Question

Find the general solution of the given nonhomogeneous differential equation:
y''+2y'+y=e^(-x)*ln(x)

Answer 1

Homogeneous part:
\[y''+2y'+y=0~~\Rightarrow~~r^2+2r+1=0~~\Rightarrow~~r=-1\]
So the homogeneous set of solution is
\[y_c=C_1e^{-x}+C_2xe^{-x}\]
For the nonhomogeneous part, use variation of parameters. You know two sample solutions, \(e^{-x}\) and \(xe^{-x}\).

Answer 2

what is Yp?

not sure how to replace ln(x) ?

Answer 3

When you're using variation of parameters, you have \(y_{p_1}=e^{-x}\) and \(y_{p_2}=xe^{-x}\).

First, check to make sure they're linearly independent:
\[W(e^{-x},xe^{-x})=\begin{vmatrix}e^{-x}&xe^{-x}\\-e^{-x}&e^{-x}(1-x)\end{vmatrix}=e^{-2x}(1-x)+xe^{-2x}=e^{-2x}\not=0\]
So yes, they're linearly independent.