problem stringlengths 75 971 | solution stringlengths 17 716 | type stringclasses 1
value | idx int64 61 108 | answer listlengths 1 3 |
|---|---|---|---|---|
Consider the following "mixing problem." A tank holds $V$ liters of salt water. Suppose that a saline solution with concentration of $c \mathrm{gm} /$ liter is added at the rate of $r$ liters/minute. A mixer keeps the salt essentially uniformly distributed in the tank. A pipe lets solution out of the tank at the same r... | The differential equation for $y(t)$ is $\boxed{y^{\prime}+r y-r x(t)=0}$. | Differential Equations (18.03 Spring 2010) | 62 | [
"y^{\\prime}+ry-rx(t)=0\n"
] |
Preamble: The following subproblems refer to a circuit with the following parameters. Denote by $I(t)$ the current (where the positive direction is, say, clockwise) in the circuit and by $V(t)$ the voltage increase across the voltage source, at time $t$. Denote by $R$ the resistance of the resistor and $C$ the capacita... | $c=\boxed{I(0)}$. | Differential Equations (18.03 Spring 2010) | 61 | [
"I(0)e^{-\\frac{t}{RC}}\n"
] |
Preamble: In the following problems, take $a = \ln 2$ and $b = \pi / 3$.
Given $a = \ln 2$ and $b = \pi / 3$, rewrite $e^{a+b i}$ in the form $x + yi$, where $x, y$ are real numbers. | Using Euler's formula, we find that the answer is $\boxed{1+\sqrt{3} i}$. | Differential Equations (18.03 Spring 2010) | 67 | [
"1 + i\\sqrt{3}\n",
"1 + i \\sqrt{3}\n"
] |
In the polar representation $(r, \theta)$ of the complex number $z=1+\sqrt{3} i$, what is $r$? | For z, $r=2$ and $\theta=\pi / 3$, so its polar coordinates are $\left(2, \frac{\pi}{3}\right)$. So $r=\boxed{2}$. | Differential Equations (18.03 Spring 2010) | 66 | [
"2\n",
"2"
] |
Subproblem 0: Find the general solution of the differential equation $\dot{x}+2 x=e^{t}$, using $c$ for the arbitrary constant of integration which will occur.
Solution: We can use integrating factors to get $(u x)^{\prime}=u e^{t}$ for $u=e^{2 t}$. Integrating yields $e^{2 t} x=e^{3 t} / 3+c$, or $x=\boxed{\frac{e^{... | When $c=0, x=\boxed{e^{t} / 3}$ is the solution of the required form. | Differential Equations (18.03 Spring 2010) | 74 | [
"\\frac{e^{t}} {3}+c e^{-2 t}\n",
"\\frac{e^{t}}{3}+ce^{-2t}\n"
] |
Find the complex number $a+b i$ with the smallest possible positive $b$ such that $e^{a+b i}=1+\sqrt{3} i$. | $1+\sqrt{3} i$ has modulus 2 and argument $\pi / 3+2 k \pi$ for all integers k, so $1+\sqrt{3} i$ can be expressed as a complex exponential of the form $2 e^{i(\pi / 3+2 k \pi)}$. Taking logs gives us the equation $a+b i=\ln 2+i(\pi / 3+2 k \pi)$. The smallest positive value of $b$ is $\pi / 3$. Thus we have $\boxed{\l... | Differential Equations (18.03 Spring 2010) | 73 | [
"\\ln 2 + i\\frac{\\pi}{3}\n",
"\\ln 2 + \\frac{i\\pi}{3}\n"
] |
Preamble: The following subproblems relate to applying Euler's Method (a first-order numerical procedure for solving ordinary differential equations with a given initial value) onto $y^{\prime}=y^{2}-x^{2}=F(x, y)$ at $y(0)=-1$, with $h=0.5$. Recall the notation \[x_{0}=0, y_{0}=-1, x_{n+1}=x_{h}+h, y_{n+1}=y_{n}+m_{n}... | $y_3 = \boxed{-0.875}$ | Differential Equations (18.03 Spring 2010) | 69 | [
"-0.875\n"
] |
Subproblem 0: Find the general solution of the differential equation $y^{\prime}=x-2 y$ analytically using integrating factors, solving for $y$. Note that a function $u(t)$ such that $u \dot{x}+u p x=\frac{d}{d t}(u x)$ is an integrating factor. Additionally, note that a general solution to a differential equation has ... | The straight line solution occurs when $c=\boxed{0}$. | Differential Equations (18.03 Spring 2010) | 68 | [
"\\frac{x}{2} - \\frac{1}{4} + ce^{-2x}",
"\\frac{x}{2} - \\frac{1}{4} + ce^{-2x}\n"
] |
An African government is trying to come up with good policy regarding the hunting of oryx. They are using the following model: the oryx population has a natural growth rate of $k$, and we suppose a constant harvesting rate of $a$ oryxes per year.
