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My professor has said that this will be an easy homework exercise. He suggested using change of variable $t = \dfrac{2}{3}x^{3/2}$, and then removing the first derivative term of the form $p(t) \dfrac {dy}{dt}$ by a transformation

\begin{equation} y(t) = w(t)e^{\frac{-1}2\int p(t)dt} \end{equation}

Whenever I try do this I get a very messy bunch of terms that have the exponential in them. Is this problem really that easy to do?

After change of variables I get:

\begin{equation*} y^{(2)} - xy = (3/2)^{2/3}t^{2/3}*{\frac{d^2y}{dt^2}} + \frac{1}2 (3/2)^{-1/3}t^{-1/3}{\frac{dy}{dt}} - (3/2)^{2/3}t^{2/3}y \end{equation*}

Then I try to use \begin{equation} y(t) = w(t)e^{\frac{-1}2\int p(t)dt} \end{equation}

by taking derivatives and substituting into the above equation.. then its just a mess of terms with exponentials. The second derivative of y(t) in that transformation is really ugly.. so I think I'm doing it wrong.

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    Jeff, it'd be nice if you show where you get lost. Write your start down and let's see what's the issue.2012-05-03
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    Jeff, you either keep $x$ or $t$, but no both. If you change variables, all $x$ should be replaced by their equivalent in $t$.2012-05-03
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    Thanks, i've subbed in t for all the x terms. I was trying to keep the x's in until as late as possible, because its representation in t was so messy..2012-05-03
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    Should I be taking derivatives of the y(t) in the transformation identity, and then substituting that into my equation?2012-05-03
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    You might want to see how it's done [here](http://books.google.com/books?hl=en&id=2szu5oErrKsC&pg=PA79).2012-05-04
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    @J.M. That might as well serve as an answer.2012-05-08
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    @Peter: it's a bit long to write out for me. If you want to, you're welcome to it.2012-05-08
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    This question has been solved perfectly. Hope that the asker has been diving enough and accept the answer at an early date.2012-09-10

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