Question

Observe that x and e^x are solutions to the homogeneous equation associated with:

Answer 1

To take advantage of the characteristic solutions $y_1(x)=x$ and $y_2(x)=e^x$, you can try the method of variation of parameters, where we look for a solution of the form

$y=y_1u_1+y_2u_2$

with the condition that

${u_1}'y_1+{u_2}'y_2=0$

$\implies{u_1}'x+{u_2}'e^x=0$ $(\mathbf 1)$

Then

$y'={y_1}'u_1+y_1{u_1}'+{y_2}'u_2+y_2{u_2}'$

$\implies y'={y_1}'u_1+{y_2}'u_2$

$y''={y_1}''u_1+{y_1}'{u_1}'+{y_2}''u_2+{y_2}'{u_2}'$

Substituting into the ODE gives

$(1-x)({y_1}''u_1+{y_1}'{u_1}'+{y_2}''u_2+{y_2}'{u_2}')+x({y_1}'u_1+{y_2}'u_2)-y_1u_1+y_2u_2=2(x-1)^2e^{-x}$

Since

$y_1=x\implies{y_1}'=1\implies{y_1}''=0$

$y_2=e^x\implies{y_2}'=e^x\implies{y_2}''=e^x$

the above reduces to

$(1-x)({u_1}'+e^x{u_2}')=2(x-1)^2e^{-x}$

${u_1}'+e^x{u_2}'=2(1-x)e^{-x}$ $(\mathbf 2)$

$(\mathbf 1)$ and $(\mathbf 2)$ form a linear system that we can solve for ${u_1}',{u_2}'$ using Cramer's rule:

${u_1}'=\dfrac{W_1(x)}{W(x)},{u_2}'=\dfrac{W_2(x)}{W(x)}$

where $W(x)$ is the Wronskian determinant of the fundamental system and $W_i(x)$ is the same determinant, but with the $i$-th column replaced with $(0,2(x-1)^2e^{-x})$.

$W(x)=\begin{vmatrix}x&e^x\\1&e^x\end{vmatrix}=e^x(x-1)$

$W_1(x)=\begin{vmatrix}0&e^x\\2(x-1)^2e^{-x}&e^x\end{vmatrix}=-2(x-1)^2$

$W_2(x)=\begin{vmatrix}x&0\\e^x&2(x-1)^2e^{-x}\end{vmatrix}=2xe^{-x}(x-1)^2$

So we have

${u_1}'=\dfrac{-2(x-1)^2}{e^x(x-1)}\implies u_1=2xe^{-x}$

${u_2}'=\dfrac{2xe^{-x}(x-1)^2}{e^x(x-1)}\implies u_2=-x^2e^{-2x}$

Then the particular solution is

$y_p=2x^2e^{-x}-x^2e^{-x}=x^2e^{-x}$

giving the general solution to the ODE,

$\boxed{y(x)=C_1x+C_2e^x+x^2e^{-x}}$