Answer:
[tex]y" + 2y' + 2y = 0[/tex]
Step-by-step explanation:
Given
[tex]y=c_1e^{-x}cosx+c_2e^{-x}sinx[/tex]
Required
Determine a homogeneous linear differential equation
Rewrite the expression as:
[tex]y=c_1e^{\alpha x}cos(\beta x)+c_2e^{\alpha x}sin(\beta x)[/tex]
Where
[tex]\alpha = -1[/tex] and [tex]\beta = 1[/tex]
For a homogeneous linear differential equation, the repeated value m is given as:
[tex]m = \alpha \± \beta i[/tex]
Substitute values for [tex]\alpha[/tex] and [tex]\beta[/tex]
[tex]m = -1 \± 1*i[/tex]
[tex]m = -1 \± i[/tex]
Add 1 to both sides
[tex]m +1= 1 -1 \± i[/tex]
[tex]m +1= \± i[/tex]
Square both sides
[tex](m +1)^2= (\± i)^2[/tex]
[tex]m^2 + m + m + 1 = i^2[/tex]
[tex]m^2 + 2m + 1 = i^2[/tex]
In complex numbers:
[tex]i^2 = -1[/tex]
So, the expression becomes:
[tex]m^2 + 2m + 1 = -1[/tex]
Add 1 to both sides
[tex]m^2 + 2m + 1 +1= -1+1[/tex]
[tex]m^2 + 2m + 2= 0[/tex]
This corresponds to the homogeneous linear differential equation
[tex]y" + 2y' + 2y = 0[/tex]
