Question

Solve the given problem by means of an eigenfunction expansion.

y"+2y= −x, y(0)=0, y'(1)=0

Answer 1

The general solution

$y(x)=c_1 \cos \sqrt{\lambda}x+c_2\sin \sqrt{\lambda}x\\ y'(x)=-c_1\sqrt{\lambda} \sin \sqrt{\lambda}x+c_2\sqrt{\lambda} \cos \sqrt{\lambda}x\\$

Determining the constants

$0=y(0)=c_1\\ 0=y'(1)= -c_1\sqrt{\lambda} \sin \sqrt{\lambda}+c_2\sqrt{\lambda} \cos \sqrt{\lambda}\\ c_1=0\\ = c_2\sqrt{\lambda} \cos \sqrt{\lambda}\\ =c_2 \not= 0 \implies c_2\sqrt{\lambda} \cos \sqrt{\lambda}=0\\ \cos \sqrt{\lambda}=0\\ cos x =0 \space \space if \space \space x= \frac{(2n-1) \pi}{2}\\ \phi(x)=k_n sin \frac{(2n-1) \pi x}{2}$

Normalizing

$1= \int _0^1r(x) \phi_n^2(x)dx \\ =k_n^2 \int _0^1 sin^2 \frac{(2n-1) \pi x}{2} dx\\ 1=k_n^2(0.5)\\ k_n=\sqrt{2}\\ \phi(x)=\sqrt{2} sin \frac{(2n-1) \pi x}{2}\\$

The general solution

$L(y)= \mu r (x)y+f(x)\\ y(x)= \sum _{n=1}^{+\infin} \frac{c_n}{\lambda_n - \mu}\phi_n(x)\\ y(x)= \sum _{n=1}^{+\infin} \frac{c_n}{ (2n-1)^2 \pi^2 /4 - 2}\sqrt{2} sin \frac{(2n-1) \pi x}{2}\\ c_n = \int_0^1 f(x)\phi_n(x)dx = \sqrt{2} \int_0^1 x sin \frac{(2n-1) \pi x}{2}dx = - \frac{4 \sqrt{2}(-1)^n}{(2n-1)^2 \pi^2}\\ y(x)= \sum _{n=1}^{+\infin} \frac{(-1)^{n+1}}{ (n-0.5)^2 \pi^2 ((n-0.5)^2)\pi^2-2}sin (n-0.5)\pi x$