Question

Y″+ 5y′ + 6y = 3δ(t − 2) − 4δ(t −5); y(0) = y′′(0) = 0

Answer 1

y(t) = 3u₂(t) [
`e^(-2t+4) - e^(-5t + 10))` ] - 4u₅(t) [
`e^(-2t+10)) - e^(-5t + 25))` ]

Explanation:

To find - y″+ 5y′ + 6y = 3δ(t − 2) − 4δ(t −5); y(0) = y′(0) = 0

Formula used -

L{δ(t − c)} =
`e^(-cs)`

L{f''(t) = s²F(s) - sf(0) - f'(0)

L{f'(t) = sF(s) - f(0)

Solution -

By Applying Laplace transform, we get

L{y″+ 5y′ + 6y} = L{3δ(t − 2) − 4δ(t −5)}

⇒L{y''} + 5L{y'} + 6L{y} = 3L{δ(t − 2)} − 4L{δ(t −5)}

⇒s²Y(s) - sy(0) - y'(0) + 5[sY(s) - y(0)] + 6Y(s) = 3
`e^(-2s)` - 4
`e^(-5s)`

⇒s²Y(s) - 0 - 0 + 5[sY(s) - 0] + 6Y(s) = 3
`e^(-2s)` - 4
`e^(-5s)`

⇒s²Y(s) + 5sY(s) + 6Y(s) = 3
`e^(-2s)` - 4
`e^(-5s)`

⇒[s² + 5s + 6] Y(s) = 3
`e^(-2s)` - 4
`e^(-5s)`

⇒[s² + 3s + 2s + 6] Y(s) = 3
`e^(-2s)` - 4
`e^(-5s)`

⇒[s(s + 3) + 2(s + 3)] Y(s) = 3
`e^(-2s)` - 4
`e^(-5s)`

⇒[(s + 2)(s + 3)] Y(s) = 3
`e^(-2s)` - 4
`e^(-5s)`

⇒Y(s) =
`(3e^(-2s) )/((s + 2)(s + 3)) - (4e^(-5s) )/((s + 2)(s + 3))`

Now,

Let

`(1)/((s+2)(s+3)) = (A)/(s+2) + (B)/(s+3) \\(1)/((s+2)(s+3)) = (A(s + 3) + B(s+2))/((s+2)(s+3))\\1 = As + 3A + Bs + 2B\\1 = (A+B)s + (3A + 2B)`

By Comparing, we get

A + B = 0 and 3A + 2B = 1

⇒A = -B

and

3(-B) + 2B = 1

⇒-B = 1

⇒B = -1

So,

A = 1

∴ we get

`(1)/((s+2)(s+3)) = (1)/(s+2) + (-1)/(s+3)`

So,

Y(s) =
`3e^(-2s)[ (1)/((s + 2)) - (1)/((s + 3))] - 4e^(-5s)[ (1)/((s + 2)) - (1)/((s + 3))]`

⇒Y(s) =
`3e^(-2s) (1)/((s + 2)) - 3e^(-2s) (1)/((s + 3)) - 4e^(-5s)(1)/((s + 2)) + 4e^(-5s)(1)/((s + 3))`

By applying inverse Laplace , we get

y(t) = 3u₂(t) [
`e^(-2(t-2)) - e^(-5(t - 2))` ] - 4u₅(t) [
`e^(-2(t-5)) - e^(-5(t - 5))` ]

⇒y(t) = 3u₂(t) [
`e^(-2t+4) - e^(-5t + 10))` ] - 4u₅(t) [
`e^(-2t+10)) - e^(-5t + 25))` ]

It is the required solution.