Write down an ordinary differential equation describing the evolution of ... | The natural growth rate is $k$, meaning that after some short time $\Delta t$ year(s) passes, we expect $k x(t) \Delta t$ new oryxes to appear. However, meanwhile the population is reduced by $a \Delta t$ oryxes due to the harvesting. Therefore, we are led to
\[
x(t+\Delta t) \simeq x(t)+k x(t) \Delta t-a \Delta t,
\]
... | Differential Equations (18.03 Spring 2010) | 64 | [
"\\frac{dx}{dt} = kx - a"
] |
If the complex number $z$ is given by $z = 1+\sqrt{3} i$, what is the magnitude of $z^2$? | $z^{2}$ has argument $2 \pi / 3$ and radius 4, so by Euler's formula, $z^{2}=4 e^{i 2 \pi / 3}$. Thus $A=4, \theta=\frac{2\pi}{3}$, so our answer is $\boxed{4}$. | Differential Equations (18.03 Spring 2010) | 65 | [
"4",
"4\n"
] |
Given the ordinary differential equation $\ddot{x}-a^{2} x=0$, where $a$ is a nonzero real-valued constant, find a solution $x(t)$ to this equation such that $x(0) = 0$ and $\dot{x}(0)=1$. | First, notice that both $x(t)=e^{a t}$ and $x(t)=e^{-a t}$ are solutions to $\ddot{x}-a^{2} x=0$. Then for any constants $c_{1}$ and $c_{2}$, $x(t)=c_{1} e^{a t}+c_{2} e^{-a t}$ are also solutions to $\ddot{x}-a^{2} x=0$. Moreover, $x(0)=c_{1}+c_{2}$, and $\dot{x}(0)=a\left(c_{1}-c_{2}\right)$. Assuming $a \neq 0$, to ... | Differential Equations (18.03 Spring 2010) | 71 | [
"\\frac{e^{at}-e^{-at}}{2a}\n"
] |
Find a solution to the differential equation $\ddot{x}+\omega^{2} x=0$ satisfying the initial conditions $x(0)=x_{0}$ and $\dot{x}(0)=\dot{x}_{0}$. | Suppose \[x(t)=a \cos (\omega t)+b \sin (\omega t)\] $x(0)=a$, therefore $a=x_{0}$. Then \[x^{\prime}(0)=-a \omega \sin 0+b \omega \cos 0=b \omega=\dot{x}_{0}\] Then $b=\dot{x}_{0} / \omega$. The solution is then $x=\boxed{x_{0} \cos (\omega t)+$ $\dot{x}_{0} \sin (\omega t) / \omega}$. | Differential Equations (18.03 Spring 2010) | 72 | [
"x_{0} \\cos (\\omega t)+\\frac{\\dot{x}_{0}}{\\omega} \\sin (\\omega t)\n"
] |
Rewrite the function $f(t) = \cos (2 t)+\sin (2 t)$ in the form $A \cos (\omega t-\phi)$. It may help to begin by drawing a right triangle with sides $a$ and $b$. | Here, our right triangle has hypotenuse $\sqrt{2}$, so $A=\sqrt{2}$. Both summands have "circular frequency" 2, so $\omega=2 . \phi$ is the argument of the hypotenuse, which is $\pi / 4$, so $f(t)=\boxed{\sqrt{2} \cos (2 t-\pi / 4)}$. | Differential Equations (18.03 Spring 2010) | 70 | [
"\\sqrt{2} \\cos (2 t-\\pi / 4)\n"
] |
Subproblem 0: Find a purely exponential solution of $\frac{d^{4} x}{d t^{4}}-x=e^{-2 t}$.
