This should be a comment, but it is too long.
I would like to start noting that the answer is not correct. Taking the solution $(t_f-t)$ and substituting on the equation we get $(NK^2-NK)/(t_f-t)$, which is not zero.
In case you want also a method of solution for some equations of this type (for more general ones, exponentiation methods give always the solution, and I am sure there are more astucious methods than this one), consider a fuction that has the form: $ y(t)g(t) $ if we derive it twice and make it be zero, we get: $ y''(t)g(t)+2y'(t)g'(t)+y(t)g''(t)=0 $ or equivalently (outside zeros and etc): $ y''(t)+\frac{2g'(t)}{g(t)}y'(t)+y(t)\frac{g''(t)}{g(t)}=0. $
Comparing it with and equation of the form: $ y''(t)+2Ay'(t)+By(t)=0. $ we have: $ \frac{g'(t)}{g(t)} = \frac{A}{t_f-t} \text{ and } \frac{g''(t)}{g(t)} = \frac{B}{(t_f-t)^2} $
Deriving the first equation, and comparing with the second, to have a solution using this method we must have: $ A^2+A=B $ in which case we solve the first equation as usual to obtain: $ g(t) = C(t-t_f)^{-A} $
The equation $(yg)''=0$ can be solved easily as a linear function, and from that we get the solution: $ y(t) = C_1(t-t_f)^A + C_2(t-t_f)^{A+1} $
But I remember, this does not solve all equations of this type. A good way to see is that it only gives two solutions that differ in degree by one!