Describe a situation where it is easier to use decimals than friction,and explain why.
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Take the Laplace transform of both sides:
L[y'' - 4y' + 8y] = L[δ(t - 1)]
I'll denote the Laplace transform of y = y(t) by Y = Y(s). Solve for Y :
(s²Y - s y(0) - y'(0)) - 4 (sY - y(0)) + 8Y = exp(-s) L[δ(t)]
s²Y - 4sY + 8Y = exp(-s)
(s² - 4s + 8) Y = exp(-s)
Y = exp(-s) / (s² - 4s + 8)
and complete the square in the denominator,
Y = exp(-s) / ((s - 2)^2 + 4)
Recall that
L⁻¹[F(s - c)] = exp(ct) f(t)
In order to apply this property, we multiply Y by exp(2)/exp(2), so that
Y = exp(-2) • exp(-s) exp(2) / ((s - 2)² + 4)
Y = exp(-2) • exp(-s + 2) / ((s - 2)² + 4)
Y = exp(-2) • exp(-(s - 2)) / ((s - 2)² + 4)
Then taking the inverse transform, we have
L⁻¹[Y] = exp(-2) L⁻¹[exp(-(s - 2)) / ((s - 2)² + 4)]
L⁻¹[Y] = exp(-2) exp(2t) L⁻¹[exp(-s) / (s² + 4)]
L⁻¹[Y] = exp(2t - 2) L⁻¹[exp(-s) / (s² + 4)]
Next, we recall another property,
L⁻¹[exp(-cs) F(s)] = u(t - c) f(t - c)
where F is the Laplace transform of f, and u(t) is the unit step function
To apply this property, we first identify c = 1 and F(s) = 1/(s² + 4), whose inverse transform is
L⁻¹[F(s)] = 1/2 L⁻¹[2/(s² + 2²)] = 1/2 sin(2t)
Then we find
L⁻¹[Y] = exp(2t - 2) u(t - 1) • 1/2 sin(2 (t - 1))
and so we end up with
y = 1/2 exp(2t - 2) u(t - 1) sin(2t - 2)
