# ODE $y+\sqrt{x^2+y^2}-xy'=0$

I've tried all my known methods: it is not exact, it has no integrating factor, bernoulli don't, homogenous but can't isolate $y/x$ because of roots and it's not linear or separable! Halp!

$y+\sqrt{x^2+y^2}-xy'=0$

Substitute $y=xv$, where $v$ is a function of $x$. You'd get: $$y'=v+xv'$$

Substituting the above, it should be simple enough to separate it to get: $$\frac{v'}{\sqrt{v^2+1}}=\frac{1}{x}$$ Integrate to get your answer.

• Isn't that the substitution when you do homogenous equations? Except that I directly divide M and N by $x^n$ where $n$ is the degree of homogenous equation. And obviously don't work because of root. – João Pedro Oct 2 '16 at 1:51
• That's right @JoãoPedro – Kugelblitz Oct 2 '16 at 1:52

Let $y(x)=xr(x)$ which gives $y'(x)=r(x)+xr'(x)$:

$$y(x)+\sqrt{x^2+y(x)^2}-xy'(x)=0\Longleftrightarrow$$ $$xr(x)+\sqrt{x^2+\left(xr(x)\right)^2}-x\left(r(x)+xr'(x)\right)=0\Longleftrightarrow$$ $$x\left(\sqrt{1+r(x)^2}-xr'(x)\right)=0\Longleftrightarrow$$ $$\int\frac{r'(x)}{\sqrt{1+r(x)^2}}\space\text{d}x=\int\frac{1}{x}\space\text{d}x$$

Now, use:

1. Substitute $u=r(x)$ and $\text{d}u=r'(x)\space\text{d}x$: $$\int\frac{r'(x)}{\sqrt{1+r(x)^2}}\space\text{d}x=\int\frac{1}{\sqrt{1+u^2}}\space\text{d}u=\ln\left|u+\sqrt{1+u^2}\right|+\text{C}=\ln\left|r(x)+\sqrt{1+r(x)^2}\right|+\text{C}$$
2. $$\int\frac{1}{x}\space\text{d}x=\ln\left|x\right|+\text{C}$$

So, we get (using $r(x)=\frac{y(x)}{x}$):

$$\ln\left|\frac{y(x)}{x}+\sqrt{1+\left(\frac{y(x)}{x}\right)^2}\right|=\ln\left|x\right|+\text{C}$$