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

y(t) =  3u₂(t) [ ] - 4u₅(t) [ ]

Step-by-step explanation:

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

Formula used -

L{δ(t − c)} =

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 - 4

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

⇒s²Y(s) + 5sY(s) + 6Y(s) = 3 - 4

⇒[s² + 5s + 6] Y(s) = 3 - 4

⇒[s² + 3s + 2s + 6] Y(s) = 3 - 4

⇒[s(s + 3) + 2(s + 3)] Y(s) = 3 - 4

⇒[(s + 2)(s + 3)] Y(s) = 3 - 4

⇒Y(s) =

Now,

Let

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

So,

Y(s) =

⇒Y(s) =

By applying inverse Laplace , we get

y(t) = 3u₂(t) [ ] - 4u₅(t) [ ]

⇒y(t) =  3u₂(t) [ ] - 4u₅(t) [ ]

It is the required solution.