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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?

FShrike
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    P must be a function of x – Jake Freeman Mar 16 '21 at 18:26
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    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: $\int x,dy + y,dx = xy + C$. – johnnyb Mar 16 '21 at 18:34
  • This is the "standard example" of a non-trivial Riccati equation. For solution methods and solution expressions in named functions see https://math.stackexchange.com/questions/446926/riccati-differential-equation, https://math.stackexchange.com/questions/2348022/riccati-d-e – Lutz Lehmann Mar 16 '21 at 18:40
  • @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? – FShrike Mar 16 '21 at 18:42
  • Also note that $R''=(P'+P^2)R$, now set $P=-y$ and compare with the equation. – Lutz Lehmann Mar 16 '21 at 18:44
  • @IdioticShrike You may like to see this and this. – V.G Mar 16 '21 at 18:45
  • @LightYagami I would if I could understand it! – FShrike Mar 16 '21 at 18:46
  • @LightYagami the second post answers me perfectly, thank you – FShrike Mar 16 '21 at 18:47

2 Answers2

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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.

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

V.G
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