MATH 2574H
3/24 Problem Solutions
Problem 1: Solve using Laplace transforms.
y00 (t) + 6y0 (t) + 13y(t) = 4e−3t , y(0) = 2, y0 (0) = 7
Solution:
L[y00 ] + 6L[y0 ] + 13L[y] = 4L[e−3t ]
Let Y = L[y]
1
s2Y − 2s − 7 + 6 (sY − 2) + 13Y = 4
s+3
4
Y s2 + 6s + 13 =
+ 2s + 19
s+3
2s
19
4
+ 2
+ 2
Y=
2
(s + 3)(s + 6s + 13) s + 6s + 13 s + 6s + 13
Now we do partial fractions with the first term on the RHS:
4
A
Bs +C
=
+
(s + 3)(s2 + 6s + 13) s + 3 s2 + 6x + 13
As2 + 6As + 13A + Bs2 +Cs + 3Bs + 3C = 4
⇒ A + B = 0, 6A + 3B +C = 0, 13A + 3C = 4
After a bit of algebra, A = 1, B = −1,C = −3
Thus,
1
−s − 3
2s
19
Y=
+
+
+
s + 3 s2 + 6s + 13 s2 + 6s + 13 s2 + 6s + 13
1
s
16
Y=
+ 2
+ 2
s + 3 s + 6s + 13 s + 6s + 13
But,
s2 + 6s + 13 = (s + 3)2 + 4
Y=
1
s
16
+
+
2
s + 3 (s + 3) + 4 (s + 3)2 + 4
So our answer is (after a little bit of algebra):
y(t) = e−3t (1 + cos (2t) + 8 sin (2t))
MATH 2574H
3/24 Problem Solutions
Problem 2: Solve using Laplace transforms.
y00 (t) + 4y0 (t) + 13y(t) = H(t − 3), y(0) = 3, y0 (0) = −4
Solution:
L[y00 ] + 4L[y0 ] + 13L[y] = L[H(t − 3)]
Let L[y] = Y
e−3s
s2Y − 3s + 4 + 4 (sY − 3) + 13Y =
s
−3s
e
Y s2 + 4s + 13 =
+ 3s + 8
s
Y=
3s
8
e−3s
+ 2
+ 2
2
s(s + 4s + 13) s + 4s + 13 s + 4s + 13
Y = e−3s
1
s(s2 + 4s + 13)
+
3s
s2 + 4s + 13
+
8
s2 + 4s + 13
Now doing partial fractions with the first term on the RHS:
Bs +C
1
A
+
=
s s2 + 4s + 13 s(s2 + 4s + 13)
As2 + 4As + 13A + Bs2 +Cs = 1
⇒ A = 1/13, B = −1/13,C = −4/13
Using this we get
e−3s
Y=
13
1
s
4
3s
8
− 2
− 2
+ 2
+ 2
s s + 4s + 13 s + 4s + 13
s + 4s + 13 s + 4s + 13
After we notice that s2 + 4s + 13 = (s + 2)2 + 9 the second two terms in this equation are relatively straightforward
to solve. For the first term (the term with the exponential, we need to FIRST compute the Laplace inverse of each
term inside the parentheses and SECOND shift all the t’s and add our Heaviside step function. Here is how we do
that:
1 −1 1
s
4
1
4e−2t
−2t
−
=
L
−
1 − e cos (3t) −
sin (3t)
13
s (s + 2)2 + 9 (s + 2)2 + 9
13
3
So our first step is done. Now we must translate. To do this, we notice the coefficient of s in the exponential. For
the coefficient −c in our exponential, we must replace all the t’s in our inverse Laplace expression by t − c AND
simultaneously multiply the resulting expression by H(t − c). Hence we get:
!
4e−2t
4e−2(t−3)
H(t − 3)
1
−2t
−2(t−3)
1 − e cos (3t) −
sin (3t) ⇒ 1 − e
cos [3(t − 3)] −
sin [3(t − 3)]
13
3
3
13
This is the first term of y(t). We now compute the inverse Laplace transforms for the other two terms in our Y
equation to get:
!
4e−2(t−3)
H(t − 3)
8
−2(t−3)
y(t) = 1 − e
cos [3(t − 3)] −
sin [3(t − 3)]
+ 3e−2t cos (3t) + e−2t sin (3t)
3
13
3