Solving an ODE using variations of parameters and Wronskian theorem.

Question

I am attempting to solve the following differential equation by trying to follow an example that my professor did in class.

$$y''+4y'+4y=\frac{e^{-2x}}{x^3}$$

$$\begin{align} y&=y_h+y_p\\ y_h&=Ae^{-2x}+Bxe^{-2x}\\ y_p&=Ae^{-x}+2e^{-x}\\ A&=2 \end{align}$$

Using variations of parameters:

$$y_1=e^{-2x},y_2=xe^{-2x}$$

$$W(y_1,y_2)=\begin{vmatrix}e^{-2x}&xe^{-2x}\\-2e^{-2x}&e^{-2x}-2xe^{-2x}\end{vmatrix}=e^{-4x}-2xe^{-4x}-\left(-2xe^{-4x}\right)=e^{-4x}$$

$$\begin{align}y_p&=-y_1\int\frac{y_2\,r\,\mathrm dx}\omega+y_2\int\frac{y_1\,r\,\mathrm dx}\omega\\ &=-e^{-2x}\int\frac{xe^{-2x}\cdot2e^{-2x}\,\mathrm dx}{e^{-4x}}+\sin x\int\frac{e^{-2x}\cdot2e^{-x}\,\mathrm dx}{e^{-4x}}\\ &=-2e^{-2x}\int2xe^x\,\mathrm dx+xe^{-2x}\int e^x\,\mathrm dx\\ &=-2e^{-2x}\left[xe^x-e^x-xe^x\right]=2e^{-x} \end{align}$$

I am just not too sure about my answer seeing as WolframAlpha gives me this: $$y(x) = c_1 e^{-2 x}+c_2 e^{-2 x} x+\frac{e^{-2 x}}{2 x}$$

I am lead to believe that my $y_h$ and $y_p$ values are incorrect, thus throwing everything else off. Hopefully someone can clear up my confusion for me.

Answer 1

Your homogeneous solution $y_h(x) = Ae^{-2x}+Bxe^{-2x}$ is correct since it does satisfy the homogeneous linear equation $y'' + 4y' + 4y = 0$. However, your particular solution $y_p(x) = Ae^{-x}+2e^{-x}$ does not satisfy $y'' + 4y' + 4y = \dfrac{e^{-2x}}{x^3}$. You need to find a correct particular solution first.