Solution: The characteristic polynomial of the homogeneous equation is given by $p(s)=$ $s^{4}-1$. Since $p(-2)=15 \neq 0$, the exponential response formula gives the solution $\frac{e^{-2 t}}{p(-2)}=\boxed{\frac{e^{-2 t}}{15}}$... | To get the general solution, we take the sum of the general solution to the homogeneous equation and the particular solution to the original equation. The homogeneous equation is $\frac{d^{4} x}{d t^{4}}-x=0$. The characteristic polynomial $p(s)=s^{4}-1$ has 4 roots: $\pm 1, \pm i$. So the general solution to $\frac{d^... | Differential Equations (18.03 Spring 2010) | 81 | [
"\\frac{e^{-2t}}{15}\n"
] |
Preamble: Consider the differential equation $\ddot{x}+\omega^{2} x=0$. \\
A differential equation $m \ddot{x}+b \dot{x}+k x=0$ (where $m, b$, and $k$ are real constants, and $m \neq 0$ ) has corresponding characteristic polynomial $p(s)=m s^{2}+b s+k$.\\
What is the characteristic polynomial $p(s)$ of $\ddot{x}+\omeg... | The characteristic polynomial $p(s)$ is $p(s)=\boxed{s^{2}+\omega^{2}}$. | Differential Equations (18.03 Spring 2010) | 82 | [
"s^{2}+\\omega^{2}\n"
] |
Preamble: The following subproblems refer to the differential equation. $\ddot{x}+4 x=\sin (3 t)$
Find $A$ so that $A \sin (3 t)$ is a solution of $\ddot{x}+4 x=\sin (3 t)$. | We can find this by brute force. If $x=A \sin (3 t)$, then $\ddot{x}=-9 A \sin (3 t)$, so $\ddot{x}+4 x=-5 A \sin (3 t)$. Therefore, when $A=\boxed{-0.2}, x_{p}(t)=-\sin (3 t) / 5$ is a solution of the given equation. | Differential Equations (18.03 Spring 2010) | 79 | [
"-0.2\n",
"-0.2"
] |
Find the general solution of the differential equation $y^{\prime}=x-2 y$ analytically using integrating factors, solving for $y$. Note that a function $u(t)$ such that $u \dot{x}+u p x=\frac{d}{d t}(u x)$ is an integrating factor. Additionally, note that a general solution to a differential equation has the form $x=x_... | In standard form, $y^{\prime}+2 y=x$, so $u=C e^{2 x}$. Then $y=u^{-1} \int u x d x=e^{-2 x} \int x e^{2 x} d x$. Integrating by parts yields $\int x e^{2 x} d x=$ $\frac{x}{2} e^{2 x}-\frac{1}{2} \int e^{2 x} d x=\frac{x}{2} e^{2 x}-\frac{1}{4} e^{2 x}+c$. Therefore, $y=\boxed{x / 2-1 / 4+c e^{-2 x}}$. | Differential Equations (18.03 Spring 2010) | 80 | [
"\\frac{x}{2} - \\frac{1}{4} + ce^{-2x}\n"
] |
Preamble: The following subproblems refer to the following "mixing problem": A tank holds $V$ liters of salt water. Suppose that a saline solution with concentration of $c \mathrm{gm} /$ liter is added at the rate of $r$ liters/minute. A mixer keeps the salt essentially uniformly distributed in the tank. A pipe lets so... | The concentration of salt at any given time is $x(t) / V \mathrm{gm} /$ liter, so for small $\Delta t$, we lose $r x(t) \Delta t / V$ gm from the exit pipe, and we gain $r c \Delta t \mathrm{gm}$ from the input pipe. The equation is $x^{\prime}(t)=r c-\frac{r x(t)}{V}$, and in standard linear form, it is
$\boxed{x^{\pr... | Differential Equations (18.03 Spring 2010) | 89 | [
"x' + \\frac{r}{V}x - rc = 0",
"x' + \\frac{r}{V}x - rc = 0\n"
] |
Rewrite the function $\cos (\pi t)-\sqrt{3} \sin (\pi t)$ in the form $A \cos (\omega t-\phi)$. It may help to begin by drawing a right triangle with sides $a$ and $b$. | The right triangle has hypotenuse of length $\sqrt{1^{2}+(-\sqrt{3})^{2}}=2$. The circular frequency of both summands is $\pi$, so $\omega=\pi$. The argument of the hypotenuse is $-\pi / 3$, so $f(t)=\boxed{2 \cos (\pi t+\pi / 3)}$. | Differential Equations (18.03 Spring 2010) | 83 | [
"2 \\cos (\\pi t+\\pi / 3)\n"
] |
Preamble: In the following problems, take $a = \ln 2$ and $b = \pi / 3$.
