nonhomogenous reduction of order

Question

I have a problem figuring out how to solve this problem. It is $$([D^2]+[2D]+[1])y=([e^x-1])^{-2}.$$
I am more worried in solving for V. I used $y=v[e^{-x}]$. I got $$v''[e^{-x}]=([e^x]-1)^{-2}.$$ Now I have to do a double integral. On the first integral I will get a natural log of $-\ln[(e^x)-1]$, but how do I get the second integral? The answer should be $$[e^{-x}](c_1 + c_2 \, x \cdot -\ln([e^{-x}]-1).$$ I am not sure how. My professor told me that I already integrated twice, but not sure how.

Answer 1

You properly arrived to $$v''=\frac{e^x}{\left(e^x-1\right)^2}$$ Integrating once gives $$v'=-\frac{1}{e^x-1}+c_1$$ So $$v=c_2+c_1x-\int\frac{dx}{e^x-1}$$ For the last integral, change variable $$x^x=t \implies x=\log(t)\implies dx=\frac {dt}t$$ which makes $$J=\int\frac{dx}{e^x-1}=\int\frac{dt}{(t-1) t}=\int\left(\frac{1}{t-1}-\frac{1}{t}\right)\,dt=\log(t-1)-\log(t)$$ Back to $x$ $$J=\log(e^x-1)-\log(e^x)=\log(e^x-1)-x$$ All of this makes $$v=c_2+c_1x-\log(e^x-1)+x=c_2+c_3x-\log(e^x-1)$$ If you prefer to see $e^{-x}$, just rewrite $$e^x-1=e^x(1-e^{-x})\implies \log(e^x-1)=x+\log(1-e^{-x})$$