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I am trying to solve the following differential equation:

$$yy''= 3(y')^{2} \\$$

I feel like there must be some substitution to turn this equation into an easier one, but I can not seem to find a substitution that works. Also, I have tried separating the equation into terms of a chain rule, but I can't seem to make that work either. Anyone have a hint or suggestion?

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We have the quotient rule for taking derivative: $\left(\dfrac{y}{y'}\right)'= \dfrac{(y')^2-yy''}{(y')^2}= 1-3 = -2\implies \dfrac{y}{y'} = -2x+C\implies \dfrac{y'}{y} = \dfrac{1}{-2x+C}$. Can you continue?

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  • $\begingroup$ So, @DeepSea after integrating I get: $$ln(y)= -.5(ln|-2x+c|)\\$$ which simplifies to $$y= d(-2x+c)^{-.5}\\$$ $\endgroup$ – Bryan Chen Feb 13 '17 at 0:38
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If you are not at an equilibrium point, you can transform to $$ \frac{y''}{y'}=3\frac{y'}{y}\implies \ln|y'|=3\ln|y|+c\implies y'=C·y^3 $$ which is separable and can be solved as $$ \frac1{y^2}=-2Cx+D. $$

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