Applying a substitution to a ODE for solving a Bessel's function

Question

I have the following ODE: $$x^2y''+xy'+4(x^4-1)y=0$$ and I am given the hint to use the substitution $t=x^2$ which should convert the problem to a standard Bessel's Function. The substitution is the part I am stuck on. Most of the solutions to similar problems jump to the final from after to substitution with out showing the steps. I know I need to let $u(t)=u(x^2)=y(x)$ and differential both sides with respect to $x$ by the chain rule. How do I explicitly carry out this substitution?

Answer 1

I think I had a mental block when I asked the question, too late at night. I believe the solution is: Let $t=x^2$. Then: $$y(x)=u(x^2)=u(t)$$ $$y'(x)=u'(x^2)2x=2xu'(x^2)=2\sqrt{t}u'(t)$$ $$y''(x)=2u'(x^2)+2xu''(x^2)2x=2u'(x^2)+4x^2u''(x^2)=2u'(t)+4tu''(t)$$ Applying the substitution to: $$x^2y''+xy'+4(x^4-1)y=0$$ we get: $$t[2u'+4tu'']+\sqrt{t}[2\sqrt{t}u']+4(t^2-1)u=0$$ $$2tu'+4t^2u''+2tu'+4(t^2-1)u=0$$ $$4t^2u''+4tu'+4(t^2-1)u=0$$ $$t^2u''+tu'+(t^2-1)u=0$$ Which is now a standard Bessel equation, as required.