 # Final Exam Review

Directions. The following are a few select review problems for the Final Exam, but please note this review does not cover all sections on your Final Exam. It is recommended you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "Video" to learn how to solve it.
1. Find the general solution of the given differential equation.
1. $$y'+2ty=2te^{-t^2}$$
2. $$2\sqrt{x}\,y'=\sqrt{1-y^2}$$
3. $$ty'+y=3t\cos t,\qquad t>0$$​

1. $$y=(t^2+C)e^{-t^2}$$
2. $$y=\sin(\sqrt{x}+C),\quad$$ equilibrium solutions at $$y=\pm1$$
3. $$y=\dfrac{3(t\sin(t)+\cos(t))+C}{t}$$

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2. Given the differential equation $y'(t)=y^3-2y^2+y$
1. Find the equilibrium solutions.
2. Graph the phase line. Classify each equilibrium solution as either stable, unstable, or semistable.
3. Sketch the graph of some solutions.
4. If $$y(t)$$ is the solution of the equation satisfying the initial condition $$y(0)=y_0$$, where $$-\infty< y_0<\infty$$, determine the behavior of $$y(t)$$ as $$t$$ increases.
5. Do any solutions 'blow up in finite time,' namely, do they admit a vertical asymptote?​

1. Equilibrium Solutions: $$y=0,$$ $$y=1,$$
2. Semistable: $$y=1,$$ Unstable: $$y=0,$$ See the video for the graph of the phase line
3. See the video
4.  \begin{cases}  y_0<0 & \Rightarrow & y\to -\infty \\ y_0=0 & \Rightarrow & y=0 \\ 0< y_0<1 & \Rightarrow & y\to 1 \\ y_0=1 & \Rightarrow & y= 1 \\ y_0>1 & \Rightarrow & y\to \infty\end{cases}
5. Yes, $$y(t)$$ blows up in finite time. See the video for an explanation.

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3. Solve $$y'=\displaystyle\frac{(1+x)e^x}{xe^x-ye^y}.$$

$$\dfrac{y^2}{2}+xe^xe^{-y}=C$$

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4. Find the solution to the given initial value problem.$y''-4y'+3y=0, \quad y(0)=3, \quad y'(0)=4$

$$y=\dfrac{5}{2}e^{t}+\dfrac{1}{2}e^{3t}$$

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5. If the Wronskian of $$f$$ and $$g$$ is $$t\cos(t)-\sin(t)$$, and if $$u=2f-3g$$, and $$v=f+g$$, find the Wronskian of $$u$$ and $$v.$$

$$W[u,v]=5W[f,g]=5(t\cos(t)-\sin(t))$$

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6. Find the general solution of the equation $$y''+2y'+y=4e^{-t}.$$

$$y(t)=e^{-t}(c_1+c_2t+2t^2)$$

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7. Solve the differential equation $t^2y''+2ty'-2y=t, \quad t>0$ assuming $$y_1(t)=t$$ is a solution to the corresponding homogeneous equation.

$$y(t)=c_1t+c_2t^{-2}+\dfrac{1}{3}t\ln(t)$$

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8. For the equation $$(x^2+1)y''-6y=0$$
1. Determine a lower bound for the radius of convergence of the series solutions of the differential equation about $$x_0=0$$.
2. Seek its power series solution about $$x_0=0$$; find the recurrence relation.
3. Find the general term of each solution $$y_1(x)$$ and $$y_2(x)$$.
4. Find the first four terms in each of two solutions $$y_1$$ and $$y_2$$. Show that $$W[y_1, y_2](0)\ne 0$$.​

1. The lower bound is 1.
2. $$\displaystyle \sum_{n=0}^{\infty}n(n-1)a_nx^n+\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n-\sum_{n=0}^{\infty}6a_nx^n=0$$ . The recurrence relation is $$a_{n+2}=-\dfrac{n-3}{n+1}a_n$$ for all $$n=0,1,2,...$$
3. $$y_1=1+3x^2+x^4+\displaystyle\sum_{k=3}^{\infty}\dfrac{3(-1)^k}{(2k-1)(2k-3)}x^{2k}, \quad \quad$$ $$y_2=x+x^3$$
4. $$y_1=1+3x^2+x^4-\dfrac{1}{5}x^6+... \quad \quad$$ $$y_2=x+x^3$$
The Wronskian is $W[y_1,y_2](0)= \begin{vmatrix} 1 & 0 \\ 0 & 1 \\ \end{vmatrix}\\ =1\neq 0$

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9. For the following equation, determine $$\phi^{\prime \prime}\left(x_{0}\right)$$ and $$\phi^{\prime \prime \prime}\left(x_{0}\right),$$ for the given point $$x_{0}$$ if $$y=\phi(x)$$ is a solution of the given initial-value problem.
$y^{\prime \prime}+x^{2} y^{\prime}+(\sin x) y=0 ; \quad y(0)=a_{0}, \quad y^{\prime}(0)=a_{1}$

