Can dy/dx=x^2+Y^2 be solved analytically?

[joqhey]

I found this initial value problem and was supposed to comment on the accuracy of Runge Kutta method. Please enlighten me on the analytic solution.

Find y(2) given the differential equation $\frac{dy}{dx}$=y$^{2}$+x$^{2}$ and the initial value y(1)=0.

Thank you.

 

[Mute]

Your equation has the form of a Riccati Equation. As per that wikipedia article, you can transform your particular equation into a linear second order ODE using the change of variables y = -u'(x)/u(x).

 

[JJacquelin]

Have a look at :
https://www.physicsforums.com/showthread.php?t=637384

 

[joqhey]

Thanks, Mute and Jacquelin for your support.

The question was initially meant to be an easy exercise. However I have been on it since without much success. With your suggestions I was able to do the following. However my result, on comparison with the Runge Kutta estimate, does not look good. I have limited knowledge on the operations on Bessel functions and could be making a mistake. I really want to know how to get around this.

Thank you, again.

 

[joqhey]

Thanks, JJacquelin, for the advice. I followed your method and have since posted a pdf of what I did. I do not know whether I messed it up since my exact solution is not even close to the Runge Kutta estimate. Please, check it out and advise me.

Thank you again.

 

[JJacquelin]

Hi !

As far as I can see with a quick look only, there are two main mistakes:
First :
BesselJ[-n,z] = (-1)^n BesselJ[n,z] is true only if n is an integer, which is not the case.
So, you have the choice :
Continue with BesselJ[1/4,z] AND BesselJ[-1/4,z]
or, if you want positive orders, use the relationship between BesselJ[-1/4,z] and BesselJ[3/4,z]and BesselJ[7/4,z], but it will be more complicated.
Anyways, in both cases, the two constants A and B must remain. You cannot eliminate one at this stage of the calculus.
Second :
At the end, it is not correct to add a new constant D, because -(df/dx)/f is not an integral.
Finally, you will obtain an expression y(x) with, into it, one parameter on the form C=(A/B), or C=(B/A), which can be determined according to y(1)=0.

 

[joqhey]

You are right, Jacquelin. Let me try that. I tried the solution suggested in the book "Handbook for Exact Solutions for Ordinary Differential Equations" in the combinations of J and Y functions

I have worked with these suggested solutions and they don't look good either. Once I'm done I shall try to post my solutions and hope for guidance.

Thank you very much for your patience.

 

[joqhey]

Heres what I have done since (attached pdf). I know I shouldn't pursue it this long, but it is a good challenge. There are two possible solutions: a combination of Bessel J and Y, and a combination of Bessel J+ and J-. I worked them both out and if I didn't bungle it again, they neither match each other nor match the numerical solution.

Thank you Physics Forums for the guidance and patience.

 

[JJacquelin]

Hi !

It seems at first sight that you have done a good job.
But, I am awfully sorry, it would take too much time and it is a too boring task to verify all these equations just after the New Year's Day !
I suggest that you first compare your result to the WolframAlpha results :
http://www.wolframalpha.com/input/?i=y'(t)=y^2+t^2,+y(1)=0
Best wishes !

 

[JJacquelin]

The solution given by WolframAlpha leads to y(2)=6.703786022295645
which is in rather good agreement with the numerical result (Runge-Kutta process).
This make think that probably there are some mistakes in your analytical calculus.
Note that, if all was correct, the slope of the curves drawn on your graph should be equal to 1 at (x=1, y=0), which clearly is not the case for the blue curve.

 

[joqhey]

I did take a look at the WolframAlpha solution. It seems that they took the solution into the more negative direction. As I'm using the basic Excel worksheet to solve the Bessel functions, I have been avoiding the differentiation that takes me to the more negative Bessel functions. This being that I have to perform reconversion to positive Bessels for the Excel worksheet and i have no calculator that can perform calculations involving Bessel functions. I shall try to solve it and try to reproduce the Wolfram solution. It however shall be more tedious.
Thanks, JJacquelin, for the time. I am truly grateful.

 

[joqhey]

I did take a look at the WolframAlpha solution. It seems that they took the solution into the more negative direction. As I'm using the basic Excel worksheet to solve the Bessel functions, I have been avoiding the differentiation that takes me to the more negative Bessel functions. This being that I have to perform reconversion to positive Bessels for the Excel worksheet and i have no calculator that can perform calculations involving Bessel functions. I shall try to solve it and try to reproduce the Wolfram solution. It however shall be more tedious.
Thanks, JJacquelin, for the time. I am truly grateful.

 

[JJacquelin]

It is not essential to reproduce the Wolfram solution. This was only to compare to the numerical results.
There is only one solution, but many manner to express it, with or without negative order for the Bessel functions.
The method that you used in your last pdf seems good. If you find where is the mistake, I am sure that all the curves on the drawing will be perfectly the same.

 

[joqhey]

I have just checked the Excel worksheet. It truncates all non-integral Bessel functions so that $J_{\frac{1}{4}}=J_{\frac{3}{4}}=J_0$. My bad. Sorry. Let me seek a different Besselcalculator. Thanks all.

 

[joqhey]

Well, guys, I did get a good calculator and was able to use my two formulae for exact solution of the Ricatti equation to reach the same result as wolfram did i.e $y\left(2\right)=6.70378602229564586367$. Great thanks to all of you. I was even able to come up with a graph that fits sufficiently well with the numerical solution. The site of the online calculator is:http://keisan.casio.com/has10/Free.cgi