I am solving the differential equation $$xy'^2 - 2yy' - x = 0$$

My attempt:

We divide both sides with $x$, obtaining:

$$y'^2 - 2y'y/x - 1 = 0$$

Then, let $t = y'$. We find:

$$\begin{cases} y' = t \\y/x = \frac{t^2 - 1}{2t}\end{cases}$$

From the second equation, we find: $$dy = dx \frac{t^2-1}{2t} + x\frac{t^2+1}{2t^2}dt$$

From the first equation, we have $ dy = tdx$


$$tdx = dx \frac{t^2-1}{2t} + x\frac{t^2+1}{2t^2}dt$$

By rearranging, we find:

$$(t- \frac{t^2-1}{2t})dx = x \frac{t^2+1}{2t}dt$$


$$\frac{dx}{x} = dt$$


$$\ln|x| = t + c$$


$$x = ce^t$$

And thus:

$$y = c\frac{t^2-1}{2t}e^t$$

Hence, the solution in parametric equations is:

$$\begin{cases} x = ce^t \\y = c\frac{t^2-1}{2t}e^t\end{cases}$$

Can someone verify whether this is correct? The answer my book gives is $$2cy - c^2x^2 + 1 = 0$$ so there is probably an easier method, but I would like to know whether my approach (and solution) is correct too. Thanks in advance.

Edit: I know I have to check the case $x = 0$ separately (as I divided by $x$), but first I want to know whether my solution is correct.

  • $\begingroup$ Quickest check is to insert your solution back into the original equation. $\endgroup$
    – Chinny84
    Apr 26, 2017 at 17:28
  • $\begingroup$ Corrected, thanks for pointing out. $\endgroup$
    – user370967
    Apr 26, 2017 at 18:59

1 Answer 1


You lost the square of $t$ in the denominator in the last term in going from the 5th to the 6th equation. Corrected you get $\ln|x|=\ln|t|+c$ or $t=Cx$.

Alternatively, multiply with $x$, add $y^2$ to get the equivalent form $$ (xy'-y)^2=x^2+y^2\iff \left(\frac yx\right)'=\pm\frac1x\sqrt{1+\frac{y^2}{x^2}} $$ which suggests to consider the substitution $\frac yx=\sinh(u)$ with $\cosh(u)u'=\pm\frac1x\cosh(u)$. Then $u(x)=\pm\ln|x|+c$, $$y=x\sinh(\pm\ln|x|+c)=\frac x2\left(Cx-\frac1{Cx}\right).$$ Which indeed is equivalent to $2Cy-C^2x^2+1=0$.


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