How to solve $y''' = y$ I'm trying to solve the following differential equation 

$ y''' = y$                   

and given conditions: 

$  y(1) = 3$, $y'(1) = 2$ and $y''(1) = 1 $        

I began by making it:

$y''' - y = 0$          

But I'm uncertain of what to do after, or if I'm even headed in the right direction. 
Please explain step by step as I'm taking an introductory course to Differential Equations!
 A: The other solution is good.  Here is another method.
First, set $y' = x$.  Then we have $x'' = y$.
From here, set $x' = z$.  This gives us $z' = y$.
Notice that we have now expressed the original ODE as a system of equations:
$$x' = z$$
$$y' = x$$
$$z' = y$$
Therefore, we can rewrite our system in the form $A \mathbf{x'} = \mathbf{x}$, where $A$ is:
$$\left( \begin{array}{ccc}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0 \end{array} \right)$$
And I'll let you proceed with the usual eigenvalues method.
A: Hint
If $y'''=y$, the characteristic equation is $r^3=1$ the roots of which being $1$,$-(-1)^{1/3}$, $(-1)^{2/3}$. Convert to trigonometric expressions and use the general formula for linear differential equations.
A: This is a linear DE with constant coefficients. The characteristic polynomial is $x^3-1$. Using the roots, we can write down the general solution.  The "natural" solution involves some complex exponentials, but we can express these in terms of sines and cosines. 
A: The equation $y'''=y$ can be restated as $(D^3-1)y=0$. Furthermore, we can factor
$$
D^3-1=(D-1)\left(D+\tfrac12+i\tfrac{\sqrt3}2\right)\left(D+\tfrac12-i\tfrac{\sqrt3}2\right)\tag{1}
$$

We can solve
$$
(D-a)u=f\tag{2}
$$
using an integrating factor. Choose $g$ so that $Dg=ag$; for example, $g(x)=e^{ax}$. Then
$$
\begin{align}
D(gu)
&=gDu-uDg\\
&=gDu-uag\\
&=g(D-a)u\\
&=gf\tag{3}
\end{align}
$$
Therefore,
$$
\begin{align}
u(x)
&=\frac1{g(x)}\int g(x)f(x)\,\mathrm{d}x\\
&=e^{-ax}\int e^{ax}f(x)\,\mathrm{d}x\tag{4}
\end{align}
$$
where the antiderivative includes a constant of integration. Thus, when $f=0$, the solution is
$$
u(x)=Ce^{ax}\tag{5}
$$

Thus, considering $(1)$ and $(5)$, $y=e^x$, $y=e^{-\frac12-i\frac{\sqrt3}2}$, and $y=e^{-\frac12+i\frac{\sqrt3}2}$ are all solutions to $y'''=y$. Taking linear combinations of these three solutions gives the general solution to be
$$
ae^x+be^{-\frac12x}\cos\left(\tfrac{\sqrt3}2x\right)+ce^{-\frac12x}\sin\left(\tfrac{\sqrt3}2x\right)\tag{6}
$$
It is a matter of solving three simultaneous equations to find $a$, $b$, and $c$ for the given initial conditions.
A: Is $e^x$ a solution? How about $e^{\omega x}$ and $e^{\omega^2 x}$, where $\omega$ is a nonreal cube root of $1$? 
If these are solutions, what would a general solution look like?
Where did these come from you ask? You have a linear homogeneous ODE. Assume that for some complex number $r$, that $y=e^{rx}$ is a solution and as a consequence, $r^3-1=0$. The three $r$-values that make this happen are $1$, $\omega$, and $\omega^2$. Then we verify that these three $e^{rx}$ actually do solve the equation.
A: What remains the same through so many derivations? Well the first thing that one thins is always $Ae^x$

We can prove it properly aswell :): $$y=Ae^{mx}$$
$$y'=Ame^{mx}$$
$$y''=Am^2e^{mx}$$
$$y'''=Am^3e^{mx}$$
$$Am^3e^{mx}-Ae^{mx}=0$$
$$m^3-1=0,m=1$$
$$y=Ae^x$$
