Find the general solution of $4y′′+ 3y′−y=e^{−x}+x$ 
Given $$4y′′+ 3y′−y = e^{−x} + x$$
Find the general solution

I identified this as a non-homogeneous linear equation
so I assumed that my solution,
$$ y = y_h  + y_p$$
$$ 4\lambda^2 + 3\lambda - 1 = 0$$
$$ \lambda = -1 \quad\text{and}\quad \lambda =1/4$$
$$ y_h  = Ae^{x/4} + Be^{-x}$$
However, I am not too sure what should I do to find $y_p$
I do know that my $G(x) = e^{-x} + x$ but am not too sure how to find $y_p$.
 A: 
I do know that my $G(x) = \color{blue}{e^{-x}} + \color{purple}{x}$ but am not too sure how to find $y_p$

Based on the form of $G(x)$ and by linearity you would propose a particular solution of the form:
$$y_p=\color{blue}{Ae^{-x}} + \color{purple}{Bx+C}$$
but because $\color{blue}{Ae^{-x}}$ is already contained in the homogeneous solution ($ y_h  = Ae^{0.25x} + \color{red}{Be^{-x}}$), you multiply with an extra factor $x$:
$$y_p=\color{blue}{A}\color{red}{x}\color{blue}{e^{-x}} + \color{purple}{Bx+C}$$
Now substitute into the differential equation to obtain a linear system in the unknowns ("undetermined coefficients") $A$, $B$ and $C$.
A: By using the method of undetermined coefficients it follows that
$y_p$ has the form
$$y_p(x)=xAe^{-x}+Bx+C$$
where $A,B,C$ are real constants to be determined by plugging $y_p$ into the ODE. 
Note that factor $x^1$ which multiplies $Ae^{-x}$ is due to the fact that $\lambda=-1$ has multiplicity $1$ in the characteristic equation (see the final remark here).
A: Let's try
$$y_p = Axe^{-x} + Bx + C $$ 
instead. So
$$y_p' = -Axe^{-x} + Ae^{-x} + B$$
$$y_p'' = Axe^{-x} -2Ae^{-x} $$
So
$$4y′′+ 3y′−y = 4Axe^{-x} -8Ae^{-x} -3Axe^{-x} +3 Ae^{-x}  -Axe^{-x} +3B - Bx - C $$
which is equivalent to 
$$4y′′+ 3y′−y= \underbrace{-5A}_{1}e^{-x}\underbrace{-B}_1x + \underbrace{(3B-C)}_0 = e^{-x} + x$$
By identification, we get
\begin{align}
A &= -\frac{1}{5} \\
B &= -1 \\
C &= -3 
\end{align}
are the corresponding values.
A: HINT
\begin{align*}
4y^{\prime\prime} + 3y^{\prime} - y & = e^{-x} + x \Longleftrightarrow 4(y^{\prime\prime} + y^{\prime}) - (y^{\prime} + y) = e^{-x} + x \Longleftrightarrow\\\\
4(y^{\prime} + y)^{\prime} - (y^{\prime} + y) & = e^{-x} + x \Longleftrightarrow 4w^{\prime} - w = e^{-x} + x\quad\text{where}\quad w = y^{\prime} + y
\end{align*}
