Question

Solve : $(x^2{D}^2+3xD+1)y=\frac{1}{(1-x{)}^2}$

Answer 1

Given,

$(x^2{D}^2+3xD+1)y=\frac{1}{(1-x{)}^2}$......(1)

Let, $x=e^z,z=\ln x,x>0,\frac{d}{dz}\equiv \theta$

$\Rightarrow xD\equiv \theta ,x^2{D}^2\equiv \theta (\theta -1)$

substituting these in equation (1)

$\left(\theta \right(\theta -1)+3\theta +1)y=\frac{1}{(1-e^z{)}^2}$

$\Rightarrow ({\theta }^2+2\theta +1)y=\frac{1}{(1-e^z{)}^2}$....(2)

The general solution is given by, $y=y_c+y_p$

$y_c=c_1{e}^{-z}+c_{2}z{e}^{-z}$

$y_p=\frac{1}{{\theta }^2+2\theta +1}\frac{1}{(1-e^z{)}^2}$

$=\frac{1}{\theta +1}\left\{\frac{1}{\theta +1}\frac{1}{(1-e^z{)}^2}\right\}$

$=\frac{1}{\theta +1}\{e^{-z}\int \frac{1}{(1-e^z{)}^2}{e}^{z}dz\}$

$=\frac{1}{\theta +1}\left\{\frac{e^{-z}}{1-e^z}\right\}$

$=e^{-z}\int \frac{e^{-z}}{1-e^z}{e}^{z}dz$

$=-e^{-z}\ln |e^{-z}-1|$

$y_p=-e^{-z}\ln |e^{-z}-1|$

$\therefore y=c_1{e}^{-z}+c_{2}z{e}^{-z}-e^{-z}\ln |e^{-z}-1|$

$\Rightarrow y=c_1{x}^{-1}+c_2\left(\ln x\right)x^{-1}-x^{-1}\ln |x^{-1}-1|$