Exams
Exam 1 - March 13, 2017
Closed Book
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Find the general solution of the ordinary differential equation
$\dfrac{d^{3}y}{dx^{3}}-y=\sin2x+x^{3}.$
(30%)
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Find $u=u(r,\theta)$ which satisfies the Lapalce equation $\nabla^{2}u=0$ in the annulus $1$ < $r$ < $2$ with the boundary conditions $u(2,\theta)=\cos\theta$ and $u(1,\theta)=\sin2\theta$.
(40%)
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Calculate the integral $\int_{0}^{2\pi}\dfrac{d\theta}{2+\sin\theta}.$
(30%)
Solutions
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$\dfrac{d^{3}y}{dx^{3}}-y=\sin2x+x^{3}.$
$m^{3}-1=0,$ $m^{3}=1=e^{2n\pi i},$ $m=e^{2n\pi i/3},$ or $m=1,e^{2\pi i/3},$ $e^{-2\pi i/3}.$
Thus the complementary solution is $Y=ae^{x}+be^{-(1-\sqrt{3}i)x/2}+ce^{-(1+\sqrt{3}i)x/2}.$
A particular solution is $\ \ \ \ \ \ y_{p}=\dfrac{1}{D^{3}-1}(\sin 2x+x^{3})=\dfrac{1}{-4D-1}\sin2x-\dfrac{1}{1-D^{3}}x^{3}$
$=-\dfrac{4D-1}{16D^{2}-1}\sin2x-(1+D^{3})x^{3}=\dfrac{8\cos2x-\sin2x}{65}-(x^{3}+6).$
The general solution of the differential equation is
$y=ae^{x}+be^{-(1-\sqrt{3}i)x/2}+ce^{-(1+\sqrt{3}i)x/2}+\dfrac{8\cos2x-\sin 2x}{65}-(x^{3}+6).$
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$u(2,\theta)=\cos\theta$ and $u(1,\theta)=\sin2\theta$.
The solution $u(r,\theta)=(ar+b/r)\cos\theta+(cr^{2}+d/r^{2})\sin2\theta,$ where $a,b,c,d$ are constants, solves the Laplace equation.
That $u(2,\theta)=\cos\theta$ gives $(2a+b/2)=1,$ $(4c+d/4)=0.$
That $u(1,\theta)=\sin2\theta$ gives $(a+b)=0,$ $(c+d)=1.$
Thus we have $a=2/3,$ $b=-2/3,$ $c=-1/15,$ $d=16/15.$
We get $u(r,\theta)=\dfrac{2}{3}(r-1/r)\cos\theta-\dfrac {1}{15}(r^{2}-16/r^{2})\sin2\theta.$
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$I=\int_{0}^{2\pi}\dfrac{d\theta}{2+\sin\theta}.$
Let $z\equiv e^{i\theta},$ then $\sin\theta=\dfrac{z-1/z}{2i},dz=ie^{i\theta}d\theta,$ or $d\theta=dz/(iz).$ Thus
$I= \oint\limits_C \dfrac{dz}{iz[2+(z-1/z)/(2i)]} = \oint\limits_C \dfrac{2dz}{z^{2}+4iz-1}$,
where $c$ is the unit circle with the origin as its center. The singularities of the integrand are
$z=(-2\pm\sqrt{3})i.$ The contour $c$ encloses the singularity with the plus sign. Thus
$I=2\pi i\dfrac{2}{2(-2+\sqrt{3})i+4i}=\dfrac{2\pi}{\sqrt{3}}.$