Solution of $y'' - (k+\pi ^2)y=0$ Give this D-E:
$$y'' - (k+\pi ^2)y=0$$ $$k>0$$
$$y(0)=0$$
$$y(1)=1$$
How can I get to this solution: 
$$y=\frac{ \sinh \sqrt{k+\pi ^2}x}{\sinh \sqrt{k+\pi ^2}} $$
What I did:
***$$r^2-(k+\pi ^2)=0 $$
$$r^2=(k+\pi ^2) $$
$$r=+-\sqrt{(k+\pi ^2}) $$
The solution then is:
$$y=Ae^{\sqrt{(k+\pi ^2})x} +Be^{-\sqrt{(k+\pi ^2})x} $$
Managed also to get : $$A=-B$$
$$B=\frac{1}{-e^{\sqrt{(k+\pi ^2}}+e^{-\sqrt{(k+\pi ^2}}}$$
And what next?***
 A: i seems your answer is correct:
$$y(x)=c_1 e^{\sqrt{k+\pi ^2} x}+c_2 e^{-\sqrt{k+\pi ^2} x}$$
and now you must pluggin $$y(0)=0$$ and $$y(1)=1$$
A: You can find $A+B=0$ and $Ae^{\sqrt{k+\pi^2}}+Be^{-\sqrt{k+\pi^2}}=1$. Substitute $B=-A$ and divide second equality by $2$, then we get
$$
A\sinh(\sqrt{k+\pi^2})=\frac{1}{2}.
$$
Thus we get $A=\frac{1}{2\sinh(\sqrt{k+\pi^2})}$ and $B=-\frac{1}{2\sinh(\sqrt{k+\pi^2})}$. Therefore
$$
Ae^{\sqrt{k+\pi^2}x}+Be^{-\sqrt{k+\pi^2}x}=\frac{1}{\sinh\sqrt{k+\pi^2}}\cdot\frac{e^{\sqrt{k+\pi^2}x}-e^{-\sqrt{k+\pi^2}x}}{2}=\frac{\sinh\sqrt{k+\pi^2}x}{\sinh\sqrt{k+\pi^2}}.
$$
A: Notice that,
$$y = A e^{ax} + Be^{-ax} = (\frac{A}{2}+\frac{B}{2})(e^{ax} + e^{-ax}) + (\frac{B}{2}-\frac{A}{2})(e^{-ax}-e^{ax}) = 2(\frac{A}{2}+\frac{B}{2})\cosh(ax) - 2(\frac{B}{2}-\frac{A}{2})\sinh(ax),$$
now let $\alpha = 2(\frac{A}{2}+\frac{B}{2})$ and $\beta = - 2(\frac{B}{2}-\frac{A}{2})$, hence the solution is
$$y = \alpha \cosh(ax) + \beta \sinh(ax),$$
it remains to find $\alpha$ and $\beta$
A: Since the functions $\sinh$ and $\cosh$ are clearly linearly independent solutions of $y''=y$ we get easily that the most general solution to $y''=a^2y$ (for $a>0$) is
$$
y=c_1\sinh(ax)+c_2\cosh(ax)
$$
Applying the initial conditions we get
$$
c_2=0,\qquad c_1\sinh a+c_2\cosh a=1
$$
so $c_1=(\sinh a)^{-1}$. Hence your solution is
$$
y=\frac{\sinh(ax)}{\sinh a}, \qquad a=\sqrt{k+\pi^2}
$$
