18.075 - Advanced Calculus for Engineers, Spring 2015

Exams

Exam 1 - March 9, 2015

Closed Book

1. Find the general solution of the ordinary differential equation

$\dfrac{d^{2}y}{dx^{2}}+4y=\sin2x+x.$

(30%)

2. 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%)

3. Calculate the integral $\int_{-\infty}^{\infty}\dfrac{\sin x}{(x-1)^{2}+9}dx.$

(40%)

Solutions

1. $\dfrac{d^{2}y}{dx^{2}}+4y=\sin2x+x.$

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,

$y=a\cos2x+b\sin2x+\dfrac{x}{4}-\dfrac{x\cos x}{4}.$

2. $\nabla^{2}u=0$ 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

$u(r,\theta)=\dfrac{1-(r/4)^{2}\cos2\theta}{2}.$

3. $I=\int_{-\infty}^{\infty}\dfrac{\sin x}{(x-1)^{2}+9}dx.$

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,

$I=$ $\dfrac{\pi e^{-3}\sin1}{3}.$