
$$
hence
$$
y^{\prime}(x)=C_{1} e^{x}-C_{2} e^{-x}
$$
Setting $x=0$ in (6) and $x=1$ in (5) and considering the boundary conditions, we obtain the following non-homogeneous linear system for finding the values of the constants ... | y(x)=\frac{\operatorname{ch}x}{\operatorname{ch}1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,498 |
Example 2. Find the eigenvalues and eigenfunctions of the boundary value problem
$$
\begin{aligned}
& y^{\prime \prime}+\lambda^{2} y=0 \quad(\lambda \neq 0) \\
& y^{\prime}(0)=0, \quad y(\pi)=0
\end{aligned}
$$ | Solution. The general solution of equation (7)
$$
y(x)=C_{1} \cos \lambda x+C_{2} \sin \lambda x
$$
hence
$$
y^{\prime}(x)=-C_{1} \lambda \sin \lambda x+C_{2} \lambda \cos \lambda x
$$
Setting $x=\pi$ in (9) and $x=0$ in (10) and taking into account the boundary conditions (8), we obtain for finding $C_{1}$ and $C_... | \lambda_n=\frac{2n+1}{2},\quadn=0,1,2,\ldots\quad\quady_n(x)=\cos\frac{2n+1}{2}x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,499 |
Example 3. Solve the boundary value problem $x^{2} y^{\prime \prime}+2 x y^{\prime}-6 y=0, y(1)=1$, $y(x)$ is bounded as $x \rightarrow 0$. | Solution. The given equation is an Euler equation. Its general solution has the form $y(x)=\frac{C_{1}}{x^{3}}+C_{2} x^{2}$ (see, for example, 1, item $4^{\circ}, \S$ 15). By the condition, the solution $y(x)$ must be bounded as $x \rightarrow 0$. This requirement will be satisfied if in the general solution we set $C_... | x^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,500 |
Example 1. Find the solutions of the equation
$$
y^{\prime \prime}-x y-2 y=0
$$
in the form of a power series. | Solution. We seek $y_{1}(x)$ in the form of a series
$$
y_{1}(x)=\sum_{k=0}^{\infty} c_{k} x^{k}
$$
then
$$
y_{1}^{\prime}(x)=\sum_{k=1}^{\infty} k c_{k} x^{k-1}, \quad y_{1}^{\prime \prime}(x)=\sum_{k=2}^{\infty} k(k-1) c_{k} x^{k-2}
$$
Substituting $y_{1}(x), y_{1}^{\prime}(x)$ and $y_{1}^{\prime \prime}(x)$ into... | y(x)=Ay_{1}(x)+By_{2}(x) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,501 |
Example 2. Find the solution of the equation
$$
y^{\prime \prime}+y=0
$$
satisfying the initial conditions
$$
\left.y\right|_{x=0}=1,\left.\quad y^{\prime}\right|_{x=0}=0
$$ | Solution. The particular solution of equation (16), satisfying the initial conditions (17), is sought in the form of a series
$$
y(x)=y(0)+\frac{y^{\prime}(0)}{1!} x+\frac{y^{\prime \prime}(0)}{2!} x^{2}+\frac{y^{\prime \prime \prime}(0)}{3!} x^{3}+\ldots
$$
Given $y(0)=1, y^{\prime}(0)=0$.
From the given equation, ... | y(x)=\cosx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,502 |
Example 3. Find the first four terms of the Taylor series expansion of the solution $y=y(x)$ of the equation $y^{\prime \prime}=e^{x y}$, satisfying the initial conditions $\left.y\right|_{x=0}=1,\left.y^{\prime}\right|_{x=0}=0$. | Solution. It is easy to see that the right-hand side of the equation, i.e., the function $e^{z y}$, can be expanded into a power series in powers of $x$ and $y$ in a neighborhood of the point (0,0), converging in the domain $-\infty < x < +\infty, -\infty < y < +\infty$ (i.e., the right-hand side is holomorphic).
