The general solution for this differential equation?

Question

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?

Answer 1

Hint: $\displaystyle\int\frac{dy}{y^3} = \int y^{-3} dy$

Answer 2

Your equation is separable and you can rewrite it as $$\frac{dy}{y^3}=\frac{3}{4}x^5 dx$$ So, integrating both sides gives $$-\frac{1}{2 y^2}=\frac{x^6}{8}+C_1$$ which, after simplifications, write $$y^2=-\frac{4}{x^6+C_2}$$ that is to say $$y=\pm\frac{2}{\sqrt{C_3 -x^6}}$$