# Transforming Differential Equation to Bessel ODE [duplicate]

Here is the question I refer to I need help solving a differential equation

I am not the author of the question, it is just we face same problem. I was given a differential equation $y'= x^2 + y^2$. Letting $y(x)=\frac{-f(x)}{f(x)}$ gives us $$f''+x^2f=0$$ This is where I got stuck. The answer in the question stated that the ODE can be transformed to ODE of Bessel kind. However, I am not sure on how to do this since the ODE does not actually has form

$$x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx}+(x^2-v^2)y = 0$$ which is a Bessel Equation. Any help appreciated.

## marked as duplicate by LutzL, user99914, José Carlos Santos, Shailesh, Parcly TaxelMar 9 '18 at 2:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Something is wrong about letting $y(x) = -\frac{f(x)}{f(x)}$? – Chee Han Mar 8 '18 at 20:52

## 1 Answer

HINT :

The change of function : $\quad y(x)=x^\alpha J_\nu(\beta x^\gamma)\quad$ leads to a generalized form of Bessel equation. Then, identification to the standard form of Bessel equation determines the parameters $\alpha,\beta,\gamma,\nu$.

http://mathworld.wolfram.com/BesselDifferentialEquation.html