Solve the differential equation $y^2y''=(y')^3$ Can anyone check my attempt? Thanks

Attempt: Let $u=y'=dy/dx$. Then

$$u'=\frac{du}{dx} =\frac{du}{dy}\frac{dy}{dx}=\frac{du}{dy}u$$

So given equation becomes a separable differential equation \begin{align*} y^2\frac{du}{dy}=u^3 \implies u^{-2}du=y^{-2}dx \implies \frac{1}{u}=\frac{1}{y}+c \implies u= \frac{y}{1+cy} \implies \frac{dy}{dx}=\frac{y}{1+cy} \implies ln|y|+cy=x+c_1 \implies y=\pm e^{c_1}e^{x-c_1y}=c_2e^{x-c_1y}\end{align*}

So the answer is $$y=c_2e^{x-c_1y}$$

  • 1
    $\begingroup$ You haven't solved for $y$ here, though. $\endgroup$ – Nitin Nov 2 '16 at 16:13


Then by integration,

$$\frac1{y'}=\frac1y+C$$ or


Integrating a second time,

$$\log y+Cy=x+C'$$ which cannot be solved analytically for $y$, except using Lambert's $W$ function.

With free redefinition of the constants, we have

$$\log Cy+Cy=\log\left(Cye^{Cy}\right)=x+C'$$ and


  • $\begingroup$ You can write the final expression more succinctly. $y=AW(Be^x)$. $\endgroup$ – Parcly Taxel Nov 2 '16 at 16:21

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.