Answer:
Shown!
Step-by-step explanation:
First let's show that the given parameters are solutions of ODE.
![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](https://tex.z-dn.net/?f=x_1%28t%29%20%3D%20e%5E%7B-t%7D%5Ccos%20t%5C%5Cx_1%27%28t%29%20%3D%20-e%5E%7B-t%7D%5Ccos%20t-e%5E%7B-t%7D%5Csin%20t%5C%5Cx_1%27%27%28t%29%20%3D%20e%5E%7B-t%7D%5Ccos%20t%2Be%5E%7B-t%7D%5Csin%20t%2Be%5E%7B-t%7D%5Csin%20t-e%5E%7B-t%7D%5Ccos%20t%3D2e%5E%7B-t%7D%5Csin%20t%5C%5C%5C%5Cx%27%27%2B2x%27%2B2x%20%3D2e%5E%7B-t%7D%5Csin%20t-2e%5E%7B-t%7D%5Ccos%20t-2e%5E%7B-t%7D%5Csin%20t%2B2e%5E%7B-t%7D%5Ccos%20t%3D0)
![[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](https://tex.z-dn.net/?f=%5Btex%5Dx_2%28t%29%20%3D%20e%5E%7B-t%7D%5Csin%20t%5C%5Cx_2%27%28t%29%20%3D%20-e%5E%7B-t%7D%5Csin%20t%2Be%5E%7B-t%7D%5Ccos%20t%5C%5Cx_2%27%27%28t%29%20%3D%20e%5E%7B-t%7D%5Csin%20t-e%5E%7B-t%7D%5Ccos%20t-e%5E%7B-t%7D%5Ccos%20t-e%5E%7B-t%7D%5Csin%20t%3D-2e%5E%7B-t%7D%5Ccos%20t%5C%5C%5C%5Cx%27%27%2B2x%27%2B2x%20%3D-2e%5E%7B-t%7D%5Ccos%20t-2e%5E%7B-t%7D%5Csin%20t%2B2e%5E%7B-t%7D%5Ccos%20t%2B2e%5E%7B-t%7D%5Csin%20t%3D0)
Is is showed that both are the solutions.
Now, let's prove it for general case by solving ODE.
![x''+ 2x'+ 2x = 0\\r^2+2r+2=0\\](https://tex.z-dn.net/?f=x%27%27%2B%202x%27%2B%202x%20%3D%200%5C%5Cr%5E2%2B2r%2B2%3D0%5C%5C)
Roots are ![r_1 = -1-i,\:r_2=-1+i](https://tex.z-dn.net/?f=r_1%20%3D%20-1-i%2C%5C%3Ar_2%3D-1%2Bi)
So the solution is as follows:
![x(t)=C_1e^{(-1-i)t}+C_2e^{(-1+i)t}](https://tex.z-dn.net/?f=x%28t%29%3DC_1e%5E%7B%28-1-i%29t%7D%2BC_2e%5E%7B%28-1%2Bi%29t%7D)
Since by Euler's Formula,
![e^{-it}=\cos t - i\sin t\\e^{it}=\cos t+i\sin t](https://tex.z-dn.net/?f=e%5E%7B-it%7D%3D%5Ccos%20t%20-%20i%5Csin%20t%5C%5Ce%5E%7Bit%7D%3D%5Ccos%20t%2Bi%5Csin%20t)
Hence,
![x(t) = Ae^{-t}\sin t+ Be^{-t}\cos t](https://tex.z-dn.net/?f=x%28t%29%20%3D%20Ae%5E%7B-t%7D%5Csin%20t%2B%20Be%5E%7B-t%7D%5Ccos%20t)
Answer:
![\huge \boxed{x=-11}](https://tex.z-dn.net/?f=%5Chuge%20%5Cboxed%7Bx%3D-11%7D)
Step-by-step explanation:
First, divide by 2 from both sides of equation.
![\displaystyle \frac{2(x+6)}{2}=\frac{-10}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B2%28x%2B6%29%7D%7B2%7D%3D%5Cfrac%7B-10%7D%7B2%7D)
Then, solve.
![\displaystyle x+6=-5](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%2B6%3D-5)
Subtract by 6 from both sides of equation.
![\displaystyle x+6-6=-5-6](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%2B6-6%3D-5-6)
Solve to find the answer.
, which is our answer.
I have to write 20 words or move but the answer is 0