Dr. Z.’s Calc5 Lecture 6 Handout:
Using the Laplace Transform to solve Systems of Linear Differential Equations
By Doron Zeilberger
Systems of ODEs
When we have one differential equation, with one unknown function y(t) and we use the Laplace
Transform method, we trade it with one algebraic equation in Y (s). When we have several
equations with several unknown functions we trade it with a system of algebraic equations in the
Laplace Transforms. We then solve the system, and at the end, transform back.
Problem 6.1: Solve the system
dx
= −x + y ,
dt
dy
= 2x ,
dt
x(0) = 0 , y(0) = 1
.
Solution: Let L{x} = X(s), L{y} = Y (s). Applying L to both equations we get
L{
dx
} = −L{x} + L{y}
dt
L{
dy
} = 2L{x}
dt
,
.
dy
Now L{ dx
dt } = sX(s) − x(0) = sX(s), L{ dt } = sY (s) − y(0) = sY (s) − 1 = sY (s) − 1
So we have the system
sX = −X + Y
sY − 1 = 2X
Cleaning up:
(s + 1)X − Y = 0 ,
−2X + sY = 1 .
From the first equation we get
Y = (s + 1)X
.
Substituting this into the second equation:
−2X + s(s + 1)X = 1
Factoring out X:
(−2 + s(s + 1))X = 1
1
Solving for X, we get
X=
1
s2 + s − 2
=
1
(s + 2)(s − 1)
.
Since Y = (s + 1)X, we have
Y =
So we found
X(s) =
s+1
(s + 2)(s − 1)
1
(s + 2)(s − 1)
,
Y (s) =
.
s+1
(s + 2)(s − 1)
.
Finally, we need to find x(t) and y(t).
x(t) = L−1 {
1
}
(s + 2)(s − 1)
,
y(t) = L−1 {
s+1
}
(s + 2)(s − 1)
By partial fractions:
1
A
B
=
+
(s + 2)(s − 1)
s+2 s−1
,
1
A(s − 1) + B(s + 2)
=
(s + 2)(s − 1)
(s + 2)(s − 1)
.
So 1 = A(s − 1) + B(s + 2). When s = 1, 1 = B · (1 + 2) so B = 31 . When s = −2, 1 = A(−2 − 1)
so A = − 13 . So
1
1
1
1
X=− ·
+ ·
.
3 s+2 3 s−1
So
x(t) = L−1 {−
1 1
1
1
1 1
+
} = − e−2t + et
3s+2 3s−1
3
3
.
As for Y (s):
s+1
A
B
=
+
(s + 2)(s − 1)
s+2 s−1
,
s+1
A(s − 1) + B(s + 2)
=
(s − 2)(s + 1)
(s + 2)(s − 1)
.
So s + 1 = A(s − 1) + B(s + 2). When s = 1, 1 + 1 = B · (1 + 2) so B = 32 . When s = −2,
−2 + 1 = A(−2 − 1) so A = 13 . So
Y =
So
y(t) = L−1 {
Ans. to 6.1: x(t) = − 31 e−2t + 13 et
1
2
1
1
·
+ ·
3 s+2 3 s−1
.
2 1
1
2
1 1
+
} = e−2t + et
3s+2 3s−1
3
3
,
y(t) = 13 e−2t + 32 et
Problem 6.2: Solve the system
1
d2 x
= (−x + y) ,
dt2
2
d2 y
1
= (x − y) ,
2
dt
2
2
.
.
x(0) = 0
,
x0 (0) = 2
;
y(0) = 0
,
y 0 (0) = 0
.
Solution: Let L{x} = X(s), L{y} = Y (s). Applying L to both equations we get
L{
d2 x
1
1
} = − L{x} + L{y}
dt2
2
2
L{
1
1
d2 y
} = L{x} − L{y}
dt2
2
2
,
,
2
2
Now L{ ddt2x } = s2 X(s) − x(0)s − x0 (0) = s2 X(s) − 2, L{ ddt2y } = s2 Y (s) − y(0)s − y 0 (0) = s2 Y (s)
So we have the system
1
1
s2 X − 2 = − X + Y
2
2
1
1
s2 Y = X − Y
2
2
Cleaning up:
1
1
(s2 + )X − Y = 2 ,
2
2
1
1
(s2 + )Y = X .
2
2
From the second equation we get
X = (2s2 + 1)Y
.
Substituting this into the first equation:
1
1
(s2 + )(2s2 + 1)Y − Y = 2 ,
2
2
Factor out Y :
1
1
(s2 + )(2s2 + 1) −
Y =2 ,
2
2
1 1
4
2
2s + 2s + −
Y =2 ,
2 2
(2s4 + 2s2 )Y = 2 ,
2s2 (s2 + 1)Y = 2 ,
So
Y =
1
s2 (s2 + 1)
Going back to X:
X=
2s2 + 1
s2 (s2 + 1)
3
Finally, we need to find x(t) and y(t).
2s2 + 1
}
x(t) = L−1 { 2 2
s (s + 1)
1
y(t) = L−1 { 2 2
}
s (s + 1)
,
By partial fractions:
X=
A
B
C
2s2 + 1
= + 2+ 2
2
2
s (s + 1)
s
s
s +1
2s2 + 1
(As + B)(s2 + 1) + Cs2
=
s2 (s2 + 1)
s2 (s2 + 1)
, so
2s2 + 1 = (As + B)(s2 + 1) + Cs2 = As3 + Bs2 + As + B + Cs2 = As3 + (B + C)s2 + As + B
Comparing coefficients, we get:
A=0 ,
B+C =2 ,
A=0 ,
B=1 ,
so A = 0, B = 1, C = 1 and we have
X=
So
1
1
+ 2
s2
s +1
1
1
x(t) = L−1 { 2 + 2
} = t + sin t
s
s +1
As for Y (s):
Y =
1
s2 (s2 + 1)
1
s2 (s2 + 1)
=
=
A
B
C
+ 2+ 2
s
s
s +1
(As + B)(s2 + 1) + Cs2
s2 (s2 + 1)
, so
1 = (As + B)(s2 + 1) + Cs2 = As3 + Bs2 + As + B + Cs2 = As3 + (B + C)s2 + As + B
Comparing coefficients, we get:
A=0 ,
B+C =0 ,
A=0 ,
B=1 ,
so A = 0,B = 1,C = −1 and we have
Y =
So
1
1
− 2
s2
s +1
.
1
1
y(t) = L−1 { 2 − 2
} = t − sin t
s
s +1
Ans. to 6.2: x(t) = t + sin t, y(t) = t − sin t.
4
.
.
.