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(y_2e^(-x)\log x)/(W(y_1,y_2))\,\mathrm dx`

`u_2=\displaystyle\int(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(xe^(-x)e^(-x)\log x)/(e^(-2x))\,\mathrm dx`

`u_1=-\displaystyle\int x\log x\,\mathrm dx`

`u_1=\frac14x^2-\frac12x^2\log x`

and


`u_2=\displaystyle\int(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(\frac14x^2-\frac12x^2\log x\right)e^(-x)+(\log x-1)x^2e^(-x)`

`y=(C_1+C_2x)e^(-x)+\frac14x^2e^(-x)(2\log x-3)`