
MHB Journeyman
#1
June 22nd, 2013,
10:02
I quote a question from Yahoo! Answers
Quote:
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.

June 22nd, 2013 10:02
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MHB Journeyman
#2
June 22nd, 2013,
10:17
Thread Author
This is the answer I have posted there:
Quote:
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
.
Does anyone know an alternative?

MHB Master
#3
June 22nd, 2013,
13:13
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
$\chi$ $\sigma$

MHB Master
#4
June 23rd, 2013,
03:01
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
$\chi$ $\sigma$

#5
January 23rd, 2020,
14:36
Where is the solution to u'/u? I need to see the detail steps to arrive at the solution.