i need help with this one
solve
$\displaystyle y'' + 9y = 0; \quad y = 3, \; y'=3 \;\textnormal{at}\; x = \pi/3$
$\displaystyle y(x) = ?$
Write down the characteristic equation. Then see what form the solution has. So $\displaystyle r^2 + 9r = 0 $ or $\displaystyle r(r+9) = 0 $, which means $\displaystyle r = 0, - 9 $ which are real solutions. Then the general solution has the form $\displaystyle y = c_{1}e^{r_{1}t} + c_{2}e^{r_{2}t} $. So $\displaystyle y' = c_{1}r_{1}e^{r_{1}t} + c_{2}r_{2}e^{r_{2}t} $. You are given the initial conditions.
$\displaystyle y(x)=C_1\cos 3x+C_2\sin 3x$
$\displaystyle y'(x)=-3C_1\sin 3x+3C_2\cos 3x$
Now, put $\displaystyle x=\frac{\pi}{3}$ and solve the system
$\displaystyle \left\{\begin{array}{ll}y\left(\frac{\pi}{3}\right )=3\\y \ '\left(\frac{\pi}{3}\right)=3\end{array}\right.$