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I want to find a series of functions converging to the solution of $$\frac{dy}{dx}=\frac{x^2}{y^2+1},y(0)=0$$. I am stuck using picard's iteration method

First iteration: $$y_1 = \frac{x^3}{3}$$

Second iteration: $$y_1 = \arctan \frac{x^3}{3}$$

Third Iteration: I am stuck here, how to find this ?? as the integral has become so tough to find :/

alkabary
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Waseem Khan
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  • I think the Picard method is not suitable here, since the integral you stumble upon looks very difficult, and I can imagine the next ones won't be easier. On the other hand, you can solve the differential equation by separating variables, but maybe you want to practice the Picard method for some reason? – mickep Jun 07 '15 at 19:50

1 Answers1

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I strongly doubt that Picard's theorem can be of concrete use here; as you have discovered yourself, integrations quickly become impossible. In general, Picard's theorem is good either in theoretical contexts (where you do not want an explicit solution, but are happy only with showing that there exists one), or in the numerical analysis of equations (which you do not seem to want here).

Your problem lends itself well to finding an explicit solution through ordinary methods: rewrite it in the equivalent form $y'(x) + y'(x) y(x)^2 = x^3$ and integrate it, obtaining $y(x) + \frac {y(x) ^3} 3 = \frac {x^3} 3 + C$, with $C$ and integration constant. Evaluating in $x=0$ gives $C=0$. Therefore, one has to solve the equation $y^3 + 3y - x^3 = 0$. Considering $x$ as a parameter in this polynomial equation and deriving with respect to $y$ one gets $y^2 + 3$, which is strictly positive, therefore this 3rd degree equation in $y$ has a single real solution. Using Cardano's formulae one finally gets

$$y(x) = \frac {\sqrt[3] {\sqrt {x^6 + 4} + x^3} } {\sqrt[3] 2} - \frac {\sqrt[3] 2} {\sqrt[3] {\sqrt {x^6 + 4} + x^3} }.$$

Alex M.
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