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

Shown!

Step-by-step explanation:

First let's show that the given parameters are solutions of ODE.

Is is showed that both are the solutions.

Now, let's prove it for general case by solving ODE.

Roots are

So the solution is as follows:

Since by Euler's Formula,

Hence,

the answer is c

step-by-step explanation:

1. d.90

2. e. none of these (80)

3. a. 0

4. b. 6

5. e. none of these (10)

6. a. 5