1. Consider the following initial value problem:
ty 0 + (t + 1)y = 2te−t
y(1) = 3
t > 0.
(a) Use Euler’s method to approximate the value of y at t = 1.1 and t = 1.2 (Show your
work).
Write the DE in the form y 0 = 2e−t −
t+1
y.
t
Using h = 0.1:
2
y(1.1) ≈ [2e−1 − (3)](0.1) + 3 = 2.47358
1
y(1.2) ≈ [2e−1.1 −
2.1
(2.47358)](0.1) + 2.47358 = 2.06792
1.1
(b) Solve the initial value problem.
y = 2e−t . Now use integrating factors:
Write the DE in the standard form y 0 + t+1
t
Z
Z
t+1
µ(t) = exp(
dt) = exp( (1 + 1/t)dt) = exp(t + ln(t)) = tet
t
d t
(te y) = (tet )(2e−t ) = 2t
dt
tet y = t2 + C
Applying the initial conditions: (1)(e1 )(3) = (1)2 + C, thus C = 3e − 1.
The solution to the IVP is then y =
t2 +3e−1
.
tet
2. Find the general solution:
(a) y 0 =
x
y(1+x2 )
This DE is separable:
x
dx
1 + x2
Now integrate both sides. On the right, you may want to use the substitution u = 1+x2 .
Then you get:
(1/2)y 2 = (1/2) ln(1 + x2 ) + C
p
Solving for y yields y = ± ln(1 + x2 ) + C (yes, the C must be inside the square root).
ydy =
(b) y 0 + y 2 sin(x) = 0
Also separable:
dy
= sin(x)dx
y2
−y −1 = − cos(x) + C
1
y =
cos(x) + C
(yes, the C must be in the denominator)
3. A vat at the soda factory has a capacity of 600 gallons. The vat initially contains 200
gallons of carbonated water and 10 lbs of corn syrup. Carbonated water containing 0.1
lb/gal of corn syrup is pumped into the vat at a rate of 10 gal/min, and the well mixed
soda is allowed to leave the vat at a rate of 8 gal/min.
(a) Find the formula for the total amount of soda in the vat after at time t minutes (i.e.
the total number of gallons of liquid in the vat at time t).
fluid goes in at 10 gal/min and out at 8 gal/min, so the net change is a gain of 2gal/min.
Since we start with 200gal, the formula for the volume is 200 + 2t.
(b) When does the vat begin to overflow?
The capacity is 600 gal, so the takn starts to overflow when 200 + 2t = 600. i.e.
t = 200min.
(c) Find the formula for the amount of corn syrup in the vat at time t minutes (up until
the time when the vat begins to overflow).
The IVP governing the amount of corn syrup (y(t)) is:
y
4y
dy
= (0.1)(10) − (8)
=1−
dt
200 + 2t
100 + t
with the initial condition y(0) = 10. We can solve the DE using integrating factors:
y0 +
4
y = 1
100 + t
Z
µ(t) = exp(
4
dt) = exp(4 ln(100 + t)) = (100 + t)4
100 + t
d
((100 + t)4 y) = (100 + t)4
dt
(100 + t)4 y = (1/5)(100 + t)5 + C
y = (1/5)(100 + t) +
10 = (1/5)(100) +
C
(100 + t)4
C
1004
C = (1004 )(−10)
y = (1/5)(100 + t) −
10(1004 )
(100 + t)4
4. Consider the following differential equations:
(1) y 0 = 2y − 1
(2)y 0 = 2 + y
(3) y 0 = y − 2
(4) y 0 = y(y + 3)
(5) y 0 = y(y − 3)
(6) y 0 = 1 + 2y
(7) y 0 = −2 − y
(8) y 0 = y(3 − y)
(9) y 0 = 1 − 2y
(10) y 0 = 2 − y.
For each of the graphs below, select the differential equation that was used to create the
given direction field (there is exactly one equation corresponding to each graph- you will
not use all of the equations on the list, and you do not need to show work).
(a)
(b)
(c)
differential equation:
differential equation:
differential equation:
5. Suppose y1 and y2 are two different solutions to the differential equation y 00 + p(t)y 0 +
q(t)y = g(t).
(a) What differential equation has the function 2y1 + y2 as a solution?
y 00 + p(t)y 0 + q(t)y = 3g(t)
(b) Prove that your answer to part (a) is correct (show work or explain).
