Answer:
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)} = ![e^{-cs}](https://tex.z-dn.net/?f=e%5E%7B-cs%7D)
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![e^{-5s}](https://tex.z-dn.net/?f=e%5E%7B-5s%7D)
⇒s²Y(s) - 0 - 0 + 5[sY(s) - 0] + 6Y(s) = 3
- 4![e^{-5s}](https://tex.z-dn.net/?f=e%5E%7B-5s%7D)
⇒s²Y(s) + 5sY(s) + 6Y(s) = 3
- 4![e^{-5s}](https://tex.z-dn.net/?f=e%5E%7B-5s%7D)
⇒[s² + 5s + 6] Y(s) = 3
- 4![e^{-5s}](https://tex.z-dn.net/?f=e%5E%7B-5s%7D)
⇒[s² + 3s + 2s + 6] Y(s) = 3
- 4![e^{-5s}](https://tex.z-dn.net/?f=e%5E%7B-5s%7D)
⇒[s(s + 3) + 2(s + 3)] Y(s) = 3
- 4![e^{-5s}](https://tex.z-dn.net/?f=e%5E%7B-5s%7D)
⇒[(s + 2)(s + 3)] Y(s) = 3
- 4![e^{-5s}](https://tex.z-dn.net/?f=e%5E%7B-5s%7D)
⇒Y(s) = ![\frac{3e^{-2s} }{(s + 2)(s + 3)} - \frac{4e^{-5s} }{(s + 2)(s + 3)}](https://tex.z-dn.net/?f=%5Cfrac%7B3e%5E%7B-2s%7D%20%7D%7B%28s%20%2B%202%29%28s%20%2B%203%29%7D%20-%20%20%5Cfrac%7B4e%5E%7B-5s%7D%20%7D%7B%28s%20%2B%202%29%28s%20%2B%203%29%7D)
Now,
Let
![\frac{1}{(s+2)(s+3)} = \frac{A}{s+2} + \frac{B}{s+3} \\\frac{1}{(s+2)(s+3)} = \frac{A(s + 3) + B(s+2)}{(s+2)(s+3)}\\1 = As + 3A + Bs + 2B\\1 = (A+B)s + (3A + 2B)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%28s%2B2%29%28s%2B3%29%7D%20%3D%20%5Cfrac%7BA%7D%7Bs%2B2%7D%20%20%2B%20%5Cfrac%7BB%7D%7Bs%2B3%7D%20%5C%5C%5Cfrac%7B1%7D%7B%28s%2B2%29%28s%2B3%29%7D%20%3D%20%5Cfrac%7BA%28s%20%2B%203%29%20%2B%20B%28s%2B2%29%7D%7B%28s%2B2%29%28s%2B3%29%7D%5C%5C1%20%3D%20As%20%2B%203A%20%2B%20Bs%20%2B%202B%5C%5C1%20%3D%20%28A%2BB%29s%20%2B%20%283A%20%2B%202B%29)
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
![\frac{1}{(s+2)(s+3)} = \frac{1}{s+2} + \frac{-1}{s+3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%28s%2B2%29%28s%2B3%29%7D%20%3D%20%5Cfrac%7B1%7D%7Bs%2B2%7D%20%20%2B%20%5Cfrac%7B-1%7D%7Bs%2B3%7D)
So,
Y(s) = ![3e^{-2s}[ \frac{1}{(s + 2)} - \frac{1}{(s + 3)}] - 4e^{-5s}[ \frac{1}{(s + 2)} - \frac{1}{(s + 3)}]](https://tex.z-dn.net/?f=3e%5E%7B-2s%7D%5B%20%5Cfrac%7B1%7D%7B%28s%20%2B%202%29%7D%20-%20%20%20%20%5Cfrac%7B1%7D%7B%28s%20%2B%203%29%7D%5D%20-%204e%5E%7B-5s%7D%5B%20%5Cfrac%7B1%7D%7B%28s%20%2B%202%29%7D%20-%20%20%20%20%5Cfrac%7B1%7D%7B%28s%20%2B%203%29%7D%5D)
⇒Y(s) = ![3e^{-2s} \frac{1}{(s + 2)} - 3e^{-2s} \frac{1}{(s + 3)} - 4e^{-5s}\frac{1}{(s + 2)} + 4e^{-5s}\frac{1}{(s + 3)}](https://tex.z-dn.net/?f=3e%5E%7B-2s%7D%20%5Cfrac%7B1%7D%7B%28s%20%2B%202%29%7D%20-%20%20%20%203e%5E%7B-2s%7D%20%5Cfrac%7B1%7D%7B%28s%20%2B%203%29%7D%20-%204e%5E%7B-5s%7D%5Cfrac%7B1%7D%7B%28s%20%2B%202%29%7D%20%2B%204e%5E%7B-5s%7D%5Cfrac%7B1%7D%7B%28s%20%2B%203%29%7D)
By applying inverse Laplace , we get
y(t) = 3u₂(t) [
] - 4u₅(t) [
]
⇒y(t) = 3u₂(t) [
] - 4u₅(t) [
]
It is the required solution.