Attempt at solving $\frac{dy}{dx} = x^2 + y^2$

Question

I read in my maths textbook that an analytic solution to $\frac{dy}{dx} = x^2 + y^2$ is not known. Nevertheless, as an exercise I tried to do it, to see where the difficulty lies.

In the same textbook, I learnt of a general method for solving ones of the ordinary linear form: $\frac{dy}{dx} + Py = Q$ where $P, Q$ are purely functions of $x$, or constant. The method involved finding $R = e^{\smallint P \space dx} $, resulting in $R\cdot\frac{dy}{dx} + RPy \equiv \frac{d}{dx}[Ry]$ and then finding $y = \frac{1}{R}\smallint RQ \space dx$. I attempted to form this for powers of two, with the following being the key equation: \begin{equation} 2y \cdot \frac{dy}{dx} \cdot R + RPy^2 \equiv \frac{d}{dx}[Ry^2] \end{equation} I thought then, that finding $R$ in the same way would result in a very valid method for solving simple quadratic differential equations. I attempted this for $\frac{dy}{dx} = x^2 + y^2$: \begin{equation}\frac{dy}{dx} - y^2 = x^2\end{equation} \begin{equation} 2y \cdot \frac{dy}{dx} - 2y \cdot y^2 = 2y \cdot x^2, P = -2y \end{equation} \begin{equation} R = e^{\int -2y \space dx} = \frac{1}{e^{2yx}} \cdot C, C = 1 \text{ for simplicity.} \end{equation} \begin{equation} 2y \cdot \frac{dy}{dx} \cdot \frac{1}{e^{2yx}} - 2y \cdot y^2 \cdot \frac{1}{e^{2yx}} = 2y \cdot x^2 \cdot \frac{1}{e^{2yx}} \end{equation} \begin{equation} \frac{d}{dx}[\frac{1}{e^{2yx}} \cdot y^2] = 2y \cdot x^2 \cdot \frac{1}{e^{2yx}} \end{equation} \begin{equation} \frac{1}{e^{2yx}} \cdot y^2 = \int 2y \cdot x^2 \cdot \frac{1}{e^{2yx}} \space dx \end{equation} which equals what, exactly? I am at a loss as to how to solve this integral. I pose four questions:

(1) What is the result of that final integral?

(2) Was my attempted method sound?

(3) What is the better approach to solving differential equations with a second power? I assume my method is not the best - if it works at all!

(4) If that final integral does have a result, then why did my textbook claim no solution exists?

Answer 1

Note that $\dfrac{\mathrm d}{\mathrm dx}\left(\dfrac{y^2}{e^{2xy}} \right)= 2y \cdot \dfrac{\mathrm dy}{\mathrm dx} \cdot \dfrac{1}{e^{2yx}} - 2y \cdot y^2 \cdot \dfrac{1}{e^{2yx}}\cdot \color{red}{\dfrac{\mathrm dy}{\mathrm dx}}\ne 2y \cdot \dfrac{\mathrm dy}{\mathrm dx} \cdot \dfrac{1}{e^{2yx}} - 2y \cdot y^2 \cdot \dfrac{1}{e^{2yx}}$

Answer 2

If that's what it really says, your textbook is wrong. The general solution of $\dfrac{dy}{dx} = x^2 + y^2$ is

$$ y(x) = - \frac{x \left(c J_{-{3}/{4}}\! \left({x^{2}}/{2}\right) +Y_{-{3}/{4}}\! \left({x^{2}}/{2}\right)\right)}{c J_{{1}/{4}}\! \left({x^{2}}/{2}\right)+Y_{{1}/{4}}\! \left({x^{2}}/{2}\right)}$$

where $J_\nu$ and $Y_\nu$ are Bessel functions of the first and second kinds, and $c$ is an arbitrary constant.