Question

Use Laplace transform methods to solve the differential equation: The initial conditions are f(0) = 0 and z-dn.net/?f=f%5E%7B%27%7D%280%29%20%3D%200" id="TexFormula1" title="f^{'}(0) = 0" alt="f^{'}(0) = 0" align="absmiddle" class="latex-formula">
$\frac{d^2 f(t)}{dt^2} + 12 \frac{df(t)}{dt} + 32f(t) = 10.e^{-2t}$

Answer 1

Answer:

Step-by-step explanation:

$\frac{d^2 y}{dt^2} +12\frac{dy}{dt} +32y = 10e^{-2t}$ where f(x) is replaced by y

Take Laplace on both sides

$s^2 Y -s(0)-0 +12s Y-0+32Y = \frac{10}{s+2} \\Y(s^2+12s+32) = \frac{10}{s+2}\\Y = \frac{10}{(s+2)(s+4)(s+8)}$

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

Let this equals

$\frac{A}{s+2} +\frac{B}{s+4} +\frac{C}{s+8} \\10 = A(s+4)(s+8) + B(s+2)(s+8) +Cs+4)(s+2)$

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

$y(t) = \frac{-5e^{-4t} }{4} + \frac{5e^{-8t} }{12} +\frac{-5e^{-2t} }{6}$