Answer:
[tex] $y(t) = C_1 e^{t} + C_2 e^{-t} + D_1 \cos t + D_2 \sin t }$ [/tex]
Step-by-step explanation:
The equation is a linear differential equation: y⁽⁴⁾- y = 0
We assume the form of the solution y(t) is [tex]$y(t)=C_{1} e^{\alpha_{1} t} + C_{2} e^{\alpha_{2} t} + C_{3} e^{\alpha_{3} t} + C_{4} e^{\alpha_{4} t} $[/tex]
where [tex]$\alpha_{i}[/tex] are the roots of the auxiliary equation.
So, use the auxiliary equation: [tex]$\alpha^4 + 0 \alpha^3 + 0 \alpha^2 + 0 \alpha -1 =0$[/tex] to find the roots; the values are : α₁ = 1, α₂ = -1, α₃ = i, α₄ = -i
Then inserting [tex]$\alpha_{i}[/tex] values in the assumed solution
⇒ [tex]$y(t)=C_1 e^{t} + C_2 e^{-t} + C_{3} e^{it} + C_{4} e^{-it} $[/tex]
Also, because the last 2 terms have complex power, the solution can be written with cosine and sine terms:
Using the Euler's formula: [tex] e^{ \pm i\theta } = \cos \theta \pm i\sin \theta[/tex], we can rewrite the solution as:
[tex]$y(t) = C_{1} e^{t} + C_{2} e^{-t} + C_{3} e^{i t} + C_{4} e^{-i t}[/tex] = [tex]C_{1} e^{t} + C_{2} e^{-t} + C_{3} ( \cos t + i \sin t ) + C_{4} ( \cos t - i \sin t ) = C_{1} e^{t} + C_{2} e^{-t} + \cos t ( C_{3} + C_{4} ) + \sin t (i C_{3} - i C_{4} ) = C_{1} e^{t} + C_{2} e^{-t} + D_{1} \cos t +D_{2} \sin t$ [/tex]
Where: [tex] $D_1 = C_3 + C_4$ and $D_2= i ( C_3 - C_4 )$ [/tex]
Finally the solution for de linear differential equation y^(4) - y =0 is:
[tex] $y(t) = C_1 e^{t} + C_2 e^{-t} + D_1 \cos t + D_2 \sin t }$ [/tex]
