Find A, B so that y = A sin(2) + B cos(2) is a solution to the differential equation 4y'' +y' + 5y = 2 cos(x).
A=
BE
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If y = A sin(x) + B cos(x), then

y' = A cos(x) - B sin(x)

y'' = -A sin(x) - B cos(x)

Substitute these into the ODE:

4y'' + y' + 5y = 2 cos(x)

4 (-A sin(x) - B cos(x)) + (A cos(x) - B sin(x)) + 5(A sin(x) + B cos(x)) = 2 cos(x)

(-4A - B + 5A) sin(x) + (-4B + A + 5B) cos(x) = 2 cos(x)

(A - B) sin(x) + (A + B) cos(x) = 2 cos(x)

Then

A - B = 0

A + B = 2

Adding these equation gives

(A - B) + (A + B) = 0 + 2

2A = 2

A = 1   →   B = 1