The given differential equation has characteristic equation
[tex]r^5 - 4r^4 + 4r^3 - r^2 + 4r - 4 = 0[/tex]
Solve for the roots [tex]r[/tex].
[tex]r^3 (r^2 - 4r + 4) - (r^2 - 4r + 4) = 0[/tex]
[tex](r^3 - 1) (r^2 - 4r + 4) = 0[/tex]
[tex](r^3 - 1) (r - 2)^2 = 0[/tex]
[tex]r^3 - 1 = 0 \text{ or } (r-2)^2=0[/tex]
The first case has the three cubic roots of 1 as its roots,
[tex]r^3 = 1 = 1e^{i0} \implies r = 1^{1/3} e^{i(0+2\pi k)/3} \text{ for } k\in\{0,1,2\} \\\\ \implies r = 1e^{i0} = 1 \text{ or } r = 1e^{i2\pi/3} = -\dfrac{1+i\sqrt3}2 \text{ or } r = 1e^{i4\pi/3} = -\dfrac{1-i\sqrt3}2[/tex]
while the other case has a repeated root of
[tex](r-2)^2 = 0 \implies r = 2[/tex]
Hence the characteristic solution to the ODE is
[tex]y_c = C_1 e^x + C_2 e^{-(1+i\sqrt3)/2\,x} + C_3 e^{-(1-i\sqrt3)/2\,x} + C_4e^{2x} + C_5xe^{2x}[/tex]
Using Euler's identity
[tex]e^{ix} = \cos(x) + i \sin(x)[/tex]
we can reduce the complex exponential terms to
[tex]e^{-(1\pm i\sqrt3)/2\,x} = e^{-x/2} \left(\cos\left(\dfrac{\sqrt3}2x\right) \pm i \sin\left(\dfrac{\sqrt3}2x\right)\right)[/tex]
and thus simplify [tex]y_c[/tex] to
[tex]y_c = C_1 e^x + C_2 e^{-x/2} \cos\left(\dfrac{\sqrt3}2x\right) + C_3 e^{-x/2} \sin\left(\dfrac{\sqrt3}2x\right) \\ ~~~~~~~~ + C_3 e^{-(1-i\sqrt3)/2\,x} + C_4e^{2x} + C_5xe^{2x}[/tex]
For the non-homogeneous ODE, consider the constant particular solution
[tex]y_p = A[/tex]
whose derivatives all vanish. Substituting this into the ODE gives
[tex]-4A = 69 \implies A = -\dfrac{69}4[/tex]
and so the general solution to the ODE is
[tex]y = -\dfrac{69}4 + C_1 e^x + C_2 e^{-x/2} \cos\left(\dfrac{\sqrt3}2x\right) + C_3 e^{-x/2} \sin\left(\dfrac{\sqrt3}2x\right) \\ ~~~~~~~~ + C_3 e^{-(1-i\sqrt3)/2\,x} + C_4e^{2x} + C_5xe^{2x}[/tex]
