Find the general solution of the following first order differential equation $\frac{dy}{dx}=\frac{8x^3+3x^4}{y^4}$ I have multiplied both sides by $y^4$
which gives me $\frac{dy}{dx}y^4=x^3(3x+8)$
Then do I integrate both sides with respect to x?
$\int\frac{dy}{dx}yx^4dx=\int{x^3(3x+8)dx}$
Am I still on the right track?
So I get $\frac{yx^5}{5}=\frac{3x^5}{5}+2x^4+c$
Do I just solve for y after this?

By separation of variables, you get $$y^4 \,dy = (3x^4 + 8x^3)\,dx$$
Now, integrate each side of the equation:
$$\int y^4 \,dy = \int (3x^4 + 8x^3)\,dx$$
$$\iff \frac 15 y^5 = \frac 35x^5 + 2x^4 + c$$
