A second order differential Equation isn't being solved by the application of $2$ Boundary Conditions.

Question

$y''+25y=0\qquad\qquad y'(0)=6\qquad\qquad y'(\pi)=-9$

What class of Problem is this?

Is it solvable?

Answer 1

$$y''+25y=0\implies y''=-25 y\implies y=A\cos5t+B\sin5t\implies$$

$$\begin{cases}6&=y'(0)&=5B\cos 0&=5B\\{}\\-9&=y'(\pi)&=5B\cos5\pi&=-5B\end{cases}$$

So it doesn't seem solvable....