Question

(D^2 +2D +1)y=e^-x log(x) solve using method variation of parameter?

Answer 1

First you'll need the complementary solutions. The characteristic equation for this ODE is

$D^2+2D+1=(D+1)^2=0\implies D=-1$

with multiplicity 2. This means two linearly independent solutions will be $y_1=e^{-x}$ and $y_2=xe^{-x}$.

Via variation of parameters, the particular solution will take the form

$y_p=y_1u_1+y_2u_2$

where

$u_1=-\displaystyle\int\frac{y_2e^{-x}\log x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\int\frac{y_1e^{-x}\log x}{W(y_1,y_2)}\,\mathrm dx$

The Wronskian is

$W(y_1,y_2)=\begin{vmatrix}e^{-x}&xe^{-x}\\-e^{-x}&e^{-x}(1-x)\end{vmatrix}=e^{-2x}(1-x)+xe^{-2x}=e^{-2x}$

So, you have

$u_1=-\displaystyle\int\frac{xe^{-x}e^{-x}\log x}{e^{-2x}}\,\mathrm dx$
$u_1=-\displaystyle\int x\log x\,\mathrm dx$
$u_1=\dfrac14x^2-\dfrac12x^2\log x$

and

$u_2=\displaystyle\int\frac{e^{-x}e^{-x}\log x}{e^{-2x}}\,\mathrm dx$
$u_2=\displaystyle\int\log x\,\mathrm dx$
$u_2=x(\log x-1)$

So the solution to the ODE is

$y=C_1y_1+C_2y_2+y_1u_1+y_2u_2$
$y=(C_1+C_2x)e^{-x}+\left(\dfrac14x^2-\dfrac12x^2\log x\right)e^{-x}+(\log x-1)x^2e^{-x}$
$y=(C_1+C_2x)e^{-x}+\dfrac14x^2e^{-x}(2\log x-3)$