# 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.

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Did you try converting the third order system into a first order system? –  Amzoti Nov 7 '13 at 14:02
As suggested, convert to 1st order ODE. Also know as state space. Define $u=y, \, v=y', \, w=y''$. From there the result will follow. –  PepeToro Nov 7 '13 at 15:34
@Amzoti I didn't try that because I haven't learned about systems of linear equations yet. is there another way to solve this? I can learn about it if I must, but I'm not sure it is required because this problem comes up before systems of linear equations. –  antifb Nov 7 '13 at 19:22
@user58533 please read the comment above. thanks –  antifb Nov 7 '13 at 19:24

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.

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exactly what I wanted. thank you –  antifb Dec 6 '13 at 18:50

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.

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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.

• $u'=-w$
• $v'=u$
• $w'=v$
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Nice answer and bonus alternative approach! +1 –  amWhy Nov 8 '13 at 2:20
How'd the talk go yesterday? And...Happy POETS Day! ;-) –  amWhy Nov 8 '13 at 17:45
@amWhy: The talk was excellent, both the prof and the students were exposed to stuff they had not seen before, so it was a success! Now, work begins on the Dec 10th talk. Happy POETS Day, hope you are having a great day! :-) –  Amzoti Nov 8 '13 at 18:15