I have the following differential equation -
$s(1-s)t = f'(t)(f(t)-st)$
Initial condition: $f(0) = 0.$
I solved it and got the solution as -
$$ f(t) = \sqrt{2 c_1+s^2-2 (s-1) s \ln (t)}+s $$
But the answer given is -
$f(t) = k(s)t,$
where
$$k(s) = \frac{s+ \sqrt{4s - 3s^2}}{2}.$$
If anyone can provide me some hint on how to proceed and reach the specified answer, I would be really grateful.
The answer given, $f(t)=k(s)t,$ where $$k(s)=\frac{s+\sqrt{4s-3s^2}}{2},$$ is correct, as you can verify by differentiating and plugging into the DE. It is evident by inspection that $k(s)t$ satisfies the IC.
As for finding the solution, guess-and-check proves to be the best method. You can see by inspection that some constant (that is, some quantity constant with respect to $t$) times $t$ might be able to solve the DE, and it certainly satisfies the IC. So, posit $f(t)=k(s)t,$ plug it into the DE, and solve for $k(s):$ \begin{align*} s(1-s)t&=k(s)(k(s)t-st)\\ s(1-s)&=k(s)(k(s)-s)\quad\text{for}\; t\not=0. \end{align*} Now solve for $k(s),$ (might need the quadratic formula) and you're finished.