Question

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

Answer 1

Answer:

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

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)} = $e^{-cs}$

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) = $\frac{3e^{-2s} }{(s + 2)(s + 3)} - \frac{4e^{-5s} }{(s + 2)(s + 3)}$

Now,

Let

$\frac{1}{(s+2)(s+3)} = \frac{A}{s+2} + \frac{B}{s+3} \\\frac{1}{(s+2)(s+3)} = \frac{A(s + 3) + B(s+2)}{(s+2)(s+3)}\\1 = As + 3A + Bs + 2B\\1 = (A+B)s + (3A + 2B)$

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

$\frac{1}{(s+2)(s+3)} = \frac{1}{s+2} + \frac{-1}{s+3}$

So,

Y(s) = $3e^{-2s}[ \frac{1}{(s + 2)} - \frac{1}{(s + 3)}] - 4e^{-5s}[ \frac{1}{(s + 2)} - \frac{1}{(s + 3)}]$

⇒Y(s) = $3e^{-2s} \frac{1}{(s + 2)} - 3e^{-2s} \frac{1}{(s + 3)} - 4e^{-5s}\frac{1}{(s + 2)} + 4e^{-5s}\frac{1}{(s + 3)}$

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.