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,

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