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Solving Systems of Differential Equation Using Laplace Transform
.

Taking the inverse transform yields

t
t
t
t
y = 1 − 2e + te , z = 2e − te .
. . . .
.
Example 5.11, Solve the initial value problem for a system of
ODE’s
{
˙ x = 2x + 3y, x(0) = 1
˙ y = 3x + 4y, y(0) = 0
. . . . .
.
Solution. We begin by applying the Laplace transform to both sides of each
equation:
{
L[ ˙x] = 2L[x] + 3L[y]
L[ ˙y] = 3L[x] + 4L[y]

Setting X(s) = L[x(t)], Y (s) = L[y(t)] and using the result that

L[f (t)] = sL[f(t)] − f(0),
′
we find that the system of ODE’s is transformed into a system of linear
algebraic equations for the functions X(s) and Y (s) :

{
sX(s) − 1 = 2X(s) + 3Y (s)
,
sY (s) = 3X(s) + 4Y (s)
which can be rearranged as

{
(s − 2)X(s) − 3Y (s) = 1
.
−3X(s) + (s − 4)Y (s) = 0

Let us apply Cramer’s Rule for solving this system of linear equations:

s − 2 −1 2

∆ = = (s − 2)(s − 4) − 3 = s − 6s + 5 = (s − 1)(s − 5),
−3 s − 4

1 −1 s − 2 1
2
1
∆ = = s − 4, ∆ = = 3.
0 s − 4 −3 0
The solutions are:
∆ 1 s − 4 ∆ 2 3
X(s) = = , Y (s) = = .
∆ (s − 1)(s − 5) ∆ (s − 1)(s − 5)

We now are going to find the inverse transform of both functions X(s)
and Y (s) which will then be solutions of the system. To do this, we use
to decompose the above expressions into sums of simple fractions. Using

the method of partial fraction expansions leads to
A B
X(s) = + ; A(s − 5) + B(s − 1) = s − 4,
. . . . s − 1 s − 5

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