We w... | y(x)=1+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,503 |
## Example 4. Solve the equation
$$
2 x^{2} y^{\prime \prime}+\left(3 x-2 x^{2}\right) y^{\prime}-(x+1) y=0
$$ | Solution. Rewrite (27) as
$$
y^{\prime \prime}+\frac{3 x-2 x^{2}}{2 x^{2}} y^{\prime}-\frac{x+1}{2 x^{2}} y=0
$$
or
$$
y^{\prime \prime}+\frac{3-2 x}{2 x} y^{\prime}-\frac{x+1}{2 x^{2}} y=0
$$
The solution $y(x)$ will be sought in the form
$$
y(x)=x^{\rho} \sum_{k=0}^{x} C_{k} x^{k} \quad\left(C_{0} \neq 0\right)
... | y(x)=Ay_{1}(x)+By_{2}(x) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,504 |
Example 6. Solve the Bessel equation
$$
x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-p^{2}\right) y=0, \quad x>0
$$
where $p$ is a given constant. | Solution. Rewrite (36) as
$$
y^{\prime \prime}+\frac{1}{x} y^{\prime}+\frac{x^{2}-p^{2}}{x^{2}} y=0
$$
Here
$$
p(x)=\frac{1}{x}, \quad q(x)=\frac{x^{2}-p^{2}}{x^{2}}
$$
so that
$$
a_{0}=\lim _{x \rightarrow 0} x p(x)=1, \quad b_{0}=\lim _{x \rightarrow 0} x^{2} q(x)=-p^{2}
$$
(see formulas (24)). The characterist... | J_{-p}(x)=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!\Gamma(k+1-p)}(\frac{x}{2})^{2k-p} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,505 |
Example 7. Reduce the equation
$$
x^{2} \frac{d^{2} y}{d x^{2}}-3 x \frac{d y}{d x}+\left(x^{4}-12\right) y=0
$$
to a Bessel equation and find its general solution. | Solution. In our case, the coefficients are $a=-3, b=-12, c=1$, $m=\overline{4, \text { therefore }}$
$$
\begin{aligned}
\alpha=\frac{a-1}{2} & =-2, \quad \beta=\frac{m}{2}=2, \quad-\frac{\alpha}{\beta}=1, \quad \frac{1}{\beta}=\frac{1}{2} \\
\gamma & =\frac{2 \sqrt{c}}{m}=\frac{1}{2}, \quad p^{2}=\frac{(a-1)^{2}-4 b}... | x^{2}[C_{1}J_{2}(\frac{x^{2}}{2})+C_{2}Y_{2}(\frac{x^{2}}{2})] | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,506 |
Example 1. Reduce the system of equations to canonical form
$$
\left\{\begin{aligned}
y_{2} y_{1}^{\prime}-\ln \left(y_{1}^{\prime \prime}-y_{1}\right) & =0 \\
e^{y_{2}^{\prime}}-y_{1}-y_{2} & =0
\end{aligned}\right.
$$ | Solution. The given system is of the third order, since $k_{1}=2, k_{2}=1$, and thus $p=3$. Solving the first equation for $y_{1}^{\prime \prime}$ and the second for $y_{2}^{\prime}$, we obtain the canonical system
$$
y_{1}^{\prime \prime}=y_{1}+e^{y_{2} y_{1}^{\prime}}, \quad y_{2}^{\prime}=\ln \left(y_{1}+y_{2}\righ... | y_{1}^{\\}=y_{1}+e^{y_{2}y_{1}^{\}},\quady_{2}^{\}=\ln(y_{1}+y_{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,508 |
Example 2. Reduce the following system of differential equations to a normal system:
$$
\left\{\begin{array}{r}
\frac{d^{2} x}{d t^{2}}-y=0 \\
t^{3} \frac{d y}{d t}-2 x=0
\end{array}\right.
$$ | Solution. Let $x=x_{1}, \frac{d x}{d t}=x_{2}, y=x_{3}$. Then we will have $\frac{d x_{1}}{d t}=x_{2}$, $\frac{d y}{d t}=\frac{d x_{3}}{d t}$, and the given system will be reduced to the following normal third-order system:
$$
\left\{\begin{array}{l}
\frac{d x_{1}}{d t}=x_{2} \\
\frac{d x_{2}}{d t}=x_{3} \\
\frac{d x_... | {\begin{pmatrix}\frac{x_{1}}{}=x_{2}\\\frac{x_{2}}{}=x_{3}\\\frac{x_{3}}{}=\frac{2x_{1}}{^{3}}\end{pmatrix}.} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,509 |
Example 3. Convert the differential equation
$$
\frac{d^{2} x}{d t^{2}}+p(t) \frac{d x}{d t}+q(t) x=0
$$
to a normal system. | Solution. Let $x=x_{1}, \frac{d x}{d t}=x_{2}$, then $\frac{d x_{1}}{d t}=x_{2}, \frac{d^{2} x}{d t^{2}}=\frac{d x_{2}}{d t}$. Substituting these expressions into the given equation, we get
$$
\frac{d x_{2}}{d t}+p(t) x_{2}+q(t) x_{1}=0
$$
The normal system will have the form
$$
\frac{d x_{1}}{d t}=x_{2}, \quad \fra... | \frac{x_{1}}{}=x_{2},\quad\frac{x_{2}}{}=-p()x_{2}-q()x_{1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 34,510 |
Example 4. Show that the system of functions $x_{1}=-\frac{1}{t^{2}}, x_{2}=-t \ln t$, defined in the interval $0<t<+\infty$, is a solution to the system of differential equations
$$
\left\{\begin{array}{l}
\frac{d x_{1}}{d t}=2 t x_{1}^{2} \\
\frac{d x_{2}}{d t}=\frac{x_{2}}{t}-1
\end{array}\right.
$$ | Solution. We have $\frac{d x_{1}}{d t}=\frac{2}{t^{3}}, \frac{d x_{2}}{d t}=-1-\ln t$. Substituting into the given system of equations the expressions for $x_{1}, x_{2}, \frac{d x_{1}}{d t}$, and $\frac{d x_{2}}{d t}$ in terms of $t$, we obtain the identities
$$
\frac{2}{t^{3}} \equiv \frac{2 t}{t^{4}} \equiv \frac{2}... | proof | Calculus | proof | Yes | Yes | olympiads | false | 34,511 |
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