Question

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Answer 1

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

$u(t) = \begin{cases}1 & \text{if }t \ge 0 \\ 0 & \text{if }t < 0\end{cases}$

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)