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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!

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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 $c_1 = b_1/4$.

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    @Kashif You're welcome.2012-10-03