Question 5.5.2: Use Laplace transforms to solve the initial-value problem y...

Taking the Laplace transform of both sides of the differential equation and applying the initial condition,

sL[y]−4+L[y] = 5L[u(t −1)]

Using the established fact that L[u(t −1)] = e^{−s}/s and letting Y (s) = L(y),

sY (s)−4+Y (s) = \frac {5e^{−s}}{s}

Solving for Y (s),

Y (s) = \frac {4}{s +1}+5e^{−s} \frac {1}{s(s +1)}           (5.5.6)

At this point, we need to use the inverse transform to solve for y(t ). Finding L^{−1}[4/(s +1)] is straightforward: by linearity and the first shifting property,^{5}

L^{−1} [\frac {4}{s +1}]= 4e^{−t}                (5.5.7)

To deal with the remaining term in (5.5.6), we note that with e^{−s} present we will need to use the second shifting property (5.5.5) in reverse. For this, it will be most useful to have the function

F(s) = \frac {1}{s(s +1)}

in a simpler form. Using its partial fraction decomposition, we observe that

F(s) = \frac {1}{s}− \frac {1}{s +1}

By (5.5.5), it now follows that

L^{−1}[5e^{−s}(\frac {1}{s}− \frac {1}{s +1})]=5u(t −1)(1−e^{−(t−1)})               (5.5.8)

Combining our work at (5.5.7) and (5.5.8) to determine y(t ) from (5.5.6), we have shown that

y(t ) = 4e^{−t} +5u(t −1)−5u(t −1)e^{−(t−1)}

A plot of this solution curve is shown in figure 5.8, where we see qualitative behavior consistent with what we would expect from the forcing function in the IVP. In particular, the forcing function is 5u(t −1), which makes the forcing function behave as if the constant function 5 is turned on at t = 1 in the initialvalue problem. For t = 0 to t = 1, we see the standard exponential decay that we would expect for the homogeneous equation y′+y = 0. But at t = 1, the solution function turns and begins to approach the equilibrium solution y = 5 that we expect in the nonhomogeneous equation y′+y =5. We note specifically that the Laplace transform has successfully handled all of this at once, including the role of the initial condition y(0)=4 and the corner in the solution function y(t) at t = 1.