Complex Roots of the Characteristic Equation I have the following IVP: $y"+6y'+18y=0$, $y(\frac{\pi}{2})=0$, $y'(\frac{\pi}{2})=6$.
Converting this into a characteristic equation:
$$r^2+6r+18=0$$
Now, solving for r:
$$r=-3\pm 3i$$
When I try to solve for the constants of integration, I get $C_1=0$, $C_2=-2e^{\frac{3\pi}{2}}$.
Plugging all of this in, I get: $y=-2e^{\frac{3\pi}{2}}*e^{-3t}*sin(3t)$, but my homework program says I'm incorrect. Anyone have any idea where I might be going wrong?

First, the form of $y(x)$ is the following:
$$y(x)=e^{-3x}\left(c_{1}\cos(3x)+c_{2}\sin(3x)\right),$$
for two constants $c_{1}$ and $c_{2}$ to be computed using your initial considitons. Then
$0=y\left(\frac{\pi}{2}\right)=e^{-\frac{3\pi}{2}}\left(c_{1}\cdot\cos\left(\frac{3\pi}{2}\right)+c_{2}\sin\left(\frac{3\pi}{2}\right)\right)=-c_{2}e^{-\frac{3\pi}{2}}$ and this implies $c_{2}=0$ because:
$$\cos\left(\frac{3\pi}{2}\right)=0 \qquad \text{and} \qquad \sin\left(\frac{3\pi}{2}\right)=-1$$
So, the solution reduces to:
$$y(x)=c_{1}e^{-3x}\cos\left(3x\right)$$
Now take the derivative of $y(x)$ :
$$ y^{\prime}(x)=(-3)c_{1}e^{-3x}\cos(3x)-3c_{1}e^{-3x}\sin(3x) $$
and we evaluate at $x=\frac{\pi}{2}$ to get:
$$6=y^{\prime}\left(\frac{3\pi}{2}\right)=-3c_{1}e^{-\frac{3\pi}{2}}(-1)$$
then $c_{1}=2e^{-\frac{3\pi}{2}}$ and this finally gives:
$$ y(x)= 2e^{\frac{3\pi}{2}}e^{-3x}\cos(3x),$$
or $$ y(x)=2e^{\left(\frac{3\pi}{2}-3x\right)}\cos(3x).$$
