Suppose x = c1e^(-t) + c2e^(3t). a. Verify that x = c1e^(-t)...

Suppose x = c1e^(-t) + c2e^(3t). a. Verify that x = c1e^(-t) + c2e^(3t) is a solution to x'' - 2x' - 3x = 0 by substituting it into the differential equation. (Enter the terms in the order given. Enter c1 as c1 and c2 as c2.) [ ] + [ ] + [ ] = 0. help (formulas) b. Find x(t) if x(0) = 4 and x'(0) = 2. x(t) = [ ] help (formulas)

Transcript

00:01 We are asked to verify that the given function satisfies the differential equation.

00:05 So we need to plug in x double prime, x prime and x in the differential equation.

00:11 But first we need to calculate x prime and x double prime.

00:23 So x prime equals negative c1 e to the negative t plus 3 c2 e to the 3t.

00:34 X' ' is the derivative of x ' and equals c1e -t plus 9c2e -3t.

00:47 Now plug in each of them in the differential equation.

00:53 X' ' gets replaced by this.

00:56 X ' gets replaced by this.

00:59 And x gets replaced by this.

01:01 And we'll get c1 e to the negative t plus 9 c2 e to the 3t this is the first term minus 2 multiplied by x prime which is negative c1 e to the negative t plus 3 c2 e to the 3t t.

01:36 This is the second term and minus 3 multiplied by x which is c1 e to the negative t plus c2 multiplied by e to the 3t.

01:55 This is the last term.

01:58 And now we just need to check that this is 0.

02:04 So we'll get c1 e to the negative t plus 9 c2 e to the 3t plus 2 c1 e to the negative t minus 6c2e3t minus 3c1e -t minus 3c2e3t then c1e -t plus 2c2c1e -t minus 3c1e -t cancel and 9c2e3t minus 6c2e3t and minus 3e3t 3c2e to the 3t cancel and we'll get 0.

02:54 So this confirms x satisfying the differential equation.

03:00 In the next part we're asked to find the particular x satisfying x of 0 equals 4 and x prime of 0 equals 2...

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