Generals 2004 II 4

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[edit] Part a

This is a well-behaved function around x = 0, so we write y\left(x\right)=\sum_{n}a_{n}x^{n}:

\sum_{n}a_{n}n\left(n-1\right)x^{n-2}+\sum_{n}a_{n}nx^{n}+\sum_{n}a_{n}x^{n}=0


Shifting the first by two:

\sum_{n}\left[a_{n+2}\left(n+2\right)\left(n+1\right)+\left(n+1\right)a_{n}\right]x^{n}=0


Then we get the recurrence relation:

a_{n+2}=\frac{a_{n}}{\left(n+2\right)}


We get solutions for a0 = 0 and a1 = 0:

y_{1}\left(x\right)=\sum_{n}\frac{a_{0}x^{2n}}{2^{n}n!};\qquad y_{2}\left(x\right)=\sum_{n}\frac{a_{1}x^{2n+1}}{\left(2n+1\right)!}


[edit] Part b

The equation is:

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


Using y˜eS:

\left(S^{\prime\prime}+\left(S^{\prime}\right)^{2}\right)+xS^{\prime}+1=0


Balance between xS^{\prime} and \left(S^{\prime}\right)^{2}, S^{\prime}=-x gives S=-\frac{1}{2}x^{2}. There is an alternate balance between 1 and xS^{\prime}, giving S^{\prime}=-1/x, or S = − logx. So the two asymptotic values are:

y\rightarrow e^{-x^{2}/2},\quad \frac{1}{x}


[edit] Part c

We use the integral:

y\left(x\right)=\int e^{ixt}f\left(t\right)dt


Plugging in:

\int\left[-t^{2}+xit+1\right]f\left(t\right)e^{ixt}dt=0


Integrating by parts:

\left.tf\left(t\right)e^{ixt}\right|_{C}-\int\left[t^{2}f\left(t\right)+tf^{\prime}\left(t\right)\right]e^{ixt}dt=0


Solving this differential equation:

-t=\frac{f^{\prime}}{f}


Integrating:

-\frac{t^{2}}{2}=\ln f


Or:

f=e^{-t^{2}/2}


Then the boundary becomes:

\left.te^{-t^{2}/2}e^{ixt}\right|_{C}=0


This goes to zero as t\rightarrow\pm\infty, as well as at t\rightarrow0. So the Fourier-Laplace solutions are:

y_{A}\left(x\right)=\int_{-\infty}^{0}e^{ixt}e^{-t^{2}/2}dt;\quad y_{B}\left(x\right)=\int_{0}^{\infty}e^{ixt}e^{-t^{2}/2}dt


[edit] Part d

We can replace t with − t in fA to get:

y_{A}\left(x\right)=-\int_{\infty}^{0}e^{-ixt}e^{-t^{2}/2}dt=\int_{0}^{\infty}e^{-ixt}e^{-t^{2}/2}dt


Then we can form:

y_{1}^{\prime}\left(x\right)=\frac{1}{2}\left(y_{B}\left(x\right)+y_{A}\left(x\right)\right)=\int_{0}^{\infty}\cos\left(xt\right)e^{-t^{2}/2}dt


y_{2}^{\prime}\left(x\right)=\frac{1}{2i}\left(y_{B}\left(x\right)-y_{A}\left(x\right)\right)=\int_{0}^{\infty}\sin\left(xt\right)e^{-t^{2}/2}dt


y_{1}^{\prime} is odd and corresponds to y1, while y_{2}^{\prime} is even and corresponds to y2.


[edit] Part e

Ask seth.

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