How to come from $\frac{d}{dt} f(t) = \frac{2t}{1-t} f(t) + \frac{1}{(1-t)^2}$ to $f(t) =\frac{1-e^{-2t}}{2(1-t)^2}$

Question

From an answer on this question, I got the differential equation

$$\frac{d}{dt} f(t) = \frac{2t}{1-t} f(t) + \frac{1}{(1-t)^2}$$

and I was even given the solution $$f(t) = \frac{1-e^{-2t}}{2(1-t)^2}.$$ It seems to be correct, however, I have trouble understanding how the solution was achieved. I tried building a power series $f(t) = \sum_{n=0}^\infty a_n t^n$ out of $f$ and finding the coefficients for it like in this tutorial, but I ended up in a huge ugly sum term

$$ a_n = \frac{(2t)^n}{n!(1-t)^n} a_0 + \sum_{k=1}^n \frac{(2t)^{n-k}}{n!/k! (1-t)^{n-k}}. $$

Not fully shure if this is right. Somehow, the taylor series of the $e$-function sticks in there, but I don't think that this will lead to the solution above. So I am stuck here and any help would be appreciated.

Answer 1

$$y=f(t)$$ $$\frac {dy}{dt} -\frac {2ty}{(1-t)}=\frac{1}{(1-t)^2} $$ $$\text{I.F}=e^{2t}(t-1)^2$$ $$y\cdot e^{2t}(t-1)^2=e^{2t}dt $$ $$y\cdot e^{2t}(t-1)^2=\frac12e^{2t} + c$$ Given $f(0)=0$ $\implies c=-\frac12$ $$f(t)=\frac{1-e^{-2t}}{2(t-1)^2} $$

Explanation for fourth line : $$\frac{dy}{dx} +Py =Q $$ where $P $ and $Q $ are function of $x $.

Multiply both sides by $e^{\int Pdx} $, we get $$\frac{dy}{dx}e^{\int Pdx} +Pye^{\int Pdx} =Qe^{\int Pdx} $$ $$\implies \frac {d}{dx}{(ye^{\int Pdx})}=Qe^{\int Pdx}$$ $$\implies {d}{(ye^{\int Pdx})}=Qe^{\int Pdx}dx$$ Integrating both sides $$(ye^{\int Pdx})=Qe^{\int Pdx}+k$$ The factor $e^{\int Pdx}$ which makes it possible for the left hand side to be integrable is called an Integrating factor.

Answer 2

Thanks to the comment of Hans Lundmark I am finally able to answer my own question. A good and understandable guide for solving linear differantial equations can be found here.

We firstable put the equation into some canonical form:

$$f^\prime(t) \ - \frac{2t}{1-t} f(t) = \frac{1}{(1-t)^2}. $$

Then we generate our integrating factor

$$ \mu(t) = e^{\int\,2t/(1-t)\ dt} = (1-t)^2 \cdot e^{2t}.$$

We multiply both sides of the equation with $\mu(t)$, do some simplification and get

$$\underbrace{f^\prime(t)}_{u^\prime} \cdot \underbrace{(1-t)^2 \cdot e^{2t} }_v + \underbrace{(-2t) \cdot (1-t) \cdot e^{2t}}_{v^\prime} \cdot \underbrace{f(t)}_u = e^{2t}.$$

We make shure that $v^\prime$ is indeed a derivate of $v$ and then we can reverse the product rule which yields

$$\left(f(t)(1-t)^2 e^t\right)^\prime = e^{2t},$$

what we can integrate on both sides:

$$f(t)(1-t)^2 e^t + C = \frac{1}{2} e^{2t}.$$

Because we were given that $f(0) = 0$, we will insert it and $t=0$ as well and we get out $C=\frac{1}{2}$. We can insert $C$ again in the equation above and bring $f(t)$ alone on one side which gives

$$ f(t) = \frac{\frac{1}{2}e^{2t} - \frac{1}{2}}{(1-t)^2 e^{2t}} = \frac{1-e^{-2t}}{2(1-t)^2}. $$

And yes, this is basically, what Aakash Kumar did, but unfortunately he did not provide an explanation for any of his steps and he twisted some literals in his answer (the denominator is $2(1-t)^2$, not $2(t-1)^2$) , so I wanted to share my solution as well.