Transforming differential equations

Question

I feel as though there is a simple answer to this question but I can't seem to understand what was done, my lecturer has an example that begins like so: $$x^2\frac{d^2y}{dx^2}+y=0,$$ then if we let $y(x) = w(t)$ with $x = \frac{1}{t},$ then our equation becomes $$ \frac{1}{t^2}(t^4\frac{d^2w}{dt^2} + 2t^3\frac{dw}{dt}) + w = 0.$$

I guess this is just some basic use of substitution and chain rule or product rule maybe? I don't really understand how we find the part in the bracket and apologise if it is very simple but could someone clarify for me!

Answer 1

Introducing $w$ to replace the $y$ seems unnecessary, but nevertheless, $$\frac {dy}{dx}=\frac{dw}{dx}=\frac{dw}{dt}\cdot\frac{dt}{dx}$$ using the Chain Rule.

Meanwhile, $$\frac{dx}{dt}=-\frac{1}{t^2}$$ $$\implies\frac{dy}{dx}=-t^2\frac{dw}{dt}$$

Now $$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(-t^2\frac{dw}{dt}\right)\frac{dt}{dx}$$ using the Chain Rule again.

Now use the product rule to differentiate the terms in the bracket and the result will follow immediately.

I hope this helps.

Answer 2

$$\frac{dw}{dt} = \frac{dy}{dx} \dot \ \frac{dx}{dt}$$

hope this helped

Answer 3

These are called "Euler's equidimensional equations". Calling tour variable $e^z$ and treating it like an exponential will do the trick. You then change again variables at the end.

Answer 4

Hint: use $t=lnx$ or simply apply Euler's equation result.

Answer 5

make the ansatz $$y(x)=x^{\lambda}$$