$dy/dx=\sqrt{x^2+y^2}$

Question

2.1k Views Asked by Bumbble Comm At 14 May 2026 - 12:35

$$\frac{dy}{dx}=\sqrt{x^2+y^2}$$

slope=distance from origin, should be simple and interesting. May have no solution!

I have tried several approaches, best:

$(\frac{dy}{dx}-y)(\frac{dy}{dx}+y)=x^2$ multiply by $e(-x) * e(+x)$ as integrating factor. Substitute $\frac{1}{2}x^2=t$.

Second approach:

$y=x\sinh(u)$ and $x=e(t)$ yields $\frac{du}{dt} + \tanh(u)=e(t)$.

Sorry, I am not yet using the proper format.

There are 1 best solutions below

**Bumbble Comm** · Answer 1 · 2018-08-19 07:42:26

Let $u=y+\sqrt{x^2+y^2}$ ,

Then $y=\dfrac{u}{2}-\dfrac{x^2}{2u}$

$\dfrac{dy}{dx}=\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}-\dfrac{x}{u}$

$\therefore\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}-\dfrac{x}{u}=u-\left(\dfrac{u}{2}-\dfrac{x^2}{2u}\right)$

$\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}-\dfrac{x}{u}=\dfrac{u}{2}+\dfrac{x^2}{2u}$

$\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}=\dfrac{u}{2}+\dfrac{x^2+2x}{2u}$

$(u^2+x^2)\dfrac{du}{dx}=u^3+(x^2+2x)u$

Let $v=u^2$ ,

Then $\dfrac{dv}{dx}=2u\dfrac{du}{dx}$

$\therefore\dfrac{u^2+x^2}{2u}\dfrac{dv}{dx}=u^3+(x^2+2x)u$

$(u^2+x^2)\dfrac{dv}{dx}=2u^4+(2x^2+4x)u^2$

$(v+x^2)\dfrac{dv}{dx}=2v^2+(2x^2+4x)v$

Let $w=v+x^2$ ,

Then $v=w-x^2$

$\dfrac{dv}{dx}=\dfrac{dw}{dx}-2x$

$\therefore w\left(\dfrac{dw}{dx}-2x\right)=2(w-x^2)^2+(2x^2+4x)(w-x^2)$

$w\dfrac{dw}{dx}-2xw=2w^2+(4x-2x^2)w-4x^3$

$w\dfrac{dw}{dx}=2w^2+(6x-2x^2)w-4x^3$

This belongs to an Abel equation of the second kind.

In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $w=\dfrac{1}{z}$ ,

Then $\dfrac{dw}{dx}=-\dfrac{1}{z^2}\dfrac{dz}{dx}$

$\therefore-\dfrac{1}{z^3}\dfrac{dz}{dx}=\dfrac{2}{z^2}+\dfrac{6x-2x^2}{z}-4x^3$

$\dfrac{dz}{dx}=4x^3z^3+(2x^2-6x)z^2-2z$