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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 $$y=c_2e^{x-c_1y}$$

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    $\begingroup$ You haven't solved for $y$ here, though. $\endgroup$ – Nitin Nov 2 '16 at 16:13
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$$\frac{y''}{y'^2}=\frac{y'}{y^2}$$

Then by integration,

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

$$\left(\frac1y+C\right)y'=1.$$

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

$$y=C\,W\left(C'e^x\right).$$

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

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