Math 215
Spring 2010
Assignment 6
You are highly recommended to own a textbook, Boyce-DiPrima or another, to see a comprehensive treatment of the subjects and more examples.
§6.3: 8, 13, 15, 23,
§6.4: 6a, 10a,
§6.5: 4a, 6a, 10a
• (§6.3: 8) Sketch the graph of the given function and express it in terms of the unit
step function uc (t).


 1, (0 ≤ t < 1 or 2 ≤ t < 3);
f (t) = −1, (1 ≤ t < 2 or 3 ≤ t < 4);


0, (t ≥ 4).
Answer.
(a) [skip] (b)
f (t) = (1 − u1 ) − (u1 − u2 ) + (u2 − u3 ) − (u3 − u4 ) = (1 − 2u1 + 2u2 − 2u3 + u4 )(t).
½
0,
• (§6.3: 13) Find the Laplace transform of the function f (t) =
(t − 2)2 ,
Answer.
Note f (t) = u2 (t)g(t − 2) with g(t) = t2 . We have
Lf = (Lg)e−2s =
2 −2s
e .
s3


0,
• (§6.3: 15) Find the Laplace transform of the function f (t) = t − π,

0,
Answer.
We have
t<2
t≥2
t<π
π ≤ t < 2π
t ≥ 2π
Note f (t) = (t − π)[uπ (t) − u2π (t)] = (t − π)uπ (t) − (t − 2π)u2π (t) − πu2π (t).
Lf = (Lt)e−πs − (Lt)e−2πs − (Lπ)e−2πs =
1 −πs
1
π
e
− 2 e−2πs − e−2πs .
2
s
s
s
• (§6.3: 23) Find the inverse Laplace transform of the function F (s) =
Answer.
F (s) = G(s)e−s with G(s) =
1 t
1 3t
2 e + 2 e , and
(s−2)
(s−1)(s−3)
=
1/2
s−1
+
1/2
s−3 .
1
L−1 F = u1 (t)[L−1 G]t→t−1 = u1 (t)[et−1 + e3t−3 ].
2
It can be rewritten as u1 (t)e2t−2 cosh(t − 1).
(s − 2)e−s
.
s2 − 4s + 3
Thus L−1 G =
• (§6.4: 6a) Solve the initial value problem
y 00 + 3y 0 + 2y = u2 (t);
Answer.
y 0 (0) = 1.
y(0) = 0,
Taking the Laplace transform of both sides of the ODE, we obtain
s2 Y (s) − sy(0) − y 0 (0) + 3[sY (s) − y(0)] + 2Y (s) =
e−2s
.
s
Applying the initial conditions,
s2 Y (s) − 1 + 3sY (s) + 2Y (s) =
e−2s
.
s
Solving for the transform,
Y (s) =
s2
1
e−2s
+
.
2
+ 3s + 2 s(s + 3s + 2)
Using partial fractions,
s2
1
1
1
=
−
,
+ 3s + 2
s+1 s+2
s(s2
1
1/2
1
1/2
=
−
+
.
+ 3s + 2)
s
s+1 s+2
Taking the inverse transform term-by-term, the solution of the IVP is
1
1
y(t) = e−t − e−2t + [ − e−(t−2) + e−2(t−2) ]u2 (t).
2
2
• (§6.4: 10a) Solve the initial value problem
5
y + y + y = g(t);
4
00
Answer.
0
y(0) = 0,
0
y (0) = 0;
½
sin t, 0 ≤ t < π
g(t) =
0,
t≥π
Rewrite
g(t) = sin t(1 − uπ (t)) = sin t − uπ (t) sin t = sin t + uπ (t) sin(t − π).
Thus Lg = s21+1 + s21+1 e−πs . Taking the Laplace transform of both sides of the ODE
and using the zero initial conditions, we obtain
5
1
s2 Y (s) + sY (s) + Y (s) = Lg = 2
(1 + e−πs ).
4
s +1
Solving for the transform,
Y (s) =
1 + e−πs
.
(s2 + s + 54 )(s2 + 1)
2
Using partial fractions,
"
4
4s + 3
=
5
2
2
2
17 s + s +
(s + s + 4 )(s + 1)
1
Completing square,
4s + 3
2
s +s+
Thus
5
4
5
4
#
−4s + 1
.
+ 2
s +1
4(s + 21 ) + 1
=
.
(s + 21 )2 + 1
"
#
4 4(s + 12 ) + 1 −4s + 1
Y (s) =
+ 2
(1 + e−πs )
17 (s + 12 )2 + 1
s +1
Taking the inverse transform term-by-term, the solution of the IVP is
i
4 h −1t
e 2 (4 cos t + sin t) − 4 cos t + sin t .
y(t) = h(t) + uπ (t)h(t − π), h(t) =
17
• (§6.5: 4a) Solve the initial value problem
y 00 − y = −20δ(t − 3);
Answer.
y(0) = 1,
y 0 (0) = 0.
Taking the Laplace transform of the DE and using the initial data,
(s2 − 1)Y = s − 20e−3s .
Thus
Y (s) =
1/2
−10
10
1/2
+
+(
+
)e−3s .
s−1 s+1
s−1 s+1
1
y(t) = (et + e−t ) + (−10et−3 + 10e−(t−3) )u3 (t).
2
It can also be written as cosh t − 20 sinh(t − 3)u3 (t).
• (§6.5: 6a) Solve the initial value problem
y 00 + 4y = δ(t − 4π);
Answer.
y(0) = 1/2,
y 0 (0) = 0.
Taking the Laplace transform of the DE and using the initial data,
(s2 + 4)Y = s/2 + e−4πs .
Thus
s/2
1
+
e−4πs .
s2 + 4 s2 + 4
1
1
1
1
y(t) = cos 2t + sin[2(t − 4π)]u4π (t) = cos 2t + sin(2t)u4π (t).
2
2
2
2
Y (s) =
3
• (§6.5: 10a) Solve the initial value problem
2y 00 + y 0 + 4y = δ(t − π/6) sin t;
y(0) = 0,
y 0 (0) = 0.
Answer.
Taking the Laplace transform of the DE and using the zero initial data
and δ(t − π/6) sin t = δ(t − π/6) sin π/6 = δ(t − π/6)/2,
1
(2s2 + s + 4)Y = e−πs/6 .
2
Thus
Y (s) = G(s)e−πs/6 ,
Since L−1 G =
√1 e−t/4 sin
31
G(s) =
√
31
4 t,
2(2s2
1
1/4
=
+ s + 4)
(s + 41 )2 +
√
31
1 −(t−π/6)/4
y(t) = uπ/6 (t) √ e
sin
(t − π/6).
4
31
4
31 .
16