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!
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:

$$\left( \begin{array}{ccc}
0 &amp; 0 &amp; 1 \\
1 &amp; 0 &amp; 0 \\
0 &amp; 1 &amp; 0 \end{array} \right)$$

And I'll let you proceed with the usual eigenvalues method.
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 :): $$y=Ae^{mx}$$
$$y'=Ame^{mx}$$
$$y''=Am^2e^{mx}$$
$$y'''=Am^3e^{mx}$$

$$Am^3e^{mx}-Ae^{mx}=0$$
$$m^3-1=0,m=1$$
$$y=Ae^x$$