Relationship between solutions of $y'''+y=0$ let $u,v,w$ be solutions of $y'''+y=0$ such that 
$$u(0)=1, u'(0)=0, u''(0)=0$$
$$v(0)=0, v'(0)=1, v''(0)=0$$
$$w(0)=0, w'(0)=0, w''(0)=1$$
show that $u'=-w, v'=u, w'=v$ without findind the solutions.

I suppose all the inequalities are solved in the same way, so one of them should be enough for me to handle the exercise.
if I could find a differential equation such that it has solutions $u'$ and $-w$, then since they share the same initial condition, I could conclude $u'=-w$. but what differential equation would that be?
I don't think I'm going in the right direction and I feel like a hint will be enough for me to solve this.
 A: There is no need to cast the ODE into a $1^{st}$ order ODE.
Since the ODE $y''' + y = 0$ is linear, homogenous and doesn't depends on $t$, if $f$ is a solution of it, so does all of its derivatives $f'$, $f''$ and linear combination of them.
In particular, $\varphi = u' + w$ is a solution of it.
By definition, it is clear $\varphi(0) = \varphi'(0) = 0$. Notice
$$\varphi''(0) = ( u' + w )''(0) = u'''(0) + w''(0) = -u(0) + w''(0) = -1 + 1 = 0$$
Since $y''' + y = 0$ is a $3^{rd}$ order linear ODE and its derivatives up to second order all vanishes, $\varphi(t)$ is identically zero for all $t$. As a result, $u' = - w$.
The other two relations $v' = u, w' = v$ can be proved in similar manner.
A: The approach is to convert this DEQ to a linear first order system using:


*

*$w=y$, so $w'=y'$

*$v=y'$, so $v'=y'' = u$

*$u'=y''' = -w$


We end up with the system
$$\begin{bmatrix}w'\\v' \\ u' \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1 \\ -1 & 0 & 0 \end{bmatrix}\begin{bmatrix}w\\v \\ u \end{bmatrix}$$
Thus, you can easily verify that $u'=-w, v'=u, w'=v$ without finding the solutions.
As an alternate to this approach (the problem does not want you doing this, but it is instructive), you can do the following.


*

*Solve the system three times using the different initial conditions.

*Case 1: $u''' + u = 0, u(0) = 1, u'(0)=0, u''(0)=0$

*Case 2: $v''' + v = 0, v(0) = 0, v'(0)=1, v''(0)=0$

*Case 3: $w''' + w = 0, w(0) = 0, w'(0)=0, w''(0)=1$


After you find each solution, verify the following relations.


*

*$u'=-w$

*$v'=u$

*$w'=v$

A: I will answer this, by petition of the OP. 
you have an ODE of the form $y'''+y=0$. It is an ODE which you can transform into a 1st order differential equation. This is done as follows.
Let $u=y$. So your "artificial" variable $u$ means the solution $y$. But still you have a way to go, because you need still more information on the derivatives of $y$. So let us define $v=y'$ and $w=y''$. This last is, for now, jus notation. So, now we have the following:
We have now three variables, $u,v,w$. So we can write as follows:
$$
\begin{bmatrix}
u\\
v\\
w
\end{bmatrix}'
$$
where $'$ means derivative on each variable. Recall how we defined $u,v,w$. 
so we have
$$
\begin{bmatrix}
u\\
v\\
w
\end{bmatrix}'=\begin{bmatrix} 0 & 1 & 0\\
0 & 0 & 1\\
-1 & 0 & 0
\end{bmatrix}\begin{bmatrix} u\\ v\\w \end{bmatrix}
$$
Note that this matrix notation is exactly the way we defined the new variables. So we have
$u'=v$
$v'=w$
$w'=-u$
Just from making the computations of the above vector expression.
Please let me know if something else is not clear. I might be skipping something as it is clear to me but not necessarily it is clear to you.
