Question

4. Show that both x1(t) = e−tcos tandx2(t) = e−tsin t are solutions to the second-order differential equation x′′+ 2x′+ 2x = 0. Show that x(t) = Ae−tsin t + Be−tcos t is a solution for any values of the constants A and B

Answer 1

Shown!

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`

`[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`

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\\`

Roots are
`r_1 = -1-i,\:r_2=-1+i`

So the solution is as follows:

`x(t)=C_1e^((-1-i)t)+C_2e^((-1+i)t)`

Since by Euler's Formula,

`e^(-it)=\cos t - i\sin t\\e^(it)=\cos t+i\sin t`

Hence,

`x(t) = Ae^(-t)\sin t+ Be^(-t)\cos t`