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  1. MHB Journeyman
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    Fernando Revilla's Avatar
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    #1
    I quote a question from Yahoo! Answers

    Quote 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.

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  3. MHB Journeyman
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    #2 Thread Author
    This is the answer I have posted there:

    Quote 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?

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    #3
    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

    $\chi$ $\sigma$

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    #4
    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


    $\chi$ $\sigma$


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    #5
    Where is the solution to -u'/u? I need to see the detail steps to arrive at the solution.

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