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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?

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    $\begingroup$ P must be a function of x $\endgroup$ Commented Mar 16, 2021 at 18:26
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    $\begingroup$ Generally speaking, you can't have mixed variables and a single differential under an integral and successfully solve it. Usually you want to separate the variables, but sometimes you can solve it without that, but it usually (always?) requires having differentials for each variable as well. Example: $\int x\,dy + y\,dx = xy + C$. $\endgroup$ Commented Mar 16, 2021 at 18:34
  • $\begingroup$ This is the "standard example" of a non-trivial Riccati equation. For solution methods and solution expressions in named functions see math.stackexchange.com/questions/446926/…, math.stackexchange.com/questions/2348022/riccati-d-e $\endgroup$ Commented Mar 16, 2021 at 18:40
  • $\begingroup$ @johnnyb would the method be viable if I found (and I assume such a thing cannot be found if the textbook’s claim is correct) some $P(x)$ that satisfied my equations? $\endgroup$ Commented Mar 16, 2021 at 18:42
  • $\begingroup$ Also note that $R''=(P'+P^2)R$, now set $P=-y$ and compare with the equation. $\endgroup$ Commented Mar 16, 2021 at 18:44

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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.

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  • $\begingroup$ Maybe the textbook was being simple? Because that looks way too complicated for A-level students. Thanks! $\endgroup$ Commented Mar 16, 2021 at 19:06
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    $\begingroup$ "Being simple" is no excuse for being wrong. If they wrote $y^3$ instead of $y^2$ they might be correct. $\endgroup$ Commented Mar 16, 2021 at 19:09
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Note that $\dfrac{\mathrm d}{\mathrm dx}\left(\dfrac{y^2}{e^{2xy}} \right)= 2y \cdot \dfrac{\mathrm dy}{\mathrm dx} \cdot \dfrac{1}{e^{2yx}} - 2y \cdot y^2 \cdot \dfrac{1}{e^{2yx}}\cdot \color{red}{\dfrac{\mathrm dy}{\mathrm dx}}\ne 2y \cdot \dfrac{\mathrm dy}{\mathrm dx} \cdot \dfrac{1}{e^{2yx}} - 2y \cdot y^2 \cdot \dfrac{1}{e^{2yx}}$

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  • $\begingroup$ Oh dear - thank you for spotting my oversight! $\endgroup$ Commented Mar 16, 2021 at 18:38

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