We have that $$\phi''(0)=0$$, $$\phi'''(0)=-a_0$$, and  $$\phi(x)=a_0(1-\dfrac{x^3}{6}+...)+a_1(x+...)$$

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10. Find the solution of the initial value problem.
1. $$y''+y=\delta(t-2\pi)\cos t,\qquad$$ $$y(0)=0,\quad$$ $$y'(0)=1$$
2. $$y''+3y'+2y=\left\{\begin{array}{ll} 1, \quad &0\leq t<10\\ 0,&10\leq t\end{array}\right.,\qquad$$ $$y(0)=0,\quad$$ $$y'(0)=0$$​

1. $$y(t)=\sin(t)+u_{2\pi}(t)\sin(t)$$
2. $$y(t)=\dfrac{1}{2}-e^{-t}+\dfrac{1}{2}e^{-2t}-u_{10}(t)\left[\dfrac{1}{2}-e^{-(t-10)}+\dfrac{1}{2}e^{-2(t-10)}\right]$$

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11. Solve the initial value problem $y''-2y'-3y=g(t), ~~~y(0)=1,~~y'(0)=-3$

$$y(t)=-\dfrac{1}{2}e^{3t}+\dfrac{3}{2}e^{-t}+\dfrac{1}{4}\displaystyle \int_{0}^tg(t-\tau)(e^{3\tau}-e^{-\tau}) d\tau$$

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12. Find the general solution of the system and the fundamental matrix. Classify the type of the critical point (0,0), and determine whether it is stable or unstable. Sketch the phase portrait.
1. $$\textbf{x}'=\begin{pmatrix} 5&-1\\3&1\end{pmatrix}\textbf{x}$$
2. $${\bf x}'=\begin{pmatrix} 1&1\\-5&-1\end{pmatrix}{\bf x}$$​

1. The general solution is $\textbf{x}(t)=C_1e^{2t}\begin{pmatrix}1\\3\\\end{pmatrix}+C_2e^{4t}\begin{pmatrix}1\\1\end{pmatrix}$and the fundamental matrix is $\begin{pmatrix}e^{2t}&e^{4t}\\3e^{2t}&e^{4t}\end{pmatrix}$The critical point $$(0,0)$$ is a nodal source and unstable. Refer to video for sketch of phase portrait.
2. The general solution is $\textbf{x}(t)=C_1\begin{pmatrix}-\cos(2t)\\\cos(2t)+2\sin(2t)\\\end{pmatrix}+C_2\begin{pmatrix}-\sin(2t)\\\sin(2t)-2\cos(2t)\end{pmatrix}$and the fundamental matrix is $\begin{pmatrix}-\cos(2t)&-\sin(2t)\\\cos(2t)+2\sin(2t)&\sin(2t)-2\cos(2t)\end{pmatrix}$The critical point $$(0,0)$$ is a center and stable. Refer to video for sketch of phase portrait.

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3. $${\bf x}'=\begin{pmatrix} 3&-2\\2&-2\end{pmatrix}{\bf x}$$

The general solution is $\mathbf{x}(t)=C_1e^{-t}\begin{pmatrix}1\\2\\\end{pmatrix}+C_2e^{2t}\begin{pmatrix}2\\1\end{pmatrix}$ and the fundamental matrix is $\begin{pmatrix}e^{-t}&2e^{2t}\\2e^{-t}&e^{2t}\end{pmatrix}$The critical point $$(0,0)$$ is a saddle point and unstable. Refer to the video for a sketch of the phase portrait.

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4. $${\bf x}'=\begin{pmatrix} 3&-4\\1&-1\end{pmatrix}{\bf x}$$

The general solution is $\mathbf{x}(t)=C_1e^{t}\begin{pmatrix}2\\1\\\end{pmatrix}+C_2e^{t}\left[t\begin{pmatrix}2\\1\end{pmatrix}+\begin{pmatrix}1\\0\end{pmatrix}\right]$ and the fundamental matrix is $\begin{pmatrix}2e^{t}&(2t+1)e^{t}\\e^{t}&te^{t}\end{pmatrix}$ The critical point $$(0,0)$$ is an improper node and unstable. Refer to the video for a sketch of the phase portrait.

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5. $${\bf x}'=\begin{pmatrix} 1&-1\\5&-3\end{pmatrix}{\bf x}$$

The general solution is \begin{align}\mathbf{x}(t)=&C_1e^{-t}\left[\begin{pmatrix}1\\2\\\end{pmatrix}\cos(t)+\begin{pmatrix}0\\1\end{pmatrix}\sin(t)\right]\\&+C_2e^{-t}\left[\begin{pmatrix}1\\2\end{pmatrix}\sin(t)+\begin{pmatrix}0\\-1\end{pmatrix}\cos(t)\right]\end{align} and the fundamental matrix is $\begin{pmatrix}e^{-t}\cos(t)&e^{-t}\sin(t)\\e^{-t}(2\cos(t)+\sin(t))&e^{-t}(2\sin(t)-\cos(t))\end{pmatrix}$ The critical point $$(0,0)$$ type is a spiral source and asymptotically stable. Refer to the video for a sketch of the phase portrait.

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