Consider the following initial value problem:
y′′−3y′−18y=sin(5t)y(0)=3,y′(0)=1
Using Y for the Laplace transform of y(t), i. e., Y=L{y(t)}, find the equation you get by taking the Laplace transform of the differential equation and solve for Y(s)=

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)

a

step-by-step explanation: