How to solve this for any constant k?

Question

I have to solve this: $$r’(t)=\frac{-3}{\lambda}(r(t))^2+ k$$ Where k is a constant I was able to solve it when k is 0 but I’m not able to solve it for any k.

Answer 1

One possible trick since the equation is separable.

Write $$-\frac 3\lambda r^2+k=-\frac 3\lambda \left(r-\frac{\sqrt{k} \sqrt{\lambda }}{\sqrt{3}}\right)\left(r+\frac{\sqrt{k} \sqrt{\lambda }}{\sqrt{3}}\right)$$ making, after partial fraction decomposition $$\frac {dr}{-\frac 3\lambda r^2+k}=\frac{ \sqrt{3\lambda }}{2 \sqrt{k}}\left(\frac{1}{3r+ \sqrt{k} \sqrt{3\lambda }}-\frac{1}{3 r- \sqrt{k} \sqrt{3\lambda }} \right)\,dr=dt$$ You will get two logarithms and back to $r(t)$, "probably" a $\tanh(.)$ function (as Mattos suggested).