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?

Let's see:
If $y=e^{rx}$, then the Chain Rule tells us that $y' = re^{rx}$ and $y''' = r^3e^{rx}$.
If $y$ is a solution, that means $y'''+y'=0$.
In other words, $r^3e^{rx} + re^{rx}=0$
If you factor, you find that $(r^3+r)e^{rx}=0$. This happens when the (characteristic) polynomial $p(r) = r^3+r = r(r^2+1)$ has roots: $r = 0,i,-i$.
From here you know that each of $y_1=e^{0x}=1$, $y_2=e^{ix}$, $y_3=e^{-ix}$ are solutions.
Alternatively, using Euler's Identity, the real valued functions $y_1=e^{0x}=1$, $y_2=\sin(x)$, $y_3=\cos(x)$ are solutions.
Using linearity, the general (real valued) solution takes the form $$y = A + B\sin(x) + C\cos(x)$$
You should be able to use the boundary conditions now to solve for the $A, B, C$.
