Respuesta :
Answer:
y(t) = 3u₂(t) [ [tex]e^{-2t+4} - e^{-5t + 10)}[/tex] ] - 4u₅(t) [ [tex]e^{-2t+10)} - e^{-5t + 25)}[/tex] ]
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)} = [tex]e^{-cs}[/tex]
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[tex]e^{-2s}[/tex] - 4[tex]e^{-5s}[/tex]
⇒s²Y(s) - 0 - 0 + 5[sY(s) - 0] + 6Y(s) = 3[tex]e^{-2s}[/tex] - 4[tex]e^{-5s}[/tex]
⇒s²Y(s) + 5sY(s) + 6Y(s) = 3[tex]e^{-2s}[/tex] - 4[tex]e^{-5s}[/tex]
⇒[s² + 5s + 6] Y(s) = 3[tex]e^{-2s}[/tex] - 4[tex]e^{-5s}[/tex]
⇒[s² + 3s + 2s + 6] Y(s) = 3[tex]e^{-2s}[/tex] - 4[tex]e^{-5s}[/tex]
⇒[s(s + 3) + 2(s + 3)] Y(s) = 3[tex]e^{-2s}[/tex] - 4[tex]e^{-5s}[/tex]
⇒[(s + 2)(s + 3)] Y(s) = 3[tex]e^{-2s}[/tex] - 4[tex]e^{-5s}[/tex]
⇒Y(s) = [tex]\frac{3e^{-2s} }{(s + 2)(s + 3)} - \frac{4e^{-5s} }{(s + 2)(s + 3)}[/tex]
Now,
Let
[tex]\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)[/tex]
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
[tex]\frac{1}{(s+2)(s+3)} = \frac{1}{s+2} + \frac{-1}{s+3}[/tex]
So,
Y(s) = [tex]3e^{-2s}[ \frac{1}{(s + 2)} - \frac{1}{(s + 3)}] - 4e^{-5s}[ \frac{1}{(s + 2)} - \frac{1}{(s + 3)}][/tex]
⇒Y(s) = [tex]3e^{-2s} \frac{1}{(s + 2)} - 3e^{-2s} \frac{1}{(s + 3)} - 4e^{-5s}\frac{1}{(s + 2)} + 4e^{-5s}\frac{1}{(s + 3)}[/tex]
By applying inverse Laplace , we get
y(t) = 3u₂(t) [ [tex]e^{-2(t-2)} - e^{-5(t - 2)}[/tex] ] - 4u₅(t) [ [tex]e^{-2(t-5)} - e^{-5(t - 5)}[/tex] ]
⇒y(t) = 3u₂(t) [ [tex]e^{-2t+4} - e^{-5t + 10)}[/tex] ] - 4u₅(t) [ [tex]e^{-2t+10)} - e^{-5t + 25)}[/tex] ]
It is the required solution.
