Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Some weeks ago I was asked to solve one ODE. I tried all methods I know, but couldn't crack this equation. Also I tried to use Matlab's dsolve function - without any result.

$y' + \frac{2x}{y} = x^2$

Does anyone have any suggestions on this?

P.S. This is not for homework or anything like this, just want to know if this equation is integrable?

share|improve this question
    
Try the same technique as I used in artofproblemsolving.com/Forum/viewtopic.php?p=2471315#p2471315. It looks like the only (but essential) difference will arise in the very end. –  fedja Dec 28 '11 at 13:46

3 Answers 3

up vote 15 down vote accepted

If you mean a (real) analytical solution with $y(0)=0$, then the answer is 'no'. If you write this as the problem of looking for integral curves of $\omega = (2x-x^2y)\ dx + y\ dy$ in the $xy$-plane, you'll immediately see that the origin is an isolated singular point of elliptic type (i.e., the eigenvalues of the linearization at $(x,y)=(0,0)$ are pure imaginary). In particular, there is no nontrivial integral curve of $\omega$ that passes through the origin.

All of the integral curves of $\omega$ near the origin spiral around it and converge inward as they go counterclockwise around the origin. To see this, just consider the vector field $$ Z = (2x - x^2y)\frac{\partial\ }{\partial y} - y\frac{\partial\ }{\partial x} $$ which is tangent to the integral curves of $\omega$, and consider the convex function $H = 2x^2 + y^2$. The derivative of $H$ with respect to $Z$ is $-2x^2y^2$, which is non-positive and vanishes only on the axes, which are not tangent to $Z$. Since $H$ is strictly decreasing along each integral curve of $Z$, the integral curves of $Z$ spiral in to the origin.

Added note about complex solutions: I neglected to mention that, over the complex numbers, there are, of course, two solutions satisfying $y(0)=0$. I doubt that there is a closed form expression for them in elementary terms, but one easily obtains that they have convergent power series expansions of the form $$ y_{\pm}(x) = {} \pm i\sqrt{2}\ x f_0(x^4) + x^3 f_1(x^4), $$ where $$ f_0(t) = 1 - \frac{t}{3\cdot2^6} + \frac{13\ t^2}{5\cdot3^2\cdot2^{13}} - \cdots $$ and $$ f_1(t) = \frac{1}{2^2} + \frac{t}{3\cdot2^9} - \frac{11\ t^2}{5\cdot3^3\cdot2^{15}} + \cdots. $$

share|improve this answer
1  
Robert, various types of engineering and computer students refer to a closed form solution as analytic or analytical. Even our own Suvrit says he uses the words that way sometimes. –  Will Jagy Dec 27 '11 at 22:10
2  
@Will: I'm aware of this, but I think that this should be discouraged whenever one encounters it. 'Analytic' has a well-defined, accepted meaning, but 'closed form' is in the eye of the beholder, I think. For example, would they accept the solution of $y' = -2xy/(1+x^2+5y^4)$ in the form $y(x)^5 + (1+x^2)y(x) = y(0)^5 + y(0)$ as being in 'closed form'? I would. Would you? –  Robert Bryant Dec 27 '11 at 22:26
1  
Robert, thanks, I was not sure. This is one of those situations where the OP is probably not going to understand, but if you have something to teach the rest of us and patience to type it in, that's good. I would prefer that the word analytic be used only in its mathematical sense, and I used to leave comments telling the OP not to use words they did not understand. I ran out of steam on that issue, though. I'm fairly happy with an implicit description of a function, as in your example. –  Will Jagy Dec 27 '11 at 22:39
1  
Sorry for my incorrect using of "analytical" word (I use it to distiguish from numerical solution; English isn't my native language, sometimes I occasionally use calque from russian words). This is not Cauchy problem, there's no any initial conditions. Just was asked to find solution in form of $y = y(x, C)$ or $F(y,x,C) = 0$. –  Maksim V. Bolonkin Dec 28 '11 at 11:33
3  
@Maksim: Many (native English speaking) people do use 'analytical' in this antique way, but I think that it would be better to say something like 'integrable in elementary terms', even though the meaning of 'elementary terms' isn't universally agreed on either. Since your equation is of Abel type, I'd say that your best hope of integrating it in elementary terms would be to compute its invariant(s) and see whether it belongs on the list of known 'integrable' Abel equations. I'm not sure that either Maple or Matlab has programmed in all of the known 'integrable' cases, so it's worth a try. –  Robert Bryant Dec 28 '11 at 16:30

If it is any help, $ y = \pm \sqrt{2} (k-x^2)^{1/2}$ are homogeneous solutions, that is, they solve $$y^\prime + \dfrac{2x}{y} = 0$$

share|improve this answer
    
I tried this way, to find solution of homogeneous equation and then use the "variation of constant" method. It's almost impossible. –  Maksim V. Bolonkin Dec 28 '11 at 11:21

Maple classifies this as an Abel equation of the second kind, class A: see http://www.maplesoft.com/support/help/Maple/view.aspx?path=odeadvisor/Abel2A

Maple does not find a closed-form solution, moreover the equation has no symmetries according to symgen.

share|improve this answer
3  
@Robert: Actually, one must be skeptical about symgen. What this means is that it didn't find any symmetries in the class of symmetries in which it looks. For a single ODE, away from singularities, there always exist symmetries in the sense that there exist nontrivial symmetries of the $xy$-plane that carry solutions curves to solution curves. Moreover, this equation does admit the involution $(x,y)\to (-x, -y)$ and, in the complex plane, the $4$-fold symmetry generated by $(x,y)\to (ix, -iy)$. –  Robert Bryant Dec 27 '11 at 21:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.