help solve the eigenvalue problem: y''+λ²y=0, y'(0)=0, y(π)=0

Question

anonymous · Answer

so$$y"=-\lambda ^{2}$$
integrating both sides we get$$\int\limits_{?}^{?}y"=-\int\limits_{?}^{?}\lambda ^{2}dx$$
so we get
$$y'=-\lambda ^{2}x+c$$
since y'(0)=0...c=0..
so$$y'=-\lambda ^{2}x$$
similarly proceed..integrate this 1st order equation..put in the given value of y and x...find c...
and you will get the required general solution..

blacksteel · Answer

RaGhavv, I suspect your answer is incorrect, since I assume you meant
$$y'' = -\lambda^2y$$

anonymous · Answer

$$y''+\lambda^2y=0$$
gives the characteristic equation
$$r^2+\lambda^2=0$$
with roots
$$r=\pm\lambda i$$
So the solution is of the form
$$y=C_1\sin\lambda x+C_2\cos \lambda x$$
Use the given boundary conditions to solve for $C_1,C_2$.

meanhacker · Answer

thank you @SithsAndGiggles

anonymous · Answer

You're welcome!

meanhacker · Answer

check out this grapher for windows by the way http://www.walterzorn.de/en/grapher/grapher_app.htm

anonymous · Answer

@blacksteel ya i wrongly thought the ques to be...$$y"+\lambda ^{2}=0$$