1. ## Difficult DiffyQ

y'' - 4y = xe^(x)

Use the method of vartiation of parameters to find the general solution of the given differential equation.

2. Originally Posted by bearej50
y'' - 4y = xe^(x)

Use the method of vartiation of parameters to find the general solution of the given differential equation.
First solve the homogeneous equation $y^{\prime\prime}-4y=0$.

The characteristic equation is $r^2-4=0\implies r=\pm 2$.

Thus, the complimentary solution is $y=c_1e^{-2x}+c_2e^{2x}$.

Now, let $y_1\!\left(x\right)=e^{-2x}$ and $y_2\!\left(x\right)=e^{2x}$.

We now need to find $u_1^{\prime}\!\left(x\right)$ and $u_2^{\prime}\!\left(x\right)$

$u_1^{\prime}\!\left(x\right)=\frac{\displaystyle\b egin{vmatrix}0 & y_2\\f\!\left(x\right)&y_2^{\prime}\end{vmatrix}}{ \displaystyle\begin{vmatrix}y_1 & y_2\\y_1^{\prime} & y_2^{\prime}\end{vmatrix}}=\frac{\displaystyle\beg in{vmatrix}0 & e^{2x}\\xe^{x} & 2e^{2x}\end{vmatrix}}{\displaystyle\begin{vmatrix} e^{-2x} & e^{2x}\\-2e^{-2x} & 2e^{2x}\end{vmatrix}}=\tfrac{1}{4}xe^{3x}$

Thus, $u_1\!\left(x\right)=\tfrac{1}{4}\int xe^{3x}\,dx$

I leave it for you to find $u_1\!\left(x\right)$ (See spoiler to check your answer).

Spoiler:
$u_1\!\left(x\right)=\tfrac{1}{36}e^{3x}\left(3x-1\right)$

$u_2^{\prime}\!\left(x\right)=\frac{\displaystyle\b egin{vmatrix}y_1 & 0\\ y_1^{\prime} & f\!\left(x\right)\end{vmatrix}}{\displaystyle\begi n{vmatrix}y_1 & y_2\\y_1^{\prime} & y_2^{\prime}\end{vmatrix}}=\frac{\displaystyle\beg in{vmatrix}e^{-2x} &0\\-2e^{-2x} & xe^{x}\end{vmatrix}}{\displaystyle\begin{vmatrix}e ^{-2x} & e^{2x}\\-2e^{-2x} & 2e^{2x}\end{vmatrix}}=\tfrac{1}{4}xe^{-x}$

Thus, $u_2\!\left(x\right)=\tfrac{1}{4}\int xe^{-x}\,dx$

I leave it for you to find $u_2\!\left(x\right)$ (See spoiler to check your answer).

Spoiler:
$u_2\!\left(x\right)=-\tfrac{1}{4}e^{-x}\left(x+1\right)$

Once you find $u_1\!\left(x\right)$ and $u_2\!\left(x\right)$, the solution to your differential equation would be $y=u_1\!\left(x\right)e^{-2x}+u_2\!\left(x\right)e^{2x}=\dots$ (See spoiler to check your answer)

Spoiler:
$y=\tfrac{1}{36}e^{x}\left(3x-1\right)-\tfrac{1}{4}e^{x}\left(x+1\right)=\boxed{-\tfrac{1}{18}e^x\left(3x+5\right)}$