Question

Use Laplace transform methods to solve the differential equation: The initial conditions are f(0) = 0 and
`f^(')(0) = 0`


`(d^2 f(t))/(dt^2) + 12 (df(t))/(dt) + 32f(t) = 10.e^(-2t)`

Answer 1

Explanation:

`(d^2 y)/(dt^2) +12(dy)/(dt) +32y = 10e^(-2t)` where f(x) is replaced by y

Take Laplace on both sides

`s^2 Y -s(0)-0 +12s Y-0+32Y = (10)/(s+2) \\Y(s^2+12s+32) = (10)/(s+2)\\Y = (10)/((s+2)(s+4)(s+8))`

We can resolve into partial fraction to get Y = L(y)

Let this equals

`(A)/(s+2) +(B)/(s+4) +(C)/(s+8) \\10 = A(s+4)(s+8) + B(s+2)(s+8) +Cs+4)(s+2) \\`

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

`y(t) = (-5e^(-4t) )/(4) + (5e^(-8t) )/(12) +(-5e^(-2t) )/(6)`