Simple Nonlinear Differential Equation

Question

I have tried in vain to search for a closed form solution (involving known functions) to the simple first order equation $\dfrac{dy}{dx}=x^2+y^2, y(0) = 0$. Can anyone help with a solution that does not have the form of an infinite series?

Answer 1

I tried to solve this using maple and here's my code:

dy:=diff(y(x),x)=x^2+y(x)^2;

dsolve({dy,y(0)=0},y(x));

And it gives me a solution in terms of Bessel function, so I don't think there's a solution in terms of elementary functions.

Answer 2

As already said in comments, as provided by a CAS, the solution of $$\dfrac{dy(x)}{dx}=x^2+y^2(x)$$ is given by $$y(x)=\frac{x^2 \left(-c_1 J_{-\frac{5}{4}}\left(\frac{x^2}{2}\right)+c_1 J_{\frac{3}{4}}\left(\frac{x^2}{2}\right)-2 J_{-\frac{3}{4}}\left(\frac{x^2}{2}\right)\right)-c_1 J_{-\frac{1}{4}}\left(\frac{x^2}{2}\right)}{2 x \left(c_1 J_{-\frac{1}{4}}\left(\frac{x^2}{2}\right)+J_{\frac{1}{4}}\left(\frac{x^2}{2}\right)\right)}$$ where appear Bessel functions of the first kind.

Where the problem happens is when $x\to 0$ since $$y(0)=-\frac{2 \Gamma \left(\frac{3}{4}\right)}{c_1 \Gamma \left(\frac{1}{4}\right)}$$ which would imply an infinite value for $c_1$ and no solution. For any $y(0)\neq 0$ there is a solution using $$c_1=-\frac{2 \Gamma \left(\frac{3}{4}\right)}{\Gamma \left(\frac{1}{4}\right)}\frac 1{y(0)}$$

Answer 3

Putting $$ y = -\dfrac{d(\ln z)}{dx} = -\dfrac{z'}{z}$$ gives $$y'=\left(\dfrac{z'}{z}\right)^2-\dfrac{z''}{z},$$ and so $$ y' = y^2 - \frac{z''}{z}$$

So in $y' = y^2 + x^2$ the $y^2$ cancels out leaving $-\dfrac{z''}{z} = x^2$, or $z''=-x^2z$, which according to Wolfram has the solutions $z = c \cdot D[-1/2, u]$, where D is the parabolic cylinder function of fractional order $1/2$ and $u = (1+i)x$ or (-1+i)x

The relation $y = -\dfrac{d(\ln z)}{dx}$ then leads to $$(1-i)y = \frac{2D[1/2, u]}{D[-1/2, u]}-u,$$ where $u$ can be $(1+i)x$. However, Wolfram says $2D[1/2, 1] - D[-1/2, 1] = 1.03133....$ and not $0$ as is required for $y(0)=0$ so there also seems to be a problem with this solution!