I have a Non-Homogeneous ODE of the form $y'' + 8y' + 17y = 5$ which I have obtained the solution $y(x) = ae^{-4x}[\cos(x) + i\sin(x)] + be^{-4x}[\cos(x)-i\sin(x)] + \frac{5}{17}$ where $a$ and $b$ are complex constants.
I am unsure of how to find a constant, $c$, which satisfies the equation $z(x) = y(x) - c$ where $c$ is a real constant and $z(x)$ is a Homogeneous Differential Equation. I originally thought that $c = \frac{5}{17}$ but that seems too simple.
How should I go about finding $c$?
On
You are correct that $c=\frac{5}{17}$, which gives the particular solution to be $y_{p}(x)=\frac{5}{17}$.
Since the RHS is a constant, you would try a constant as your particular solution $y_{p}(x)=c$ and substituting gives $c=\frac{5}{17}$. See here.
Note your solution can be simplified further, since $a$ and $b$ are arbitrary constants we can write $$y(x) = c_{1}e^{-4x}\cos(x)+c_2e^{-4x}\sin(x) + \frac{5}{17}$$
where $c_1=a+b$ and $c_2=i(a-b)$.
$$y'' + 8y' + 17y = 5$$ $$y'' + 8y' + 17y - 5=0$$ $$y'' + 8y' + 17(y-\dfrac 5 {17}) = 0$$ Since $(y-\dfrac 5{17})'=y'$: $$(y-\dfrac 5 {17})'' + 8(y-\dfrac 5 {17})' + 17(y-\dfrac 5 {17}) = 0$$ $$z'' + 8z' + 17z = 0$$ Where $z=y-\dfrac 5 {17}$.