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In this case, I have not posted a link there.How would you go about solving the differential equation dy/dx = x^2 + y^2?

- Thread starter Fernando Revilla
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- Thread starter
- #1

In this case, I have not posted a link there.How would you go about solving the differential equation dy/dx = x^2 + y^2?

- Thread starter
- #2

This is the answer I have posted there:

Does anyone know an alternative?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.

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

$\chi$ $\sigma$

The solution of the ODE...

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

can be found in Polyanin A.D. & Zaitzev V.F.

$$ u(x) = \sqrt{x}\ \{c_{1}\ J_{\frac{1}{4}} (\frac{x^{2}}{2}) + c_{2}\ Y_{\frac{1}{4}} (\frac{x^{2}}{2})\ \}\ (2)$$

... 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

$\chi$ $\sigma$

- Jan 23, 2020

- 1

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