Subproblem 0: Given $a = \ln 2$ and $b = \pi / 3$, rewrite $e^{a+b i}$ in the form $x + yi$, where $x, y$ are real numbers.
Solution: Using Euler's formula, we find that the answer is $\boxed{1+\sqrt{3} i}$.
Final answer: The final answer is... | $e^{n(a+b i)}=(1+\sqrt{3} i)^{n}$, so the answer is $\boxed{-2+2 \sqrt{3} i}$. | Differential Equations (18.03 Spring 2010) | 93 | [
"1+\\sqrt{3}i\n"
] |
Preamble: The following subproblems refer to the exponential function $e^{-t / 2} \cos (3 t)$, which we will assume is a solution of the differential equation $m \ddot{x}+b \dot{x}+k x=0$.
What is $b$ in terms of $m$? Write $b$ as a constant times a function of $m$. | We can write $e^{-t / 2} \cos (3 t)=\operatorname{Re} e^{(-1 / 2 \pm 3 i) t}$, so $p(s)=m s^{2}+b s+k$ has solutions $-\frac{1}{2} \pm 3 i$. This means $p(s)=m(s+1 / 2-3 i)(s+1 / 2+3 i)=m\left(s^{2}+s+\frac{37}{4}\right)$. Then $b=\boxed{m}$, | Differential Equations (18.03 Spring 2010) | 100 | [
"m\n"
] |
If the complex number $z$ is given by $z = 1+\sqrt{3} i$, what is the real part of $z^2$? | $z^{2}$ has argument $2 \pi / 3$ and radius 4 , so by Euler's formula, $z^{2}=4 e^{i 2 \pi / 3}=-2+2 \sqrt{3} i$. Thus $a = -2, b = 2\sqrt 3$, so our answer is \boxed{-2}. | Differential Equations (18.03 Spring 2010) | 98 | [
"-2"
] |
Find a purely sinusoidal solution of $\frac{d^{4} x}{d t^{4}}-x=\cos (2 t)$. | We choose an exponential input function whose real part is $\cos (2 t)$, namely $e^{2 i t}$. Since $p(s)=s^{4}-1$ and $p(2 i)=15 \neq 0$, the exponential response formula yields the solution $\frac{e^{2 i t}}{15}$. A sinusoidal solution to the original equation is given by the real part: $\boxed{\frac{\cos (2 t)}{15}}$... | Differential Equations (18.03 Spring 2010) | 92 | [
"\\frac{\\cos(2t)}{15}\n"
] |
Preamble: The following subproblems refer to the differential equation. $\ddot{x}+4 x=\sin (3 t)$
Subproblem 0: Find $A$ so that $A \sin (3 t)$ is a solution of $\ddot{x}+4 x=\sin (3 t)$.
Solution: We can find this by brute force. If $x=A \sin (3 t)$, then $\ddot{x}=-9 A \sin (3 t)$, so $\ddot{x}+4 x=-5 A \sin (3 t)... | To find the general solution, we add to $x_{p}$ the general solution to the homogeneous equation $\ddot{x}+4 x=0$. The characteristic polynomial is $p(s)=s^{2}+4$, with roots $\pm 2 i$, so the general solution to $\ddot{x}+4 x=0$ is $C_{1} \sin (2 t)+C_{2} \cos (2 t)$. Therefore, the general solution to $\ddot{x}+4 x=\... | Differential Equations (18.03 Spring 2010) | 101 | [
"-0.2\n"
] |
Find a solution to $\dot{x}+2 x=\cos (2 t)$ in the form $k_0\left[f(k_1t) + g(k_2t)\right]$, where $f, g$ are trigonometric functions. Do not include homogeneous solutions to this ODE in your solution. | $\cos (2 t)=\operatorname{Re}\left(e^{2 i t}\right)$, so $x$ can be the real part of any solution $z$ to $\dot{z}+2 z=e^{2 i t}$. One solution is given by $x=\operatorname{Re}\left(e^{2 i t} /(2+2 i)\right)=\boxed{\frac{\cos (2 t)+\sin (2 t)}{4}}$. | Differential Equations (18.03 Spring 2010) | 78 | [
"\\frac{\\cos(2t) + \\sin(2t)}{4}\n",
"\\frac{\\cos(2t)+\\sin(2t)}{4}\n"
] |
Find the general solution of $x^{2} y^{\prime}+2 x y=\sin (2 x)$, solving for $y$. Note that a general solution to a differential equation has the form $x=x_{p}+c x_{h}$ where $x_{h}$ is a nonzero solution of the homogeneous equation $\dot{x}+p x=0$. Additionally, note that the left hand side is the derivative of a pro... | We see that $\left(x^{2} y\right)^{\prime}=x^{2} y^{\prime}+2 x y$. Thus, $x^{2} y=-\frac{1}{2} \cos (2 x)+c$, and $y=\boxed{c x^{-2}-\frac{\cos (2 x)}{2 x^{2}}}$. | Differential Equations (18.03 Spring 2010) | 63 | [
"cx^{-2} - \\frac{\\cos(2x)}{2x^2}\n",
"y=cx^{-2}-\\frac{\\cos(2x)}{2x^{2}}\n"
] |
Preamble: The following subproblems refer to the damped sinusoid $x(t)=A e^{-a t} \cos (\omega t)$.
