How to Solve the ODE $y'' - n^2y = 0,~y'(\pi)=0$

Question

can you help me with the following question? Solve the ODE:

$y'' - n^2y = 0, y'(\pi)=0$

I know

$$y''-k^2y=0\to y(x)=a\cdot \exp(kx)+b\cdot \exp(-kx)=A\cosh(kx)+B\sinh(kx)$$

But I am now stuck.

By the way, the answer is $c\cdot\cosh(n(x-\pi))$

thanks:)

Answer 1

As the equation is autonomous, you are free to introduce a shift without changing the form of the solution. So you could also solve to $$ y(x)=a\,e^{n(x-\pi)}+b\,e^{-n(x-\pi)}=A\cosh(n(x-\pi))+B\sinh(n(x-\pi)). $$ Now you get from the initial condition that $B=0$.

Answer 2

If

$$y(x)=Ae^{nx}+Be^{-nx}$$

then

$$y'(x)=Ane^{nx}-Bne^{-nx}.$$

Then $y'(\pi)=0\implies Ane^{n\pi}=Bne^{-n\pi}$ gives you a relation between $A$ and $B$.