I have to find the differential of $(y-xy')^2=x^2+y^2$. Now, I have solved homogeneous equations but this is different because there are two $y'$. I know how to prove that it is a homogeneous equation of degree zero, so we can skip that, but how to solve this? Some hints would be highly appreciated.

  • $\begingroup$ There are two $y'$... So, make it one. $\endgroup$ – Ma Ming May 7 '13 at 11:10
  • $\begingroup$ Use the change of variable $z=y/x$ and everything will go smoothly. $\endgroup$ – Did May 7 '13 at 17:06

Use the fact that

$$\left ( \frac{y}{x}\right )' = \frac{y-x y'}{x^2}$$

From your equation, I get

$$\frac{y-x y'}{x^2} = \frac{1}{x} \sqrt{1+\left ( \frac{y}{x}\right )^2}$$

Let $u = y/x$. Then this equation is equivalent to

$$\frac{du}{\sqrt{1+u^2}} = \frac{dx}{x}$$

Integrating both sides, I get

$$\log{(u + \sqrt{1+u^2})} = \log{x} + C$$

where $C$ is a constant of integration. Then

$$u + \sqrt{1+u^2} = A x$$

where $A = e^C$. Finishing off the algebra, and using the definition of $u$, I get

$$y(x) = \frac12 A x^2 - \frac{1}{2 A}$$

You should verify that this solution does indeed satisfy the original differential equation.



$$(y-xy')^2=x^2+y^2\Leftrightarrow y-xy'=\sqrt{x^2+y^2}\Leftrightarrow y' = \frac{y}{x}-\sqrt{1+\left(\frac{y}{x}\right)^2}.$$

Let me know if you'd like me to elaborate further.


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