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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.

  • 0
    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

1 Answers 1

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http://eqworld.ipmnet.ru/en/solutions/ode/ode0202.pdf and http://eqworld.ipmnet.ru/en/solutions/ode/ode0207.pdf can provide hints for you.

Let $s=x^n$, where $n$ is a constant,

Then $\dfrac{dy}{dx}=\dfrac{dy}{ds}\dfrac{ds}{dx}=nx^{n-1}\dfrac{dy}{ds}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(nx^{n-1}\dfrac{dy}{ds}\right)=nx^{n-1}\dfrac{d}{dx}\left(\dfrac{dy}{ds}\right)+n(n-1)x^{n-2}\dfrac{dy}{ds}=nx^{n-1}\dfrac{d}{ds}\left(\dfrac{dy}{ds}\right)\dfrac{ds}{dx}+n(n-1)x^{n-2}\dfrac{dy}{ds}=nx^{n-1}\dfrac{d^2y}{ds^2}nx^{n-1}+n(n-1)x^{n-2}\dfrac{dy}{ds}=n^2x^{2n-2}\dfrac{d^2y}{ds^2}+n(n-1)x^{n-2}\dfrac{dy}{ds}$

$\therefore n^2x^{2n-2}\dfrac{d^2y}{ds^2}+n(n-1)x^{n-2}\dfrac{dy}{ds}-xy=0$

$n^2x^{2n-3}\dfrac{d^2y}{ds^2}+n(n-1)x^{n-3}\dfrac{dy}{ds}-y=0$

$n^2s^{\frac{2n-3}{n}}\dfrac{d^2y}{ds^2}+n(n-1)s^{\frac{n-3}{n}}\dfrac{dy}{ds}-y=0$

The suitable choice is $n=\frac{3}{2}$

$\therefore\frac{9}{4}\dfrac{d^2y}{ds^2}+\frac{3}{4}s^{-1}\dfrac{dy}{ds}-y=0$

$9s\dfrac{d^2y}{ds^2}+3\dfrac{dy}{ds}-4sy=0$

Let $y=s^ku$, where $k$ is a constant,

Then $\dfrac{dy}{ds}=s^k\dfrac{du}{ds}+ks^{k-1}u$

$\dfrac{d^2y}{d^2s}=s^k\dfrac{d^2u}{ds^2}+ks^{k-1}\dfrac{du}{ds}+ks^{k-1}\dfrac{du}{ds}+k(k-1)s^{k-2}u=s^k\dfrac{d^2u}{ds^2}+2ks^{k-1}\dfrac{du}{ds}+k(k-1)s^{k-2}u$

$\therefore 9s\left(s^k\dfrac{d^2u}{ds^2}+2ks^{k-1}\dfrac{du}{ds}+k(k-1)s^{k-2}u\right)+3\left(s^k\dfrac{du}{ds}+ks^{k-1}u\right)-4ss^ku=0$

$9s^{k+1}\dfrac{d^2u}{ds^2}+18ks^k\dfrac{du}{ds}+9k(k-1)s^{k-1}u+3s^k\dfrac{du}{ds}+3ks^{k-1}u-4s^{k+1}u=0$

$9s^{k+1}\dfrac{d^2u}{ds^2}+(18k+3)s^k\dfrac{du}{ds}-(4s^{k+1}+3k(3k-2)s^{k-1})u=0$

$9s^2\dfrac{d^2u}{ds^2}+(18k+3)s\dfrac{du}{ds}-(4s^2+3k(3k-2))u=0$

The suitable choice is $k=\frac{1}{3}$

$\therefore 9s^2\dfrac{d^2u}{ds^2}+9s\dfrac{du}{ds}-(4s^2+1)u=0$

Let $t=ms$, where $m$ is a constant,

Then $\dfrac{du}{ds}=\dfrac{du}{dt}\dfrac{dt}{ds}=m\dfrac{du}{dt}$

$\dfrac{d^2u}{ds^2}=\dfrac{d}{ds}\left(m\dfrac{du}{dt}\right)=\dfrac{d}{dt}\left(m\dfrac{du}{dt}\right)\dfrac{dt}{ds}=m\dfrac{d^2u}{dt^2}m=m^2\dfrac{d^2u}{dt^2}$

$\therefore 9\left(\frac{t}{m}\right)^2m^2\dfrac{d^2u}{dt^2}+9\frac{t}{m}m\dfrac{du}{dt}-\left(4\left(\frac{t}{m}\right)^2+1\right)u=0$

$9t^2\dfrac{d^2u}{dt^2}+9t\dfrac{du}{dt}-\left(\frac{4t^2}{m^2}+1\right)u=0$

$t^2\dfrac{d^2u}{dt^2}+t\dfrac{du}{dt}-\left(\frac{4t^2}{9m^2}+\frac{1}{9}\right)u=0$

$t^2\dfrac{d^2u}{dt^2}+t\dfrac{du}{dt}+\left(\frac{4t^2}{9(mi)^2}-\frac{1}{9}\right)u=0$

The suitable choice is $m=\frac{2i}{3}$

$\therefore t^2\dfrac{d^2u}{dt^2}+t\dfrac{du}{dt}+\left(t^2-\frac{1}{9}\right)u=0$

  • 0
    I think there is a typo in your procedure: $ 9s^{k+1}\dfrac{d^2u}{ds^2}+(18k+3)s^k\dfrac{du}{ds}-(4s^{k+1} \color{red}{-} 3k(3k-2)s^{k-1})u=0 $2015-12-26