Third Order Differential Equations I am having trouble solving the third order differential equation 
$y'''+y'=0$
It was given to me in a quiz (which I got wrong) with boundary conditions
$y(0) = 0$
$y'(0)=2$
$y(\pi)=6$
I know that the obvious trial solution is $y=Ae^{rx}$ but I kind of get stuck after here. I have seen in many solutions to other problems when there are complex roots of the complementary equation that this can be expressed as a combination of sine and cosine (I imagine using Euler's Formula). Is that what happens here?

Set $z=y'$ and so $z''+z=0$ whose general solution is $z(t)=A\cos{t}+B\sin{t}$. One has that $y$ is an antidervative of $z$ so $z(t)=A\sin{t}-B\cos{t}+C$ and now we can plug the boundary conditions
$$\begin{align} -B+C&=0\\A&=2\\B+C&=6\end{align}$$
And our solution is
$$z(t)=2\sin{t}-3\cos{t}+3$$
