Generals 2002 I 2

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We have the equation:

$\frac{d^{3}y}{dx^{3}}+x^{2}y-\sinh\left(x\right)=0$

The homogenous equation is:

$\frac{d^{3}y}{dx^{3}}+x^{2}y=0$

Taking $y ˜ e S$ :

$S^{\prime\prime\prime}+3S^{\prime}S^{\prime\prime}+\left(S^{\prime}\right)^{3}+x^{2}=0$

Taking a balance between $\left(S^{\prime}\right)^{3}$ and $x 2$ , $S^{\prime}=-e^{i\theta}x^{2/3}$ with $\theta=0,\,2\pi/3,\,4\pi/3$ . Then $S=-\frac{3}{5}e^{i\theta}x^{5/3}$ . Now trying $S=-\frac{3}{5}e^{i\theta}x^{5/3}+g$ :

$\left(\frac{2}{9}e^{i\theta}x^{-4/3}+g^{\prime\prime\prime}\right)+3\left(-e^{i\theta}x^{2/3}+g^{\prime}\right)\left(-\frac{2}{3}e^{i\theta}x^{-1/3}+g^{\prime\prime}\right)+\left(-e^{i\theta}x^{2/3}+g^{\prime}\right)^{3}+x^{2}=0$

Expanding:

$\frac{2}{9}e^{i\theta}x^{-4/3}+g^{\prime\prime\prime}+2e^{2i\theta}x^{1/3}-3e^{i\theta}x^{2/3}g^{\prime\prime}-2e^{i\theta}x^{-1/3}g^{\prime}+3e^{2i\theta}x^{4/3}g^{\prime}-3e^{i\theta}x^{2/3}\left(g^{\prime}\right)^{2}+\left(g^{\prime}\right)^{3}=0$

Balance between $2e^{2i\theta}x^{1/3}<math> and <math>3e^{2i\theta}x^{4/3}g^{\prime}$ gives $g^{\prime}=-\frac{2}{3x}$ , or $g=-\frac{2}{3}\ln x$ . The homogenous solution can then be written:

$y_{homog}=Ae^{-3x^{5/3}/5}+\left[B\sin\left(\frac{3\sqrt{3}x^{5/3}}{10}\right)+C\cos\left(\frac{3\sqrt{3}x^{5/3}}{10}\right)\right]e^{3x^{5/3}/10}$

The inhomogenous equation is:

$\frac{d^{3}y}{dx^{3}}+x^{2}y-\sinh\left(x\right)=0$

For large $x$ , the last homogenous terms dominate (since they go like $e^{x^{5/3}}$ ), unless $B = C = 0$ . Then we guess a balance between $x 2 y$ and $e x / 2$ , giving:

$y_{part}=\frac{e^{x}}{2x^{2}}$

Which does dominate. The general solution is then:

$y=Ae^{-3x^{5/3}/5}+\left[B\sin\left(\frac{3\sqrt{3}x^{5/3}}{10}\right)+C\cos\left(\frac{3\sqrt{3}x^{5/3}}{10}\right)\right]e^{3x^{5/3}/10}+\frac{1}{2x^{2}}e^{x}$

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Category: Generals Problems

Generals 2002 I 2

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