Answer:
Shown!
Step-by-step explanation:
First let's show that the given parameters are solutions of ODE.
[tex]x_1(t) = e^{-t}\cos t\\x_1'(t) = -e^{-t}\cos t-e^{-t}\sin t\\x_1''(t) = e^{-t}\cos t+e^{-t}\sin t+e^{-t}\sin t-e^{-t}\cos t=2e^{-t}\sin t\\\\x''+2x'+2x =2e^{-t}\sin t-2e^{-t}\cos t-2e^{-t}\sin t+2e^{-t}\cos t=0[/tex]
[tex][tex]x_2(t) = e^{-t}\sin t\\x_2'(t) = -e^{-t}\sin t+e^{-t}\cos t\\x_2''(t) = e^{-t}\sin t-e^{-t}\cos t-e^{-t}\cos t-e^{-t}\sin t=-2e^{-t}\cos t\\\\x''+2x'+2x =-2e^{-t}\cos t-2e^{-t}\sin t+2e^{-t}\cos t+2e^{-t}\sin t=0[/tex]
Is is showed that both are the solutions.
Now, let's prove it for general case by solving ODE.
[tex]x''+ 2x'+ 2x = 0\\r^2+2r+2=0\\[/tex]
Roots are [tex]r_1 = -1-i,\:r_2=-1+i[/tex]
So the solution is as follows:
[tex]x(t)=C_1e^{(-1-i)t}+C_2e^{(-1+i)t}[/tex]
Since by Euler's Formula,
[tex]e^{-it}=\cos t - i\sin t\\e^{it}=\cos t+i\sin t[/tex]
Hence,
[tex]x(t) = Ae^{-t}\sin t+ Be^{-t}\cos t[/tex]
