General and particular solution of differential equation

Question

1) I need to find, in implicit form, the general solution of the differential equation $$\frac{dy}{dx}=\frac{2y^4e^{2x}}{3(e^{2x}+7)^2}$$

2) I then need to find the corresponding particular solution (in implicit form) that satisfies the initial condition $y=2$ and $x=0$.

3) I then need to find the explicit form of this particular solution.

For the first part I came up with $$-\frac{3}{y^4}\frac{ dy}{dx}= \frac{-2e^{2x}}{(e^{2x}+7)^2}$$ which is $$\frac{d}{dx} (y^{-3})=\frac{d}{dx}\left(\frac{1}{e^{2x}+7}\right)$$ then $y^{-3}=\frac{1}{e^{2x}+7} +c$

For part 2) i got $c=0$ so the particular solution would be $y^{-3}=\frac{1}{e^{2x}+7}.$

However I am confused as to how to do the 3rd part as the answer I got for part 2 seems to be in explicit form. I am not sure if I did the first part correctly even so need quite a bit of help!

Answer 1

Hint

You have a mistake in the second integration. Assuming typo's, the antiderivative of $$-\frac{2 e^{2 x}}{\left(e^{2 x}+7\right)^2}$$ is $$\frac{1}{e^{2 x}+7}+C$$ I am sure you can take from here.

Answer 2

Hint

You properly arrived at $$y^{-3}=\frac{1}{e^{2x}+7}$$ So, take the reciprocals which gives now $$y^3=e^{2 x}+7$$ Raise lhs and rhs to power $\frac{1}{3}$ and you get it.