MAT 275 Written Homework #8 SOLUTIONS
6.3, 6.4, 7.1 Due: April 18
Solve the following problems, showing any necessary work.
1. [2 points] Let
g(t) = (0 if t < 1
t−1 if 1 ≤t < 3
2 if 3 ≤t
Write g(t) as one formula, using the Heaviside (step) function uc(t), and then find its Laplace transform
G(s).
Solution:
g(t) = (u1(t)−u3(t))(t−1) + u3(t)·2=u1(t)·(t−1) + u3(t)·(3 −t)
L(g(t)) = e−1·s· L((t+ 1) −1) + e−3·sL(3 −(t+ 3)) = e−s· L(t)−e−3s· L(t)
=e−s·1
s2−e−3s·1
s2
2. Consider the initial value problem
y00 + 4y0+ 3y=f(t)
y(0) = 10
y0(0) = −4
where f(t) = 0 if t < 2
30 if t≥2.
a. [1 point] Find the Laplace transform of the solution; that is, find a formula for Y(s).
Solution: First, take the Laplace transform of f(t). Since f(t) = 30u2(t),L(f(t)) = 30e−2s·1
s. Then
L(y00 )+4L(y0)+3L(y) = L(f(t))
(s2Y−s·y(0) −y0(0)) + 4(sY −y(0)) + 3Y=30e−2s
s
s2Y−10s+ 4 + 4sY −40 + 3Y=30e−2s
s
s2Y+ 4sY + 3Y=30e−2s
s+ 10s+ 36
(s2+ 4s+ 3)Y=30e−2s
s+ 10s+ 36
Y=
30e−2s
s+ 10s+ 36
s2+ 4s+ 3 =30
s(s2+ 4s+ 3) ·e−2s+10s+ 36
s2+ 4s+ 3.
=⇒
1
b. [1 point] Your formula for Y(s) should be of the form p(s) + e−2sq(s)
r(s), where p(s), q(s), and r(s)
are polynomials. Write Y(s) in the form p(s)
r(s)+e−2s·q(s)
r(s), find Partial Fraction Decompositions
for p(s)
r(s)and q(s)
r(s), and then find a formula for y(t).
Solution: The answer to part (a) is already in the right form; the partial fraction decompositions of
30
s(s2+ 4s+ 3) and 10s+ 36
s2+ 4s+ 4 need to be calculated, however. Note that s2+4s+3 = (s+1)(s+3),
so
30
s(s2+ 4s+ 3) =A
s+B
s+ 1 +C
s+ 3
30 = A(s+ 1)(s+ 3) + Bs(s+ 3) + Cs(s+ 1)
30 = 3Aso A= 10 [substitute s= 0]
30 = 2Bso B= 15 [substitute s=−1]
30 = 6Cso C= 5 [substitute s=−3]
30
s(s2+ 4s+ 3) =10
s+15
s+ 1 +5
s+ 3
Similarly,
10s+ 36
s2+ 4s+ 3 =A
s+ 1 +B
s+ 3
10s+ 36 = A(s+ 3) + B(s+ 1)
26 = 2Aso A= 13 [substitute s=−1]
6 = −2Bso B=−3 [substitute s=−3]
10s+ 36
s2+ 4s+ 3 =13
s+ 1 −3
s+ 3
Then
y(t) = L−130
s(s2+ 4s+ 3) ·e−2s+10s+ 36
s2+ 4s+ 3
=L−1e−2s·10
s+15
s+ 1 +5
s+ 3+L−113
s+ 1 −3
s+ 3
Since L−110
s+15
s+ 1 +5
s+ 3= 10 + 15e−t+ 5e−3t,
y(t) = u2(t)·10 + 15e−(t−2) + 5e−3(t−2)+ 13e−t−3e−3t
=u2(t)·10 + 15e2−t+ 5e6−3t+ 13e−t−3e−3t
=⇒
2
3. [1 point] Consider two interconnecting tank. Tank 1 initially contains 75 L (liters) of water and 200 g
of sugar, while Tank 2 initially contains 25 L of water and 100 g of sugar. Water containing 5 g/L of
sugar is poured into Tank 1 at a rate of 4 L/m while the mixture flowing into Tank 2 contains a sugar
concentration of 3 g/L and is flowing at a rate of 1.5 L/min. The mixture flows from Tank 1 to Tank 2
at a rate of 4 L/min. The mixture drains from Tank 2 at a rate of 5 L/min, of which some flows back
into Tank 1 at a rate of 1 L/min, while the remainder leaves the tank. Let Y1and Y2, respectively, be
the amount of sugar in each tank at time t. Set up BUT DO NOT SOLVE an initial value problem
that models the flow process.
Solution: This is similar to Written Homework #2, Problem #3, except that there are two tanks.
First, the volume of water at time tneeds to be determined for each tank. Tank 1 starts off
with 75 L of water, and every minute, 5 L flows into Tank 1 (from outside, and from Tank 2), and
4 L flows out. Hence, each minute sees a net 1 L of water, and the volume of Tank 1 is thus 75+tL.
Similarly, Tank 2 starts out with 25 L, and every minute, 5.5 L flows into Tank 2, and 5 L
flows out, so 0.5 L flows in every minute, and the volume of Tank 2 is 25 + 0.5tL.
Now, set up the differential equations:
dY1
dt =rate in
from outside+rate in
from Tank 2−rate
out
=4 L
min ·5 g
1 L +1 L
min ·Y2g
25 + 0.5tL−4 L
min ·Y1g
75 + tL
dY2
dt =rate in
from outside+rate in
from Tank 1−rate
out
=1.5 L
min ·3 g
1 L +4 L
min ·Y1g
75 + tL−5 L
min ·Y2g
25 + 0.5tL
The initial value problem is then
Y0
1= 20 + Y2
25 + 0.5t−4Y1
75 + t
Y0
2= 4.5 + 4Y1
75 + t−5Y2
25 + 0.5t
Y1(0) = 200
Y2(0) = 100
3
