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Math 3D - Homework 7 Solutions Paul Macklin March 6, 2005
Page 232, #4 (5 points): Determine the Laplace transform of eat^ sin(bt).
Solution: Let us consider the following related problem:
L
eateibt
= L
e(a+ib)t
s − (a + ib)
(s − a) − ib = 1 (s − a) − ib
(s − a) + ib (s − a) + ib
= s − a (s − a)^2 + b^2
Also, L
eateibt
= L
eat^ cos(bt)
eat^ sin(bt)
By matching real and imaginary parts, we have
L(eat^ sin(bt)) = b (s − a)^2 + b^2
Page 232, #24 (15 points): Solve the initial-value problem
y′′^ − 3 y′^ + 2y = e−t; y(t 0 ) = 1, y′(t 0 ) = 0
by the method of Laplace transforms. Hint: Let ϕ(t) = y(t + t 0 ).
Solution: Let ϕ(t) = y(t + t 0 ). Then notice that
ϕ′(t) = y′(t + t 0 ) and ϕ′′(t) = y′′(t + t 0 ).
Also, notice that the differential equation states that for all t,
y′′(t) − 3 y′(t) + 2y(t) = e−t.
In particular, this holds for t = t + t 0 :
y′′(t + t 0 ) − 3 y′(t + t 0 ) + 2y(t + t 0 ) = e−te−t^0.
Let α = e−t^0. Then if we use the substitution ϕ(t) = y(t + t 0 ), we have
ϕ′′^ − 3 ϕ′^ + 2ϕ = αe−t.
Notice that the initial conditions for ϕ are
ϕ(0) = y(0 + t 0 ) = 1 and ϕ′(0) = y′(0 + t 0 ) = 0.
Let us solve this with Laplace transforms. If Φ = L(ϕ), then
L(ϕ′′^ − 3 ϕ′^ + 2ϕ) = L(ϕ′′) − 3 L(ϕ′) + 2L(ϕ)
s^2 Φ − sϕ(0) − ϕ′(0)
− 3 (sΦ − ϕ(0)) + 2Φ = s^2 Φ − s − 3 sΦ + 3 + 2Φ = Φ
s^2 − 3 s + 2
- 3 − s = L(αe−t) = α 1 s + 1
Therefore,
Φ = s − 3 (s − 2)(s − 1)
(s + 1)(s − 2)(s − 1)
If we rewrite the first part by partial fractions, we have
s − 3 (s − 2)(s − 1)
s − 1
s − 2
and so L−^1
s − 3 (s − 2)(s − 1)
= 2et^ − e^2 t.
Similarly, 1 (s + 1)(s − 1)(s − 2)
s + 1
s − 1
s − 2
and so L−^1
α
(s + 1)(s − 1)(s − 2)
αe−t^ −
αet^ +
αe^2 t.
Thus,
ϕ(t) = L−^1 (Φ(s)) = 2 et^ − e^2 t^ +
αe−t^ −
αet^ +
αe^2 t
= 2 et^ − e^2 t^ +
e−t^ −
et^ +
e^2 t
e−t^0.
Now, notice that because ϕ(t) = y(t + t 0 ) for all t, this holds in particular for t = t − t 0 :
ϕ(t − t 0 ) = y(t − t 0 + t 0 ) = y(t).
Thus,
y(t) = ϕ(t − t 0 ) = 2 et−t^0 − e2(t−t^0 )^ +
e−t+t^0 −
et−t^0 +
e^2 t−^2 t^0
e−t^0
= 2 et−t^0 − e2(t−t^0 )^ +^1 6 e−t^ − 1 2 et−^2 t^0 +^1 3 e^2 t−^3 t^0.
Scoring part value Do the substitution for each derivative 2 points Treat e−t^ consistently 2 points Do the Laplace transform 5 points Solve for Φ 1 point Partial fractions 2 points Invert transformation 2 points Solve for y(t) 1 point
Page 236, #10 (5 points): Find the inverse Laplace transform of
1 s(s^2 + 4)
Solution: We need to write 1 s(s^2 + 4) =^
A
s +^
Bs + C s^2 + 4.
After cross-multiplying, we have 1 = A(s^2 + 4) + (Bs + C)s,
Let us rewrite 1 (s^2 + 1)
(s − 1)^2 + 6
) (^) = As^ +^ B s^2 + 1 +^
Cs + D (s − 1)^2 + 6.
Then for any real s, 1 = (As + B)
(s − 1)^2 + 6
If we multiply this out and group powers of s, we have
s^3 (A + C) + s^2 (− 2 A + B + D) + s(7A − 2 B + C) + 1(7B + D) = 1.
Thus, we obtain a system of equations:
A
B
C
D
whose solution is A =
, B =
, C =
, D =
Thus,
1 (s^2 + 1)
(s − 1)^2 + 6
s s^2 + 1 +
s^2 + 1 −^
(^ s^ + 1 (s − 1)^2 + 6
s s^2 + 1 +
s^2 + 1 −^
(^ s^ −^ 1 + 2 (s − 1)^2 + 6
s s^2 + 1
s^2 + 1
(^ s^ −^1 (s − 1)^2 + 6
(s − 1)^2 + 6
whose inverse transform is
y(t) =
cos(t) +
sin(t) −
et^ cos(
6 t) −
et^ sin(
6 t)
= 1 20
cos(t) + 3 sin(t)
et
cos(
6 t) + √^2 6
sin(
6 t)
Scoring part value Laplace transform both sides 3 points Partial Fractions 4 points Inverse transform 3 points
