1
$\begingroup$

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!

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – João Pedro Oct 2 '16 at 1:51
  • 1
    $\begingroup$ That's right @JoãoPedro $\endgroup$ – Kugelblitz Oct 2 '16 at 1:52
0
$\begingroup$

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 $r(x)=\frac{y(x)}{x}$):

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.