Question

Solve the differential equation

`y {}^((5)) -4y {}^((4)) +4y'''-y''+4y'-4y=69`
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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) = -\frac{1+i\sqrt3}2 \text{ or } r = 1e^(i4\pi/3) = -\frac{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(\frac{\sqrt3}2x\right) \pm i \sin\left(\frac{\sqrt3}2x\right)\right)`

and thus simplify
`y_c` to

`y_c = C_1 e^x + C_2 e^(-x/2) \cos\left(\frac{\sqrt3}2x\right) + C_3 e^(-x/2) \sin\left(\frac{\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 = -\frac{69}4`

and so the general solution to the ODE is

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