# General solution of this ODE

Let $\alpha$, $\beta$ and $v$ be a positive real numbers consider the ordinary differential equation $$x^2 u''(x) + \bigg(\alpha^2 \beta^2 x^{2\beta} + \frac{1}{4} - v^2 \beta ^2 \bigg)u(x) = 0, x>0$$ For an unknown function u(x). Show that one solution of this ODE is $$u(x) = \sqrt x f(\alpha x^{\beta})$$ Where $f(y)$ is a solution of Bessel's Differential equation of order $v$.

So far I have tried the following:

Use the co-ordinate transformation $y = (\alpha x^{\beta})$.

Define

$v(y) = v(\alpha x^{\beta}) = u(x)$

Then by chain rule

$u'(x) = v'(\alpha x^{\beta})(\alpha \beta x^{\beta -1})$

$u''(x) = v''(\alpha x^{\beta})(\alpha \beta x^{\beta -1})^2 + v'(\alpha x^{\beta})(\alpha \beta x^{\beta -1})$

Tried plugging in to the original equation and really stuck from there. Not sure

how to prove one solution of this ODE is in a form of $u(x) = \sqrt x f(\alpha x^{\beta})$.

Any help will be appreciated.

• It would help to write out ahead of time the transformed Bessel equation satisfied by f, but you will essentially have to plug u into the ODE as it stands – Triatticus Nov 14 '16 at 5:05

As Triatticus commented, replace $u(x)$ by $\sqrt x f(a x^{b})$ in the differential equation. Using classical rules of differentiation, you should have (before simplifications)$$u'(x)=a b x^{b-\frac{1}{2}} f'\left(a x^b\right)+\frac{f\left(a x^b\right)}{2 \sqrt{x}}$$ Conitnue to get $u''(x)$.
After some basic simplifications and replacing, you should arrive to $$f\left(a x^b\right) \left(v^2-a^2 x^{2 b}\right)-a x^b \left(a x^b f''\left(a x^b\right)+f'\left(a x^b\right)\right)$$ Now, replace $a x^b$ by $y$ to get $$f(y)(v^2-y^2)-y(f'(y)+yf''(y))=0$$ that is to say $$y^2f''(y)+yf'(y)+(y^2-v^2)f(y)=0$$ Just have a look here.