Question 5.5.3: Use the Laplace transform to solve the initial-value problem...

Taking the Laplace transform of both sides of (5.5.9) and writing Y (s) = L[y(t )], we observe that

s^{2}Y (s)−sy(0)−y′(0)+2(sY (s)−y(0))+5Y (s) = \frac {e^{−2s}}{s}

Substituting the given initial conditions and factoring on the left, we have

Y (s)(s^{2} +2s +5) = s +2+\frac {e^{−2s}}{s}

Solving for Y (s), we can write

Y (s) = Y_{1}(s)+Y_{2}(s) = \frac {s +2}{s^{2} +2s +5}+e^{−2s} \frac {1}{s(s^{2} +2s +5)}                      (5.5.10)

It remains for us to determine the function y(t ) whose transform is Y(s). By linearity, it helps for us to break the function Y (s) into the simplest pieces we can; we begin by determining the inverse transform of Y_{1}(s). Because of shifting properties of the transform (and because of the fact that we cannot factor s^{2} +2s +5 in an effort to apply partial fractions), it is useful to complete the square in expressions such as s^{2} +2s +5. We instead write (s +1)^{2}+4, and seek to identify other parts of the expression that involve (s +1). Separating the numerator (s +2) into (s +1)+1, we can express the first term in (5.5.10) as

Y_{1}(s) = \frac {s +2}{s^{2} +2s +5}= \frac {s +1}{(s +1)^{2} +4}+ \frac {1}{(s +1)^{2} +4}               (5.5.11)

Recalling that L[cos2t] = s/(s^{2} +4) and L[sin2t] = 2/(s^{2} +4), we know

L^{−1}[s/(s^{2} +4)] = cos2t    and    L^{−1}[2/(s^{2} +4)] = sin2t

The inverse of the first shifting property, L^{−1}[F(s +1)] = e^{−t} f (t ), now implies that

L^{−1} [ \frac {s +1}{(s +1)^{2} +4}+ \frac {1}{(s +1)^{2} +4}]= e^{−t} cos2t + \frac {1}{2} e^{−t} sin2t            (5.5.12)

Hence, the first term Y_{1}(s) in (5.5.10) comes from taking the Laplace transform of the function y_{1}(t ) = e^{−t} cos2t + \frac {1}{2} e^{−t} sin2t .

From (5.5.10), it remains for us to find the function yy_{2}(t ) whose Laplace transform is

Y_{2}(s) = e^{−2s} \frac {1}{s(s^{2} +2s +5)}

Using a partial fraction decomposition on the rational part of the function, we have

e^{−2s} \frac {1}{s(s^{2} +2s +5)}=\frac {1}{5}e^{−2s} (\frac {1}{s} − \frac {s +2}{s^{2} +2s +5})

Observe that we have already determined the inverse transform of the function (s + 2)/(s^{2} + 2s + 5) above at (5.5.12). Here, we must deal with the additional presence of the constant 1/5, the multiplier e^{−2s} , and the basic function 1/s. Recalling the inverse second shifting property, L^{−1}[e^{−as}F(s)] =u(t −a)f (t −a), and (5.5.12), we observe that

L^{−1}[e^{−2s} (\frac {1}{s} − \frac {s +2}{s^{2} +2s +5})]

= u(t −2)[1−e^{−(t−2)}(cos 2(t −2)+ \frac {1}{2} sin 2(t −2))]                  (5.5.13)

Combining (5.5.10), (5.5.12), and (5.5.13), we have shown that the solution y(t ) to the initial-value problem is

y(t ) = e^{−t} cos2t + \frac {e^{−t}}{2}sin2t + \frac {1}{5}u(t −2) [1−e^{−(t−2)} cos 2(t −2)+ \frac {e^{−(t−2)}}{2}sin 2(t −2)]

A plot of the function y(t ) is shown in figure 5.9. Here, we see evidence of the qualitative behavior we expect: until the unit step function turns on, the homogeneous equation should show damped oscillations so that y(t ) → 0. But once the step function turns on, the forcing function makes the equation nonhomogeneous with a constant forcing function, making y = 1/5 the stable equilibrium solution to which y(t ) tends.