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I am looking for ideas to help me solve this equation: $$y''=y^5+y^2y' $$ I thought of multiplying this equation by $y'$ then integrate, but it does not seem to work.

I thought of switching the variable $y=\frac{1}{u}$ but it got worse.

Any idea would be helpfull, thank you in advance.

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    i don't really know, my math teacher told me to solve it. – 2017-01-16
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    The question is exactly states as follows: "Solve: $y''=y^5+y^2y'$. " – 2017-01-16

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There's a standard trick for when independent variable doesn't appear in the second-order equation. Let $z = dy/dx$. Then $y'' = dz/dx = (dz/dy)(dy/dx) = (dz/dy) z.$ Plug these into your equation, and $y$ becomes the new independent variable in a first order equation:

$$\frac{dz}{dy}z = y^5+y^2z.$$

If you can solve this equation for $z = \mbox{ crud },$ then you write $y' = \mbox{ crud }$ and try to solve that. (But how to solve that first equation is beyond me. Maple gives a very ugly, implicit answer.)