To show that y = 2e^x cos x − e^x sin x is a solution to y′′ + 2y′ + 2y = 0, we need to substitute y into the differential equation and verify that it satisfies the equation.
First, we find the first and second derivatives of y with respect to x:
y' = (2e^x cos x - e^x sin x)' = 2e^x (cos x - sin x) + 2e^x cos x - e^x sin x
y'' = (2e^x (cos x - sin x) + 2e^x cos x - e^x sin x)' = 4e^x cos x - 2e^x sin x + 2e^x (cos x - sin x) + 2e^x (-sin x) - e^x cos x
Simplifying this expression gives:
y'' + 2y' + 2y = (4e^x cos x - 2e^x sin x + 2e^x cos x - 2e^x sin x) + 2(2e^x (cos x - sin x) + 2e^x cos x - e^x sin x) + 2(2e^x cos x - e^x sin x)
= 4e^x cos x - 4e^x sin x + 4e^x cos x - 4e^x sin x + 4e^x cos x - 2e^x sin x
= 12e^x cos x - 10e^x sin x
Since y'' + 2y' + 2y = 0 if and only if 12e^x cos x - 10e^x sin x = 0, we can see that the function y = 2e^x cos x − e^x sin x satisfies the differential equation. Therefore, y = 2e^x cos x − e^x sin x is a solution to y′′ + 2y′ + 2y = 0