I'm trying to solve this differential equation when $y(0)=0$.
$$\frac {dy}{dt}=k\left(n_1-{\frac{y}{2}}\right)^2\left(n_2-{\frac{y}{2}}\right)^2\left(n_3-{\frac{3y}{4}}\right)^3$$
Where $$k=6.22*10^{-19}$$ $$n_1=2$$ $$n_2=2$$ $$n_3=3$$
So far I've tried moving everything on one side $$\frac{1}{\left(n_1-{\frac{y}{2}}\right)^2\left(n_2-{\frac{y}{2}}\right)^2\left(n_3-{\frac{3y}{4}}\right)^3}dx=k *dt$$ but I don't know where to go from there. Should I turn them into $\frac{1}{u^2v^2w^3}dy=k*dt$ and go ahead, or...?
With the provided constants your DE can be written as $$ y'=-k\left(\frac12\right)^3\left(\frac34\right)^3(y-4)^7 $$ which has a solution $$ (y-4)^{-6}=(y_0-4)^{-6}+c(t-t_0) $$ where $C$ assembles all the above constant factors and the $6$ from the differentiation of $(y-4)^{-6}$.