Example 1.26
Solve the following initial value problem:
y + y = 0, y(0) = 1, y (0) = 1. (1.153)
By taking Laplace transforms of both sides of the ODE, we use
L{y(x)} = Y (s),
2 (1.154)
L{y (x)} = s L{y(x)} − sy(0) − y (0)
= s2Y (s) − s − 1,
obtained upon using the initial conditions. Substituting this into the ODE
gives
s 1
Y (s) = s2 + 1 + s2 + 1 . (1.155)
To determine the solution y(x) we then take the inverse Laplace transform
L−1 to both sides of the last equation to find
L−1{Y (s)} = L−1 2 s + L−1 2 1 . (1.156)
s + 1 s + 1
This in turn gives the solution by
y(x) = cos x + sin x, (1.157)
obtained upon using the table of Laplace transforms. Notice that we obtained
the particular solution for the differential equation. This is due to the fact
that the initial conditions were used in the solution.