Answer:
Step-by-step explanation:
[tex]\frac{d^2 y}{dt^2} +12\frac{dy}{dt} +32y = 10e^{-2t}[/tex] where f(x) is replaced by y
Take Laplace on both sides
[tex]s^2 Y -s(0)-0 +12s Y-0+32Y = \frac{10}{s+2} \\Y(s^2+12s+32) = \frac{10}{s+2}\\Y = \frac{10}{(s+2)(s+4)(s+8)}[/tex]
We can resolve into partial fraction to get Y = L(y)
Let this equals
[tex]\frac{A}{s+2} +\frac{B}{s+4} +\frac{C}{s+8} \\10 = A(s+4)(s+8) + B(s+2)(s+8) +Cs+4)(s+2) \\[/tex]
Solving we get
s=-9
24C = 10 or C =5/12
s=-2: A=10/12=5/6
s=-4: B = -5/4
Taking inverse Laplace
[tex]y(t) = \frac{-5e^{-4t} }{4} + \frac{5e^{-8t} }{12} +\frac{-5e^{-2t} }{6}[/tex]