What is the spacing between successive maxima of $x(t)$? Assume that $\omega \neq 0$. | The extrema of $x(t)=A e^{-a t} \cos (\omega t)$ occur when $\dot{x}(t)=0$, i.e., $-a \cos (\omega t)=\omega \sin (\omega t)$. When $\omega \neq 0$, the extrema are achieved at $t$ where $\tan (\omega t)=-a / \omega$. Since minima and maxima of $x(t)$ are alternating, the maxima occur at every other $t \operatorname{su... | Differential Equations (18.03 Spring 2010) | 84 | [
"\\frac{2\\pi}{\\omega}\n"
] |
Subproblem 0: Find a purely sinusoidal solution of $\frac{d^{4} x}{d t^{4}}-x=\cos (2 t)$.
Solution: We choose an exponential input function whose real part is $\cos (2 t)$, namely $e^{2 i t}$. Since $p(s)=s^{4}-1$ and $p(2 i)=15 \neq 0$, the exponential response formula yields the solution $\frac{e^{2 i t}}{15}$. A ... | To get the general solution, we take the sum of the general solution to the homogeneous equation and the particular solution to the original equation. The homogeneous equation is $\frac{d^{4} x}{d t^{4}}-x=0$. The characteristic polynomial $p(s)=s^{4}-1$ has 4 roots: $\pm 1, \pm i$. So the general solution to $\frac{d^... | Differential Equations (18.03 Spring 2010) | 76 | [
"\\frac{\\cos (2 t)}{15}\n",
"\\frac{\\cos (2 t)}{15}+C_{1} e^{t}+C_{2} e^{-t}+C_{3} \\cos (t)+C_{4} \\sin (t)\n"
] |
Find a solution of $\ddot{x}+4 x=\cos (2 t)$, solving for $x(t)$, by using the ERF on a complex replacement. The ERF (Exponential Response Formula) states that a solution to $p(D) x=A e^{r t}$ is given by $x_{p}=A \frac{e^{r t}}{p(r)}$, as long as $\left.p (r\right) \neq 0$). The ERF with resonance assumes that $p(r)=0... | The complex replacement of the equation is $\ddot{z}+4 z=e^{2 i t}$, with the characteristic polynomial $p(s)=s^{2}+4$. Because $p(2 i)=0$ and $p^{\prime}(2 i)=4 i \neq 0$, we need to use the Resonant ERF, which leads to $z_{p}=\frac{t e^{2 i t}}{4 i}$. A solution of the original equation is given by $x_{p}=\operatorna... | Differential Equations (18.03 Spring 2010) | 94 | [
"\\frac{t}{4}\\sin(2t)\n"
] |
Preamble: The following subproblems refer to a spring/mass/dashpot system driven through the spring modeled by the equation $m \ddot{x}+b \dot{x}+k x=k y$. Here $x$ measures the position of the mass, $y$ measures the position of the other end of the spring, and $x=y$ when the spring is relaxed.
