<s>[INST] The general solution for this differential equation? Find the general solution of this differential equation:
$$ \frac{dy}{dx} = \frac{3x^5 y^3}{4} $$
Here's what I've done so far:
$ dy=\frac{3x^5 y^3 dx}{4} $
$ 4dy = 3x^5 y^3 dx $
$ \frac{4dy}{y^3} = 3x^5 dx $
Integrating both sides:
$ \int \frac{4dy}{y^3} = \int 3x^5 dx $
$ \int \frac{4dy}{y^3} = \frac{3x^6}{6} + C $
$ 4\int \frac{dy}{y^3} = \frac{x^6}{2} + C $
To solve the left side, let $ u=y^3 $, therefore $ du=3y^2 dy $, so:
$ \frac{4}{3} \int \frac{3y^2 dy}{y^5} = \frac{x^6}{2} + C $
$ \frac{4}{3} \int \frac{du}{y^5} = \frac{x^6}{2} + C $
$ \frac{4}{3} \int \frac{du}{y^3 y^2} = \frac{x^6}{2} + C $
$ \frac{4}{3} \int \frac{du}{u * y^2} = \frac{x^6}{2} + C $
I'm not sure what to do next. How do I get rid of that $ y^2 $? Or did I do something wrong?
 [/INST] Hint: $\displaystyle\int\frac{dy}{y^3} = \int y^{-3} dy$
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