Answer:
y = -21/697 sin(4t) + 16/697 cos(4t) + C₁e^(-t) + C₂e^(5t)
Explanation:
y" − 4y' − 5y = sin(4t)
This is a second-order, non-homogeneous equation with constant coefficients. The general solution is the sum of the complementary and particular solutions.
First, find the complementary solution:
y" − 4y' − 5y = 0
s² − 4s − 5 = 0
(s − 5) (s + 1) = 0
s = -1 or 5
y = C₁e^(-t) + C₂e^(5t)
Next, find the particular solution. Since g(t) = sin(4t):
y = A sin(4t) + B cos(4t)
y' = 4A cos(4t) − 4B sin(4t)
y" = -16A sin(4t) − 16B cos(4t)
Substituting:
y" − 4y' − 5y = sin(4t)
-16A sin(4t) − 16B cos(4t) − 4(4A cos(4t) − 4B sin(4t)) − 5(A sin(4t) + B cos(4t)) = sin(4t)
-16A sin(4t) − 16B cos(4t) − 16A cos(4t) + 16B sin(4t) − 5A sin(4t) − 5B cos(4t) = sin(4t)
(16B − 21A) sin(4t) − (16A + 21B) cos(4t) = sin(4t)
Match the coefficients:
16B − 21A = 1, 16A + 21B = 0
Solve the system of equations:
A = -21/16 B
16B − 21 (-21/16 B) = 1
16B + 441/16 B = 1
697B/16 = 1
B = 16/697
A = -21/697
Therefore, the general solution is:
y = -21/697 sin(4t) + 16/697 cos(4t) + C₁e^(-t) + C₂e^(5t)