Exams
Exam 1 - March 9, 2015
Closed Book
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Find the general solution of the ordinary differential equation
$\dfrac{d^{2}y}{dx^{2}}+4y=\sin2x+x.$
(30%)
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Find $u$ which satisfies the Laplace equation $\nabla^{2}u=0$ inside the circle $x^{2}+y^{2}=16$ in the two-dimensional plane as well as the boundary condition $u=\sin^{2}\theta$ on the circle, where $\theta$ is the polar angle.
(30%)
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Calculate the integral $\int_{-\infty}^{\infty}\dfrac{\sin x}{(x-1)^{2}+9}dx.$
(40%)
Solutions
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$\dfrac{d^{2}y}{dx^{2}}+4y=\sin2x+x.$
We have $m^{2}+4=0,$ or $m=\pm2i.$
Thus
$Y=a\cos2x+b\sin2x.$
A particular solution is
$y_{p}=\dfrac{1}{D^{2}+4}(\sin2x+x).$
Now
$\dfrac{1}{4+D^{2}}x=\dfrac{1}{4}x,$
$\dfrac{1}{D^{2}+4}\sin2x=\operatorname{Im}\dfrac{1}{D^{2}+4}e^{2ix}=\operatorname{Im}\dfrac{1}{(4i)(D-2i)}e^{2ix}=\operatorname{Im}\dfrac{xe^{2ix}}{(4i)}=-\dfrac{x\cos x}{4}.$
Therefore,
$y=a\cos2x+b\sin2x+\dfrac{x}{4}-\dfrac{x\cos x}{4}.$
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$\nabla^{2}u=0$ inside the circle $x^{2}+y^{2}=16$ and $u=\sin^{2}\theta$ on the circle.
Since
$\sin^{2}\theta=\dfrac{1-\cos2\theta}{2},$
therefore
$u(r,\theta)=\dfrac{1-(r/4)^{2}\cos2\theta}{2}.$
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$I=\int_{-\infty}^{\infty}\dfrac{\sin x}{(x-1)^{2}+9}dx.$
Let
$J\equiv$ $\int_{-\infty}^{\infty}\dfrac{e^{ix}}{(x-1)^{2}+9}dx,$
then
$I=\operatorname{Im}J.$
The integrand of $J$ has simple poles at $1\pm3i.$ Closing the contour upstairs, we get
$J=2\pi i\dfrac{e^{-3+i}}{2(3i)}=\dfrac{\pi e^{-3+i}}{3}.$
Therefore,
$I=$ $\dfrac{\pi e^{-3}\sin1}{3}.$