In this system, regard ... | The equation is $\ddot{x}+3 \dot{x}+4 x=4 A \cos t$, with the characteristic polynomial $p(s)=s^{2}+3 s+4$. The complex exponential corresponding to the input signal is $y_{c x}=A e^{i t}$ and $p(i)=3+3 i \neq 0$. By the Exponential Response Formula, $z_{p}=\frac{4 A}{p(i)} e^{i t}=\boxed{\frac{4 A}{3+3 i} e^{i t}}$. | Differential Equations (18.03 Spring 2010) | 85 | [
"\\frac{4A}{3+3i}e^{it}\n"
] |
Preamble: The following subproblems refer to a circuit with the following parameters. Denote by $I(t)$ the current (where the positive direction is, say, clockwise) in the circuit and by $V(t)$ the voltage increase across the voltage source, at time $t$. Denote by $R$ the resistance of the resistor and $C$ the capacita... | When $V$ is constant, the equation becomes $R \dot{I}+\frac{1}{C} I=0$, which is separable. Solving gives us
\[
I(t)=\boxed{I(0) e^{-\frac{t}{R C}}
}\]. | Differential Equations (18.03 Spring 2010) | 86 | [
"I(0)e^{-\\frac{t}{RC}}\n",
"I(0) e^{-\\frac{t}{R C}}\n"
] |
For $\omega \geq 0$, find $A$ such that $A \cos (\omega t)$ is a solution of $\ddot{x}+4 x=\cos (\omega t)$. | If $x=A \cos (\omega t)$, then taking derivatives gives us $\ddot{x}=-\omega^{2} A \cos (\omega t)$, and $\ddot{x}+4 x=\left(4-\omega^{2}\right) A \cos (\omega t)$. Then $A=\boxed{\frac{1}{4-\omega^{2}}}$. | Differential Equations (18.03 Spring 2010) | 77 | [
"\\frac{1}{4-\\omega^{2}}\n",
"\\frac{1}{4-\\omega^2}\n"
] |
Find the general (complex-valued) solution of the differential equation $\dot{z}+2 z=e^{2 i t}$, using $C$ to stand for any complex-valued integration constants which may arise. | Using integrating factors, we get $e^{2 t} z=e^{(2+2 i) t} /(2+2 i)+C$, or $z=\boxed{\frac{e^{2 i t}}{(2+2 i)}+C e^{-2 t}}$, where $C$ is any complex number. | Differential Equations (18.03 Spring 2010) | 108 | [
"\\frac{e^{2it}}{2+2i} + Ce^{-2t}\n"
] |
Subproblem 0: For $\omega \geq 0$, find $A$ such that $A \cos (\omega t)$ is a solution of $\ddot{x}+4 x=\cos (\omega t)$.
Solution: If $x=A \cos (\omega t)$, then taking derivatives gives us $\ddot{x}=-\omega^{2} A \cos (\omega t)$, and $\ddot{x}+4 x=\left(4-\omega^{2}\right) A \cos (\omega t)$. Then $A=\boxed{\frac... | Resonance occurs when $\omega=\boxed{2}$. | Differential Equations (18.03 Spring 2010) | 75 | [
"\\frac{1}{4-\\omega^{2}}\n"
] |
Find a solution of $\ddot{x}+3 \dot{x}+2 x=t e^{-t}$ in the form $x(t)=u(t) e^{-t}$ for some function $u(t)$. Use $C$ for an arbitrary constant, should it arise. | $\dot{x}=\dot{u} e^{-t}-u e^{-t}$ and $\ddot{x}=\ddot{u} e^{-t}-2 \dot{u} e^{-t}+u e^{-t}$. Plugging into the equation leads to $e^{-t}(\ddot{u}+\dot{u})=t e^{-t}$. Cancelling off $e^{-t}$ from both sides, we get $\ddot{u}+\dot{u}=t$. To solve this equation for $u$, we use the undetermined coefficient method. However, ... | Differential Equations (18.03 Spring 2010) | 97 | [
"\\left(\\frac{t^{2}}{2} - t + C\\right) e^{-t}\n",
"\\left(\\frac{t^{2}}{2} - t + C\\right)e^{-t}\n"
] |
Subproblem 0: Find the general (complex-valued) solution of the differential equation $\dot{z}+2 z=e^{2 i t}$, using $C$ to stand for any complex-valued integration constants which may arise.
Solution: Using integrating factors, we get $e^{2 t} z=e^{(2+2 i) t} /(2+2 i)+C$, or $z=\boxed{\frac{e^{2 i t}}{(2+2 i)}+C e^{... | When $C=0, z=\boxed{\frac{e^{2 i t}}{(2+2 i)}}$. | Differential Equations (18.03 Spring 2010) | 87 | [
"\\frac{e^{2it}}{2+2i} + Ce^{-2t}\n",
"\\frac{e^{2it}}{2+2i}+Ce^{-2t}\n"
] |
Preamble: In the following problems, take $a = \ln 2$ and $b = \pi / 3$.
