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I try to solve the time dependent differential equation $$ty''(t)-y'(t)-4t^3y(t)=0$$

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    $\begingroup$ Hint: $y' = dy/dt = dy/dx \cdot dx/dt$ $\endgroup$
    – user66081
    Jan 31, 2015 at 10:32
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    $\begingroup$ Further to the insightful hint above..and for clarity it is easier just to let $y\left(x^2\right) = y(x)$ and similarly for the derivatives. $\endgroup$
    – Chinny84
    Jan 31, 2015 at 10:37
  • $\begingroup$ What is the transformation rule for $y''(t)$? $\endgroup$ Jan 31, 2015 at 15:36
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    $\begingroup$ $\frac{d^2y}{dt^2} = \frac{d}{dt}\frac{dy}{dt} = \frac{d}{dt}\left(\frac{dy}{dx}\frac{dx}{dt}\right) = \frac{dx}{dt}\frac{d}{dt}\left(\frac{dy}{dx}\right) + \frac{dy}{dx}\frac{d}{dt}\left(\frac{dx}{dt}\right) = \left(\frac{dx}{dt}\right)^2\left(\frac{d^2y}{dx^2}\right) + \frac{dy}{dx}\left(\frac{d^2x}{dt^2}\right)$ $\endgroup$
    – Winther
    Jan 31, 2015 at 15:42

1 Answer 1

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change of variable $x = t^k, t = x^{1/k}$ where $k$ will be fixed later. $$\frac{dy}{dt} = kt^{k-1}\frac{dy}{dx}, \ \frac{d^2y}{dt^2}=k(k-1)t^{k-2}\frac{dy}{dx}+k^2t^{2k-2}\frac{d^2y}{dx^2} $$

now we can put these in

$\begin{align} t\frac{d^2y}{dt^2} -\frac{dy}{dt} -4t^3y &= t\left(k(k-1)t^{k-2}\frac{dy}{dx}+k^2t^{2k-2}\frac{d^2y}{dx^2} \right) -kt^{k-1}\frac{dy}{dx} -4t^3 y\\ &=k^2t^{2k-1}\frac{d^2y}{dx^2} +\left(k(k-1) - k \right)t^{k-1}\frac{dy}{dx} -4t^3 y\\ &=4t^3 \left( \frac{d^2y}{dx^2}-y\right) \text{ if $k = 2$ } \end{align}$

with $k= 2,$ your equation becomes much simpler $$ \frac{d^2y}{dx^2}-y = 0$$

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