Respuesta :
Answer:
y(t) = (191/102)e^(-3t) + (205/183)e^(6t) + (15/2074)cos(5t) + (43/2074)sin(5t)
Step-by-step explanation:
Given the differential equation:
y'' - 3y' - 18y = sin(5t).........................(1)
y(0) = 3
y'(0) = 1
To find this, by taking the Laplace transform of the differential equation, we know that
L{y(t)} = Y.
L{y''} = s²Y - sy(0) - y'(0)
L{y'} = sY - y(0)
y(0) = 3
y'(0) = 1
Taking the Laplace transform of both sides (1), we have have the differential equation to be:
[s²Y - sy(0) - y'(0)] - 3[sY - y(0)] - 18[Y] = 5/(s² + 25)
s²Y - 3s - 1 - 3sY + 9 - 18Y = 5/(s² + 25)
(s² - 3s - 18)Y - 3s + 8 = 5/(s² + 25)
(s² - 3s - 18)Y = 3s - 8 + 5/(s² + 25)
(s + 3)(s - 6)Y = 3s - 8 + 5/(s² + 25)
Divide both sides by (s + 3)(s - 6) to have:
Y = 3s/(s + 3)(s - 6) - 8/(s + 3)(s - 6) + 5/(s²+ 25)(s + 3)(s - 6)
Resolving each of the fractions on the right hand side into partial fractions, we have
Y = 1/(s + 3) + 2/(s - 6) - (8/9)[1/(s - 6) - 1/(s + 3)] + (5/2074)[3s/(s² + 25) - 43/(s² + 25)] + (5/549)[1/(s - 6)] - (5/306)[1/(s + 3)]
Next, we take the Inverse Laplace Transform of the last equation to obtain the required solution:
y(t) = e^(-3t) + 2e^(6t) - (8/9)[e^(6t) - e^(-3t)] + (5/2074)[3cos(5t) + (43/5)sin(5t)] + (5/549)e^(6t) - (5/306)e^(-3t)
= (1 + 8/9 - 5/306)e^(-3t) + (2 - 8/9 + 5/549)e^(6t) + (15/2074)cos(5t) + (43/2074)sin(5t)
= (191/102)e^(-3t) + (205/183)e^(6t) + (15/2074)cos(5t) + (43/2074)sin(5t)
