1
$\begingroup$

Please excuse my moment of ignorance while I reboot my education in math. I am taking an optimal controls course and it has been quite some time since I've worked with calculus.

On to my question...

I have a 2nd order homogeneous differential equation that represents an optimal solution as follows:

$$tx''(t)-3x'(t) = 0$$

For those who are not familiar with controls notation, this equation can be written in classical $x$-$y$ format as such:

$$xy''-3y' = 0$$

What is throwing me off here is the non-constant coefficient before $y''$. I do not remember how to go about solving this and wikipedia is only confusing me further.

Thanks to all in advance for your help!

$\endgroup$

1 Answer 1

3
$\begingroup$

This is actually a first order differential equation in disguise. Let $u = y'$ to get:

$$ x u' - 3 u = 0 $$

This is a separable equation:

$$ \frac{u'}{u} = \frac{3}{x} $$

Integrate:

$$ \log u = 3 \log x + a_1 = \log x^3 + a_1 $$

Therefore:

$$ u = \exp(\log x^3 + a_1) = b_1 x^3 $$

Where $b_1 = \pm \exp(a_1)$.

Given that $u = y'$, we have:

$$ y' = b_1 x^3 $$

Integrate to get the solution:

$$ y = c_1 x^4 + c_2 $$

Where $c_1 = b_1/4$.

$\endgroup$
6
  • $\begingroup$ Hi Ayman. Thanks for the edit and response! Can you explain the part between step 2 and 3? $\endgroup$
    – Kashif
    Oct 2, 2012 at 19:55
  • $\begingroup$ @Kashif We just integrate both sides. You know that the integral of $u'/u$ is $\log u$, right? $\endgroup$ Oct 2, 2012 at 20:11
  • $\begingroup$ No, I don't remember that. I remember that the derivative of $log(u)$ is $1/u$ so then the integral of $u'/u$ would be something else entirely, I think $\endgroup$
    – Kashif
    Oct 2, 2012 at 20:37
  • $\begingroup$ @Kashif The derivative of $\log(u)$ is $1/u$ if $u$ is an independent variable. If $u$ is a function of $x$, then the derivative of $\log(u)$ is $u'/u$ by the chain rule. $\endgroup$ Oct 3, 2012 at 0:40
  • $\begingroup$ Ah, okay. I see it now. Thank you! $\endgroup$
    – Kashif
    Oct 3, 2012 at 0:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .