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I have to integrate:

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

It is supposed to be an homogeneous differential equation, but I can't see the variable change to make it happens. I tried:

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

And then $z = y/x$, but I don't think that that is the path.

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  • $\begingroup$ Unless I misread something, my edit should provide the correct rewrite when dividing through by $x$... $\endgroup$
    – abiessu
    Mar 19 at 14:34
  • $\begingroup$ You're almost there ["I think" - because of my next comment]. Moving the $y$ over to the LHS gives you $xy'-y$, which is the numerator of your $z'=(y/x)'$, so... $\endgroup$
    – peter a g
    Mar 19 at 14:48
  • $\begingroup$ [Proceeding that way does not give you a homogeneous equation, I think, but certainly one that is separable.] $\endgroup$
    – peter a g
    Mar 19 at 14:49
  • $\begingroup$ looks ok to me, carry on with the sub $\endgroup$
    – Sine of the Time
    Mar 19 at 15:04

2 Answers 2

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Since one is not supposed to answer in comments:

If $z=y/x$, then $$ z'= {xy'-y\over x^2}.$$ Rearranging the original equation, and dividing by $x^2$, one gets

$$ {xy'-y\over x^2}= {1\over x}\sqrt{1-\left({y \over x}\right)^2}$$ Therefore $$ z'= {1\over x}\sqrt{1-z^2},$$ which is doable (separable).

I hope this helps, although this is not homogeneous, as had been requested.

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  • $\begingroup$ I solved the equation separating variables and the solutions match the professor ones, so I assume that there is a mistake in the exercise sentence telling that it is homogeneous $\endgroup$
    – Abel Gutiérrez
    Mar 19 at 16:35
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$$xy' = \sqrt{x^2-y^2}+y$$ $$xdy= (\sqrt{x^2-y^2}+y)dx$$ $$f(x,y)dy=g(x,y)dx$$ is homogeneous if $f$ and $g$ are homogeneous functions of the same degree: $$f(tx,ty)=tx=tf(x,y)$$ $$g(tx,ty)=\sqrt{t^2x^2-t^2y^2}+ty=t (\sqrt{x^2-y^2}+y)=tg(x,y)$$ The two functions are homogeneous ones of degree $1$ and the substitution $y=tx$ gives a separable DE.

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