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.

  • $\begingroup$ Did you try converting the third order system into a first order system? $\endgroup$ – Amzoti Nov 7 '13 at 14:02
  • $\begingroup$ 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. $\endgroup$ – PepeToro Nov 7 '13 at 15:34
  • $\begingroup$ @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. $\endgroup$ – antifb Nov 7 '13 at 19:22
  • $\begingroup$ @user58533 please read the comment above. thanks $\endgroup$ – 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.

  • $\begingroup$ exactly what I wanted. thank you $\endgroup$ – 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




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.


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.

  • $u'=-w$
  • $v'=u$
  • $w'=v$
  • $\begingroup$ Nice answer and bonus alternative approach! +1 $\endgroup$ – Namaste Nov 8 '13 at 2:20
  • $\begingroup$ How'd the talk go yesterday? And...Happy POETS Day! ;-) $\endgroup$ – Namaste Nov 8 '13 at 17:45
  • $\begingroup$ @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! :-) $\endgroup$ – Amzoti Nov 8 '13 at 18:15

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