# Solve $ty''-y'-4t^3y=0$

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

• Hint: $y' = dy/dt = dy/dx \cdot dx/dt$ Jan 31, 2015 at 10:32
• 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. Jan 31, 2015 at 10:37
• What is the transformation rule for $y''(t)$? Jan 31, 2015 at 15:36
• $\frac{d^2y}{dt^2} = \frac{d}{dt}\frac{dy}{dt} = \frac{d}{dt}\left(\frac{dy}{dx}\frac{dx}{dt}\right) = \frac{dx}{dt}\frac{d}{dt}\left(\frac{dy}{dx}\right) + \frac{dy}{dx}\frac{d}{dt}\left(\frac{dx}{dt}\right) = \left(\frac{dx}{dt}\right)^2\left(\frac{d^2y}{dx^2}\right) + \frac{dy}{dx}\left(\frac{d^2x}{dt^2}\right)$ Jan 31, 2015 at 15:42

change of variable $x = t^k, t = x^{1/k}$ where $k$ will be fixed later. $$\frac{dy}{dt} = kt^{k-1}\frac{dy}{dx}, \ \frac{d^2y}{dt^2}=k(k-1)t^{k-2}\frac{dy}{dx}+k^2t^{2k-2}\frac{d^2y}{dx^2}$$
\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{ ifk = 2} \end{align}
with $k= 2,$ your equation becomes much simpler $$\frac{d^2y}{dx^2}-y = 0$$