For y 5y 6y 2et cos5t give the particular solution Yt use

For y\" + 5y\' + 6y = 2e-t cos(5t) give the particular solution, Y(t), used in the method of undetermined coefficients.

Solution

Let y = e^-t(A cos 5t + B sin 5t)

y\' = e^-t(-5A sin 5t + 5B cos 5t) - e^-t(A cos 5t + B sin 5t)

y\'\' = e^-t(-25A cos 5t - 25B sin 5t) - 2e^-t(-5A sin 5t + 5B cos 5t) + e^-t(A cos 5t + B sin 5t)

Put them into the equation,

e^-t(-25A cos 5t - 25B sin 5t) - 2e^-t(-5A sin 5t + 5B cos 5t) + e^-t(A cos 5t + B sin 5t)

+ 5[e^-t(-5A sin 5t + 5B cos 5t) - e^-t(A cos 5t + B sin 5t)]

+ 6e^-t(A cos 5t + B sin 5t) = 2e^-t cos 5t

then

-25A cos 5t - 25B sin 5t + 10A sin 5t - 10B cos 5t + A cos 5t + B sin 5t

- 25A sin 5t + 25B cos 5t - 5A cos 5t - 5B sin 5t

+ 6A cos 5t + 6B sin 5t = 2cos 5t

so

-23A + 15B = 2

-15A - 23B =0.

so A = -23/377, B = 15/377

so y = e^t (-23 cos 5t + 15 sin 5t)/377