Inverse differential operator and inverse function We were given a second order differential equation to solve. I found the complementary function (CF) of the solution but I'm stuck with the calculation of the particular integral which is given below:
$$PI= \frac{1}{(D^2 +2D+1)}\frac{1}{(1- e^x)^2}$$  where $$D=\frac{d}{dx}$$
How do I solve this? I tried expanding it using the binomial expansion but I'm not sure if that is the right approach.
 A: $$y''+2y'+y=\frac{1}{(1- e^x)^2}$$
Multiply by $e^x$:
$$(ye^x)''=\frac{e^x}{(1- e^x)^2}$$
$$(ye^x)''=\left( \frac{1}{1- e^x}\right)'$$
Integrate twice.

For this DE you posted:
$$(s^2 D'^2+3sD'+1)t=\frac{1}{(1-s)^2}$$
$$s^2 t''+3st'+t=\frac{1}{(1-s)^2}$$
$$(s^2 t''+2st')+(st'+t)=\frac{1}{(1-s)^2}$$
$$(s^2 t')'+(st)'=\frac{1}{(1-s)^2}$$
$$(s^2 t'+st)'=\left (\frac{1}{1-s} \right)'$$
Integrate. Then you get a first order DE that you can integrate with any method you know.
$$s^2 t'+st=\frac{1}{1-s} +C_1$$
$$(s t)'=\frac{1}{s(1-s)} +\dfrac {C_1}s$$
Integrate.

For the inverse operaor method:
$$PI= \frac{1}{(D^2 +2D+1)}\frac{1}{(1- e^x)^2}$$
$$PI= \frac{1}{(D+1)^2}\frac{1}{(1- e^x)^2}$$
$$PI= \frac{1}{(D+1)}\dfrac 1 {(D+1)}\frac{1}{(1- e^x)^2}$$
$$PI= \frac{1}{(D+1)}\dfrac 1 {(D+1)}\frac{e^{-x}}{e^{-x}(1- e^x)^2}$$
$$PI= \frac{1}{(D+1)}e^{-x}\dfrac 1 {D}\frac{e^{x}}{(1- e^x)^2}$$
Note that $\dfrac 1 D= \int $.
$$PI= \frac{1}{(D+1)}\frac{e^{-x}}{1- e^x}$$
$$PI=e^{-x} \frac{1}{D}\frac{1}{1- e^x}$$
You should be able to continue to integrate this Jasmine.
$$PI=e^{-x} \int \frac{dx}{1- e^x}=-e^{-x} \int \frac{dx}{e^x-1}$$
Substitute $u=e^x \implies du =udx$
