Question

Solve the differential equation0%2B4y%27%27%27-y%27%27%2B4y%27-4y%3D69" id="TexFormula1" title="y {}^{(5)} -4y {}^{(4)} +4y'''-y''+4y'-4y=69" alt="y {}^{(5)} -4y {}^{(4)} +4y'''-y''+4y'-4y=69" align="absmiddle" class="latex-formula">
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Answer 1

The given differential equation has characteristic equation

$r^5 - 4r^4 + 4r^3 - r^2 + 4r - 4 = 0$

Solve for the roots $r$.

$r^3 (r^2 - 4r + 4) - (r^2 - 4r + 4) = 0$

$(r^3 - 1) (r^2 - 4r + 4) = 0$

$(r^3 - 1) (r - 2)^2 = 0$

$r^3 - 1 = 0 \text{ or } (r-2)^2=0$

The first case has the three cubic roots of 1 as its roots,

$r^3 = 1 = 1e^{i0} \implies r = 1^{1/3} e^{i(0+2\pi k)/3} \text{ for } k\in\{0,1,2\} \\\\ \implies r = 1e^{i0} = 1 \text{ or } r = 1e^{i2\pi/3} = -\dfrac{1+i\sqrt3}2 \text{ or } r = 1e^{i4\pi/3} = -\dfrac{1-i\sqrt3}2$

while the other case has a repeated root of

$(r-2)^2 = 0 \implies r = 2$

Hence the characteristic solution to the ODE is

$y_c = C_1 e^x + C_2 e^{-(1+i\sqrt3)/2\,x} + C_3 e^{-(1-i\sqrt3)/2\,x} + C_4e^{2x} + C_5xe^{2x}$

Using Euler's identity

$e^{ix} = \cos(x) + i \sin(x)$

we can reduce the complex exponential terms to

$e^{-(1\pm i\sqrt3)/2\,x} = e^{-x/2} \left(\cos\left(\dfrac{\sqrt3}2x\right) \pm i \sin\left(\dfrac{\sqrt3}2x\right)\right)$

and thus simplify $y_c$ to

$y_c = C_1 e^x + C_2 e^{-x/2} \cos\left(\dfrac{\sqrt3}2x\right) + C_3 e^{-x/2} \sin\left(\dfrac{\sqrt3}2x\right) \\ ~~~~~~~~ + C_3 e^{-(1-i\sqrt3)/2\,x} + C_4e^{2x} + C_5xe^{2x}$

For the non-homogeneous ODE, consider the constant particular solution

$y_p = A$

whose derivatives all vanish. Substituting this into the ODE gives

$-4A = 69 \implies A = -\dfrac{69}4$

and so the general solution to the ODE is

$y = -\dfrac{69}4 + C_1 e^x + C_2 e^{-x/2} \cos\left(\dfrac{\sqrt3}2x\right) + C_3 e^{-x/2} \sin\left(\dfrac{\sqrt3}2x\right) \\ ~~~~~~~~ + C_3 e^{-(1-i\sqrt3)/2\,x} + C_4e^{2x} + C_5xe^{2x}$