Setting up the particular solution to $y''+4y'=e^{-4x}+3$ $y''+4y'=e^{-4x}+3$
The characteristic polynomial is $r^2+4r=0 \to r = 0,-4$
the complementary solution is: $y_c = C_1+C_2e^{-4x}$
The part that is throwing me off is the addition of 3. I can't even set that up as a polynomial?
 A: In examples like this I suggest that you find two separate particular solutions corresponding to each of the "different" parts of the right-hand side, and then you can combine them thanks to the principle of superposition.


*

*For $y''+4y'=e^{-4x}$, a particular solution can be set up as $y_{p1}=\cdots$.

*For $y''+4y'=3$, a particular solution can be set up as $y_{p2}=\cdots$.

*And then the general solution will be $y=y_c+y_{p1}+y_{p2}$.


If your question is not about the general strategy, but about that term of "$3$" in particular, i.e. about what I called $y_{p2}$ above, then yes — it is a polynomial of degree zero. Since $r=0$ is a root of the characteristic equation, this particular solution should be sought in the form $y_{p2}=Ax$, where "$A$" is a generic polynomial of degree zero, and an extra $x$ is there because $r=0$ is a root of multiplicity one.
A: In this particular case, and similar cases, there is a second derivative with constant coefficients equaling a constant. This suggests a linear polynomial of the order of the highest derivative. This leads to the form (the exponential term being equal to a known solution so other rules apply):
$$f(x) = a_{0} + a_{1} \, x + a_{2} \, x^2 + b \, x \, e^{-4 x}.$$ 
This solution is to satisfy
$$f'' + 4 \, f' = e^{-4 x} + 3.$$
Now:
\begin{align}
3 + e^{-4 x} &= \frac{d}{dx}(a_{1} + 2 a_{2} x + b (-4 x + 1) \, e^{-4 x}) +  4 \, (a_{1} + 2 a_{2} x + b (-4 x + 1) \, e^{-4 x}) \\
&= 2 a_{2} + b (16 x - 8) \, e^{-4 x} + 4 a_{1} + 8 a_{2} x + b (-16 x + 4) \, e^{-4 x} \\
&= 4 a_{1} + 2 a_{2} + 8 a_{2} \, x - 4 b e^{-4 x}
\end{align}
leading to $4 a_{1} = 3$, $a_{2} = 0$, $-4 b = 1$, and 
$$f(x) = a_{0} + \frac{3 \, x}{4} - \frac{x}{4} \, e^{-4 x}.$$
Since the general solution also contains a constant the $a_{0}$ constant in the particular solution can be removed to present the total solution of
$$y(x) = c_{0} + \frac{3 \, x}{4} + \left( c_{1} - \frac{x}{4} \right) \, e^{-4 x}.$$
A second method
By integration one can show that:
$$y' + 4 y = - \frac{1}{4} \, e^{-4 x} + 3 \, x + b_{1}.$$
Now this has equation has the form
\begin{align}
e^{- 4 x} \, \frac{d}{dx} (e^{4 x} \, y) &= - \frac{1}{4} \, e^{-4 x} + 3 \, x + b_{1} \\
\frac{d}{dx} (e^{4 x} \, y) &= - \frac{1}{4} + 3 \, x \, e^{4 x} + b_{1} \, e^{4 x} \\
e^{4 x} \, y &= - \frac{x}{4} + 3 \, \left( \frac{x}{4} - \frac{1}{16} \right) \, e^{4 x} + \frac{b_{1}}{4} \, e^{4 x} + b_{0} \\
y(x) &= \frac{4 b_{1} - 3}{16} + \frac{3 x}{4} + \left(b_{0} - \frac{x}{4}\right) \, e^{- 4 x}.
\end{align} 
This solution is in agreement with the first solution.
A: Here is a direct way to solve that ODE : \begin{aligned} y''+4y'&=\mathrm{e}^{-4x}+3\\ \iff \ \ \ \ \ \ \ \ \ y'+4y&=-\frac{1}{4}\,\mathrm{e}^{-4x}+3x+c_{1}\\ \iff \left(y'+4y\right)\mathrm{e}^{4x}&=-\frac{1}{4}+3x\,\mathrm{e}^{4x}+c_{1}\,\mathrm{e}^{4x}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ y\,\mathrm{e}^{4x}&=-\frac{x}{4}+3\int{x\,\mathrm{e}^{4x}\,\mathrm{d}x}+c_{2}\,\mathrm{e}^{4x}\\ y\,\mathrm{e}^{4x}&=-\frac{x}{4}+\frac{3}{16}\,\mathrm{e}^{4x}\left(4x-1\right)+c_{3}+c_{2}\,\mathrm{e}^{4x}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y&=-\frac{x\,\mathrm{e}^{-4x}}{4}+\frac{3x}{4}+c_{3}\,\mathrm{e}^{-4x}+c_{4}\end{aligned}
A: $y''+4y'=e^{-4x}+3~~~(1)$
Let $y''_1+4y'_1=3~~~~(2)$ and $y''_2+4y'_2=e^{-4x}~~~~(3)$
Adding (2) and (3), we get
$$(y_1+y_2)''+4(y_1+y_2)'=e^{4x}+3.$$
So the solution of (1) is $y_1+y_2$.
Let $y_1=Ax^2+Bx+C$, then
$$(2) \implies 2A+4(2Ax+B)=3 \implies A=0, B=3/4 \implies y_1=3x/4+C$$
$$(3) \implies  e^{4x}y''_2+4e^{4x}y'_2=1$$
$$ \implies \frac{d(e^{4x}y'_2)}{dx}=1 \implies \int d(y'_2e^{4x})=\int dx+C= \implies y'_2=xe^{-4x}+C_1 e^{-4x}\implies y_2=\int[xe^{-4x}+C_1e^{-4x}]dx+C_2$$
$$\implies y_2(x)=-e^{4x}(1/16+x/4)-C_1 e^{-4x}/4+C_2$$
Finally, the total solution of (1) is
$$y(x)=3x/4+C-e^{-4x}(1/16+x/4)-C_1 e^{-4x}/4+C_2$$
$$\implies 3x/4-e^{-4x}(1/16+x/4)-C_1 e^{-4x}/4+C'_2$$
