The general solution of the given differential equation. y'' + 2y' + 5y = 3 sin 2t is y = e^(-t) * [ A cos(2t) + B sin(2t) ] + (-12/17) * cos(2t) + (3/17)*sin(2t)
y'' + 2y' + 5y = 3sin(2t)
Auxiliary equation
m^2 + 2m + 5 = 0
m = -1 + 2i, -1 - 2i
yc = e^(-t) * [ A cos(2t) + B sin(2t) ]
let yp = C*cos(2t) + D*sin(2t)
yp' = -2C*sin(2t) +2 D*cos(2t)
yp'' = -4C*cos(2t) - 4D*sin(2t)
yp'' + 2yp' + 5yp = 3sin(2t)
= -4C*cos(2t) - 4D*sin(2t) + 2 * [-2C*sin(2t) +2 D*cos(2t)] + 5 * [C*cos(2t) + D*sin(2t)] = 3sin(2t)
= -4C*cos(2t) - 4D*sin(2t) - 4C*sin(2t) + 4 D*cos(2t) + 5C*cos(2t) + 5D*sin(2t) = 3sin(2t)
= - 4C*sin(2t) + 4 D*cos(2t) + C*cos(2t) + D*sin(2t) = 3sin(2t)
Compare like terms
D - 4C = 3
4D + C = 0 => C = -4D
From D -4C = 3
D - 4*(-4D) = 3
17D = 3
D = 3/17
C = -12/17
Therefore, the general solution of the given differential equation. y'' + 2y' + 5y = 3 sin 2t is y = e^(-t) * [ A cos(2t) + B sin(2t) ] + (-12/17) * cos(2t) + (3/17)*sin(2t).
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