Question

Verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval I of definition for each solution. y'' − 4y' + 13y = 0; y = e2x cos 3x When y = e2x cos 3x, y=?

Answer 1

To verify that the given function y = e2x cos 3x is an explicit solution of the differential equation y'' − 4y' + 13y = 0, we can substitute the function into the equation and check if it holds true.

Step-by-step explanation:

To verify that the given function y = e2x cos 3x is an explicit solution of the differential equation y'' − 4y' + 13y = 0, we need to substitute this function into the differential equation and check if it holds true.

First, let's find the first and second derivatives of y:

1. y' = (d/dx)(e2x cos 3x) = 2e2x cos 3x - 3e2x sin 3x
2. y'' = (d^2/dx^2)(e2x cos 3x) = 4e2x cos 3x - 12e2x sin 3x - 9e2x cos 3x - 6e2x sin 3x

Now, substitute these derivatives into the differential equation:

4e2x cos 3x - 12e2x sin 3x - 9e2x cos 3x - 6e2x sin 3x - 4(2e2x cos 3x - 3e2x sin 3x) + 13e2x cos 3x = 0

Simplifying, we get:

13e2x cos 3x - 13e2x cos 3x = 0

Since the equation holds true, we can conclude that y = e2x cos 3x is an explicit solution of the given differential equation.

Answer 2

y = e^(2x) cos 3x is a explicit solution of y'' − 4y' + 13y = 0

Step-by-step explanation:

y'' − 4y' + 13y = 0 (1)

Verify that y = e^(2x) cos 3x is a explicit solution. We derive this solution:

y'=2e^(2x) cos 3x-3e^(2x) sin 3x (2)

y''=4e^(2x) cos 3x - 6e^(2x) sin 3x -6e^(2x) sin 3x -9e^(2x) cos 3x

y''=-5e^(2x) cos 3x - 12e^(2x) sin 3x (3)

In order to verify, we replace (2) (3) in (1):

`-5e^(2x) cos 3x - 12e^(2x) sin 3x -4[2e^(2x) cos 3x-3e^(2x) sin 3x ]+13{e^}2x} cos 3x=(-5-8+13)e^(2x) cos 3x +(-12+12)e^(2x) sin 3x=0`