Can I use the method of seperation variables to solve the following ODE Can I use the method of seperation variables to solve the following ODE
$$F'(x)=\frac{a}{(F(x))^2}+c$$
where $a,c>0$
If without $c>0$, I can solve it , but now how can I solve this equation?
 A: Rearrange to get $$\frac{F^2}{a+cF^2} dF = dx$$ which is a separable equation. Integrate both sides to get $$x = \int \frac{F^2}{a+cF^2} dF = \frac 1a \int \frac{F^2}{1+\frac caF^2} dF.$$ Let $\tan u = \sqrt{\frac ca} F$, then $\sec^2u \,du = \sqrt{\frac{c}{a}} \, dF$ so we get $$x = \frac 1a \int \frac{\frac ac \tan^2u}{1+\tan^2u} \sqrt{\frac ac}\sec^2 u \, du = \frac 1c \sqrt{\frac ac}\int \frac{\tan^2u}{\sec^2 u} \sec^2 u \, du = \frac 1c \sqrt{\frac ac}\int \tan^2u \, du$$ and thus $$x = \frac 1c \sqrt{\frac ac} \left ( \tan u - u\right)+K = \frac 1c \sqrt{\frac ac} \left ( \sqrt{\frac ca}F - \tan^{-1} \left( \sqrt{\frac ca}F\right) \right) +K$$ Therefore the solution is given implicitly by 
$$x = \frac 1c F - \sqrt{\frac{a}{c^3}}\tan^{-1} \left( \sqrt{\frac ca}F\right)+K$$ where $K$ is our integration constant.
A: $$x = \int \frac{F^2}{a+cF^2} dF = \frac 1a \int \frac{F^2}{1+\frac caF^2} dF.$$
Susbstitute $u=\sqrt{\frac c a}F$
$$x= \sqrt{\frac a {c^3}} \int \frac{u^2}{1+u^2} du$$
$$x= \sqrt{\frac a {c^3}}( \int du -\int \frac{1}{1+u^2} du)$$
$$x= \sqrt{\frac a {c^3}}( u -\arctan (u))+C$$
$$x= \frac F c - \sqrt{\frac a {c^3}}\arctan (F\sqrt{\frac  c a})+C$$
