Question 5.5.1: Use Laplace transforms to solve the initial-value problem y...

We begin by taking the Laplace transform of both sides of the differential equation. Using the linearity of the transform,

L[y′]−2L[y] = 5L[1]

By theorem 5.3.4 and the familiar transform of the function f (t ) = 1, it follows that

sL[y]−y(0)−2L[y] = \frac {5}{s}

Using the given fact that y(0) = 4 and denoting L[y] = Y (s),

sY (s)−2Y (s) = 4+\frac {5}{s}                 (5.5.1)

Note particularly that (5.5.1) is now an algebraic equation in the unknown function Y (s). Solving for Y (s), we find

Y (s) = \frac {4s +5}{s(s −2)}

At this point, we recall that Y (s) = L[y], where y(t ) is the original unknown function we seek as the solution to the stated IVP. Solving the IVP has now been reduced to finding the function y(t ) that has Laplace transform Y (s). That is,we seek y(t ) = L^{−1}[Y (s)].

With a bit of algebraic rearrangement and insight, we can find the function y(t ). In particular, using a partial fraction decomposition, we can show that

Y (s) = \frac {4s +5}{s(s −2)}=−\frac{5/2}{s}+ \frac{13/2}{s −2}          (5.5.2)

Recalling that L[1] = 1/s and L[e^{2t}] = 1/(s −2), (5.5.2) implies

y(t )=−\frac {5}{2}+ \frac {13}{2} e^{2t}

This is precisely the solution we would find to the IVP were we to use an integrating factor or separation of variables to solve the differential equation.