MHB Riccati's equation and Bessel functions

[Fernando Revilla]

I quote a question from Yahoo! Answers

How would you go about solving the differential equation dy/dx = x^2 + y^2?

In this case, I have not posted a link there.

 

[Fernando Revilla]

Re: Riccati's equation an Bessel functions

This is the answer I have posted there:

You should specify the exact meaning of 'solving' here. Although we have a
Riccati's equation it is not a trivial problem to find the general solution. It can
be expressed in terms of the $J_n$ Bessel functions of the first kind. Have a
look here.

Does anyone know an alternative?

 

[chisigma]

Re: Riccati's equation an Bessel functions

The non linear first order Riccati ODE...

$$ y^{\ '} = x^{2} + y^{2}\ (1)$$

... can be transformed into a linear second order ODE with the substitution...

$$y = - \frac{u^{\ '}}{u} \implies y^{\ '} = - \frac{u^{\ ''}}{u} + (\frac{u^{\ '}}{u})^{2}\ (2)$$

... so that we have to engage the ODE...

$$u^{\ ''} + x^{2}\ u =0\ (3)$$

At first the (3) may seem ‘simple’ but of course it isn’t... an attempt will be made in next post...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Re: Riccati's equation an Bessel functions

The solution of the ODE...

$$u^{\ ''} + x^{2}\ u = 0\ (1)$$

can be found in Polyanin A.D. & Zaitzev V.F. Handbook of Exact Solutions for Ordinary Differential Equations, 2nd edition...$$ u(x) = \sqrt{x}\ \{c_{1}\ J_{\frac{1}{4}} (\frac{x^{2}}{2}) + c_{2}\ Y_{\frac{1}{4}} (\frac{x^{2}}{2})\ \}\ (2)$$... where $J_{\frac{1}{4}} (*)$ and $Y_{\frac{1}{4}} (*)$ are Bessel function of the first and second type, $c_{1}$ and $c_{2}$ arbitrary constants. Now computing $y= - \frac{u^{\ '}}{u}$ leads us to the solution of the Riccati's equation...Kind regards $\chi$ $\sigma$

 

[kjpatterson21]

Where is the solution to -u'/u? I need to see the detail steps to arrive at the solution.