Solve y''' = y with given conditions? I'm given the differential equation:
$$y''' = y$$
which solves to:
$$y(x) = c_1e^x + e^{-x/2}\left(c_2\cos\left(\frac{\sqrt{3}x}{2}\right) + c_3\sin\left(\frac{\sqrt{3}x}{2}\right)\right)$$
But I'm given the following initial conditions: 
$$y(0) = 1, y'(0) = y''(0) = 0$$
 I don't understand how I would do that because I ended up with an answer $c_1 = 0$, $c_2 = 0$, and $c_3 = 0$, which can't possibly right. Any way to do this, and if so, what is the final answer?
 A: $y(0) = 1 \implies 1 = c_1 + c_2$ so $c_1 = c_2 = c_3 = 0$ can't possibly be correct.
A: Your initial conditions lead to
$$\begin{eqnarray}
c_1+c_2=1\\
c_1-\frac {c_2}2+\frac{\sqrt3}2{c_3}=0\\
c_1-\frac {c_2}2-\frac{\sqrt3}2{c_3}=0
\end{eqnarray}$$
The difference of the two last equations gives $c_3=0$. Their sum gives $2c_1-c_2=0$, and subtracting from $2c_1+2c_2=2$, you get $c_2=\frac{2}{3}$. Then the first equation gives $c_1=\frac13$.
A: We solve the initial value problem using the complex exponential version of the general solution.
Let $\omega=\frac{-1+\sqrt{3}i}{2}$ be one of the non-real roots of $x^3-1=0$. Then the other is $\omega^2$, and the general solution is $pe^x+qe^{\omega x}+re^{\omega^2 x}$. From the initial conditions we have $p+q+r=1$, $p+\omega q+\omega^2  r=0$ and $p+\omega^2q+\omega r=0$. 
From the last two equations, by subtraction we get $q=r$. The first two equations now say that $p+2a=1$ and $p+(\omega+\omega^2)q=0$. But $\omega+\omega^2=-1$, so we get $p=q$ and therefore $p=q=r=\frac{1}{3}$. (Actually, the answer can be written down without any calculation if we are accustomed to cube roots of unity.)
If we want to, we can now express the  $\frac{1}{3}(e^{\omega x}+e^{\omega^2 x})$ part in terms of $e^{-x/2}$ and sines and cosines. But we end up with something  less attractive than $\frac{1}{3}\left(e^{x}+e^{\omega x}+e^{\omega^2 x}\right)$! 