Subproblem 0: Given $a = \ln 2$ and $b = \pi / 3$, rewrite $e^{a+b i}$ in the form $x + yi$, where $x, y$ are real numbers.
Solution: Using Euler's formula, we find that the answer is $\boxed{1+\sqrt{3} i}$.
Final answer: The final answer is... | $e^{n(a+b i)}=(1+\sqrt{3} i)^{n}$, so the answer is $\boxed{-8}$. | Differential Equations (18.03 Spring 2010) | 91 | [
"-8"
] |
Find the polynomial solution of $\ddot{x}-x=t^{2}+t+1$, solving for $x(t)$. | Since the constant term of the right-hand side is nonzero, the undetermined coefficients theorem asserts that there is a unique quadratic polynomial $a t^{2}+b t+c$ satisfying this equation. Substituting this form into the left side of the equation, we see that $a=-1,-b=1$, and $2 a-c=1$, so $b=-1$ and $c=-3$. Finally,... | Differential Equations (18.03 Spring 2010) | 90 | [
"-t^2-t-3\n"
] |
Preamble: The following subproblems consider a second order mass/spring/dashpot system driven by a force $F_{\text {ext }}$ acting directly on the mass: $m \ddot{x}+b \dot{x}+k x=F_{\text {ext }}$. So the input signal is $F_{\text {ext }}$ and the system response is $x$. We're interested in sinusoidal input signal, $F_... | Set $F_{\mathrm{cx}}=e^{i \omega t}$. The complex replacement of the equation is $\ddot{z}+\frac{1}{4} \dot{z}+2 z=e^{i \omega t}$, with the characteristic polynomial $p(s)=s^{2}+\frac{1}{4} s+2.$ Given that $p(i \omega)=-\omega^{2}+\frac{\omega}{4} i+2 \neq 0$, so by the exponential response formula, $z_{p}=e^{i \omeg... | Differential Equations (18.03 Spring 2010) | 88 | [
"\\frac{2-\\omega^{2}-\\frac{\\omega i}{4}}{\\omega^{4}-\\frac{63}{16} \\omega^{2}+4}\n",
"\\frac{2-\\omega^{2}-\\frac{\\omega}{4} i}{\\omega^{4}-\\frac{63}{16} \\omega^{2}+4}\n"
] |
Preamble: The following subproblems refer to the differential equation $\ddot{x}+b \dot{x}+x=0$.\\
Subproblem 0: What is the characteristic polynomial $p(s)$ of $\ddot{x}+b \dot{x}+x=0$?
Solution: The characteristic polynomial is $p(s)=\boxed{s^{2}+b s+1}$.
Final answer: The final answer is s^{2}+b s+1. I hope it i... | To exhibit critical damping, the characteristic polynomial $s^{2}+b s+1$ must be a square, i.e., $(s-k)^{2}$ for some $k$. Multiplying and comparing yields $-2 k=b$ and $k^{2}=1$. Therefore, $b$ could be either one of $=-2, 2$. When $b=-2, e^{t}$ is a solution, and it exhibits exponential growth instead of damping, so ... | Differential Equations (18.03 Spring 2010) | 107 | [
"2\n"
] |
Preamble: In the following problems, take $a = \ln 2$ and $b = \pi / 3$.
Subproblem 0: Given $a = \ln 2$ and $b = \pi / 3$, rewrite $e^{a+b i}$ in the form $x + yi$, where $x, y$ are real numbers.
Solution: Using Euler's formula, we find that the answer is $\boxed{1+\sqrt{3} i}$.
Final answer: The final answer is... | $e^{n(a+b i)}=(1+\sqrt{3} i)^{n}$, so the answer is $\boxed{-8-8 \sqrt{3} i}$. | Differential Equations (18.03 Spring 2010) | 105 | [
"-8-8\\sqrt{3}i\n",
"-8-8\\sqrt{3}i",
"-8-8 \\sqrt{3} i"
] |
Preamble: The following subproblems refer to the exponential function $e^{-t / 2} \cos (3 t)$, which we will assume is a solution of the differential equation $m \ddot{x}+b \dot{x}+k x=0$.
Subproblem 0: What is $b$ in terms of $m$? Write $b$ as a constant times a function of $m$.
