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)