How to find general solution to this differential equation (if it exists): $$ xyy'=x^4+y^4 ?$$ I do not know how to even approach it since I never dealt with nonlinear equations. Only thing that I notice is that $x$ and $y$ are symetric in equation: $$ y'=\dfrac{x^4+y^4}{xy}$$ so I expect (maybe) some kind of symetric function for y(x).

Thanks for any help.

  • $\begingroup$ Try an indefinite integral on both sides of the equation. It still won't be easy but there are tools to help. Good luck $\endgroup$ – poetasis Jan 12 at 17:19
  • $\begingroup$ @poetasis In the link you sent there is step where $y$ is taken out of integral, why is that posible if $y$ itself is function of $x$? $\endgroup$ – Thom Jan 12 at 17:23
  • $\begingroup$ I didn't do the calculation; the tool did and it's been 40 years since I got my degree so I'm rusty on this stuff. I wish I could help you more. $\endgroup$ – poetasis Jan 12 at 17:25
  • $\begingroup$ Couple of comments: 1. It isn't actually symmetric, as you can see by rearranging into differential form: $(x^4+y^4)\,dx-xy\,dy=0.$ 2. The solution, if it exists, must have positive slope in the first and third quadrants, and negative slope in the second and fourth quadrants. $\endgroup$ – Adrian Keister Jan 12 at 17:45
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    $\begingroup$ The definition of a new dependent variable $z := \frac{1}{2} y^2$ yields a Riccati equation for $z$: $z' = x^3 + \frac{4}{x} z^2$. $\endgroup$ – Christoph Jan 12 at 17:49

Observing that $y y' = \left(\frac{1}{2}y^2\right)'$ we define the new dependent variable $z := \frac{1}{2} y^2$, $y^4 = 4 z^2$. We then obtain a Riccati equation for $z$: $z' = x^3 + \frac{4}{x} z^2$. This Riccati equation (like any Riccati equation) can be reduced to a second-order linear ordinary differential equation by writing \begin{equation} z = - \frac{x}{4} \frac{u'}{u}, \quad z^2 = \frac{x^2}{16} \frac{(u')^2}{u^2}, \quad z' = - \frac{1}{4} \frac{u'}{u} - \frac{x}{4} \frac{u'' u - (u')^2}{u^2}, \end{equation} which yields $x^2 u'' + x u' + 4 x^4 u = 0$. With the definition of a new independent variable \begin{equation} \xi := x^2, \quad \frac{d}{dx} = 2 \xi^{1/2} \frac{d}{d\xi}, \quad \frac{d^2}{dx^2} = 2 \frac{d}{d\xi} + 4 \xi \frac{d^2}{d\xi^2}, \end{equation} we obtain the Bessel differential equation \begin{equation} \xi^2 \frac{d^2 u}{d \xi^2} + \xi \frac{d u}{d\xi} + \xi^2 u = 0, \end{equation} with fundamental solutions $J_0(\xi)$, $Y_0(\xi)$ (zeroth-order Bessel functions).


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