Solution: We can write $e^{-t / 2} ... | Having found that $p(s)=m(s+1 / 2-3 i)(s+1 / 2+3 i)=m\left(s^{2}+s+\frac{37}{4}\right)$ in the previous subproblem, $k=\boxed{\frac{37}{4} m}$. | Differential Equations (18.03 Spring 2010) | 104 | [
"m\n"
] |
What is the smallest possible positive $k$ such that all functions $x(t)=A \cos (\omega t-\phi)$---where $\phi$ is an odd multiple of $k$---satisfy $x(0)=0$? \\ | $x(0)=A \cos \phi$. When $A=0$, then $x(t)=0$ for every $t$; when $A \neq 0$, $x(0)=0$ implies $\cos \phi=0$, and hence $\phi$ can be any odd multiple of $\pi / 2$, i.e., $\phi=\pm \pi / 2, \pm 3 \pi / 2, \pm 5 \pi / 2, \ldots$ this means $k=\boxed{\frac{\pi}{2}}$ | Differential Equations (18.03 Spring 2010) | 102 | [
"\\frac{\\pi}{2}\n"
] |
Find a purely exponential solution of $\frac{d^{4} x}{d t^{4}}-x=e^{-2 t}$. | The characteristic polynomial of the homogeneous equation is given by $p(s)=$ $s^{4}-1$. Since $p(-2)=15 \neq 0$, the exponential response formula gives the solution $\frac{e^{-2 t}}{p(-2)}=\boxed{\frac{e^{-2 t}}{15}}$. | Differential Equations (18.03 Spring 2010) | 99 | [
"\\frac{e^{-2t}}{15}\n"
] |
Given the ordinary differential equation $\ddot{x}-a^{2} x=0$, where $a$ is a nonzero real-valued constant, find a solution $x(t)$ to this equation such that $x(0) = 1$ and $\dot{x}(0)=0$. | First, notice that both $x(t)=e^{a t}$ and $x(t)=e^{-a t}$ are solutions to $\ddot{x}-a^{2} x=0$. Then for any constants $c_{1}$ and $c_{2}$, $x(t)=c_{1} e^{a t}+c_{2} e^{-a t}$ are also solutions to $\ddot{x}-a^{2} x=0$. Moreover, $x(0)=c_{1}+c_{2}$, and $\dot{x}(0)=a\left(c_{1}-c_{2}\right)$. Assuming $a \neq 0$, to ... | Differential Equations (18.03 Spring 2010) | 95 | [
"\\frac{1}{2}(e^{at} + e^{-at})\n",
"\\frac{1}{2} (e^{at} + e^{-at})\n"
] |
Find the general solution of the differential equation $\dot{x}+2 x=e^{t}$, using $c$ for the arbitrary constant of integration which will occur. | We can use integrating factors to get $(u x)^{\prime}=u e^{t}$ for $u=e^{2 t}$. Integrating yields $e^{2 t} x=e^{3 t} / 3+c$, or $x=\boxed{\frac{e^{t}} {3}+c e^{-2 t}}$. | Differential Equations (18.03 Spring 2010) | 96 | [
"x = \\frac{e^{t}}{3} + ce^{-2t}\n",
"\\frac{e^{t}}{3} + c e^{-2t}\n"
] |
Rewrite the function $\operatorname{Re} \frac{e^{i t}}{2+2 i}$ in the form $A \cos (\omega t-\phi)$. It may help to begin by drawing a right triangle with sides $a$ and $b$. | $e^{i t}=\cos (t)+i \sin (t)$, and $\frac{1}{2+2 i}=\frac{1-i}{4}$. the real part is then $\frac{1}{4} \cos (t)+$ $\frac{1}{4} \sin (t)$. The right triangle here has hypotenuse $\frac{\sqrt{2}}{4}$ and argument $\pi / 4$, so $f(t)=\boxed{\frac{\sqrt{2}}{4} \cos (t-\pi / 4)}$. | Differential Equations (18.03 Spring 2010) | 106 | [
"\\frac{\\sqrt{2}}{4} \\cos (t-\\pi / 4)\n"
] |
Preamble: The following subproblems refer to the differential equation $\ddot{x}+b \dot{x}+x=0$.\\
What is the characteristic polynomial $p(s)$ of $\ddot{x}+b \dot{x}+x=0$? | The characteristic polynomial is $p(s)=\boxed{s^{2}+b s+1}$. | Differential Equations (18.03 Spring 2010) | 103 | [
"s^{2}+b s+1\n",
"s^2 + bs + 1",
"s^{2}+bs+1\n"
] |
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