(2y1 + y2 )00 + p(t)(2y1 + y2 )0 + q(t)(2y1 + y2 )
2y100 + y200 + 2p(t)y10 + p(t)y20 + 2q(t)y1 + q(t)y2
2(y100 + p(t)y10 + q(t)y1 ) + (y200 + p(t)y20 + q(t)y2 )
2(g(t)) + g(t)
=
=
=
= 3g(t)
6. Consider the differential equation ay 00 + by 0 + cy = 0.
(a) If b2 − 4ac = 0, what is the general solution to this differential equation?
C1 e(−b/2a)t + C2 te(−b/2at)
(b) Under the assumption of part (a), what further restrictions must we place on the
coefficients a, b, and c in order to guarantee that all solutions tend to zero as t → ∞?
need (−b/2a) < 0. So a and b must have the same sign. No restrictions on c.
7. Use the method of undetermined coefficients to find the general solution:
y 00 + 2y 0 + y = 2e−t .
Homogeneous solution:
r2 + 2r + 1
(r + 1)2
r
yc
=
=
=
=
0
0
−1
C1 e−t + C2 te−t
Particular solution:
yp = At2 e−t
yp0 = 2Ate−t − At2 e−t
yp00 = 2Ae−t − 4Ate−t + At2 e−t
2Ae−t − 4Ate−t + At2 e−t + 4Ate−t − 2At2 e−t + At2 e−t
2A
A
yp
=
=
=
=
Overall, the general solution is y = C1 e−t + C2 te−t + t2 e−t .
2e−t
2
1
t2 e−t
8. A mass weighing 2 lbs. stretches a spring 6 in. Suppose the mass is pulled down an
additional 3 in. and then released. Assume that there is no damping and the system is
acted on by an external force of 2 cos(t) lbs. at time t seconds.
(a) What is the differential model for the position of the object (i.e. what is the differential equation and what are the initial conditions? Do not solve).
m = 2/32, L = 0.5f t, mg = kL so 2 = 0.5k and k = 4
(2/32)u00 + 4u = 2 cos(t)
u(0) = 0.25, u0 (0) = 0.
(b) What is the frequency at which resonance occurs?
p
p
√
k/m = 4/(2/32) = 64 = 8.
(c) What is the frequency of the steady-state solution? (note: you should be able to
determine this without actually finding the steady-state solution)
Frequency of the steady-state sol. mathces frequency of forcing term = 1.
(d) Suppose that we now eliminate the external force and attach the system to a damper
with damping coefficient γ. What value of γ will cause the system to be critically
damped?
need γ 2 − 4mk = 0 so γ 2 = 4(2/32)(4), γ = 1.
9. Consider the function f (t) = (t + 5)u2 (t) − (t − 2)u3 (t).
(a) Sketch f (t) on the axes below.


0≤t<2
0,
First write f as f (t) = t + 5, 2 ≤ t < 3


7,
3≤t
See me if you’re not sure how to make the sketch from here.
(b) Find the Laplace transform of f (t) (you may refer to the chart).
L{(t + 5)u2 (t) − (t − 2)u3 (t)} = L{(t + 5)u2 (t)} − L{(t − 2)u3 (t)}
= L{(t − 2 + 7)u2 (t)} − L{(t − 3 + 1)u3 (t)}
= L{(t − 2)u2 } + 7L{u2 } − L{(t − 3)u3 } − L{u3 }
= e−2s s12 + 7e−2s 1s − e−3s s12 − e−3s 1s
10. Use the Laplace transform to solve the following initial value problem (you may refer to
the chart):
y 00 + y = f (t)
y(0) = 0
y 0 (0) = 1
(
1, 0 ≤ t < π/2
where f (t) =
0, t ≥ π/2.
First, f (t) = 1 − uπ/2 (t), so L{f (t)} =
1
s
−
e(−π/2)s
.
s
Then we have s2 L{y} − sy(0) − y 0 (0) + L{y} =
(s2 + 1)L{y} = 1 +
L{y} =
1
s
−
e(−π/2)s
.
s
1 e(−π/2)s
−
s
s
1
e(−π/2)s
1
+
−
s2 + 1 s(s2 + 1) s(s2 + 1)
y = sin(t) + L−1 {
(−π/2)s
1
−1 e
}
−
L
{
}
s(s2 + 1)
s(s2 + 1)
Now use partial fractions:
1
+ 1)
1
A+B
C
A
1
s(s2 + 1)
s(s2
=
=
=
=
=
=
A Bs + C
+ 2
s
s +1
2
A(s + 1) + (Bs + C)s
0
0
1
1
s
− 2
s s +1
Then L−1 { s(s21+1) } = 1 − cos(t).
All together: y = sin(t) + 1 − cos(t) − uπ/2 (t)(1 − cos(t − π/2)).