# Exercises 1

```Exercises 1
More exercises are available in Elementary Differential Equations. If you have a problem to solve any of
them, feel free to come to office hour.
1
Problem 1
Write down a differential equation of the form
t → ∞.
dy
dt
= ay + b whose solutions have the required behavior as
1. All solutions approach y = 3.
2. All other solution diverge from y = 2.
1.1
Solution
b
at
We know that the general solution of dy
dt = ay + b is y = − a + ce . If a &lt; 0, the exponential term goes to
b
zero when t goes to infinity ⇒ y tends to a . If a &gt; 0, the exponential term goes to infinity when t goes to
infinity ⇒ y diverges from ab .
1. We want y to be something like y = 3 + cet with a &lt; 0. If we choose a = −1 and c = 1, we get:
dy
= 3 − y.
dt
(1)
2. We want y to be something like y = 2 + ceat with a &gt; 0. If we choose a = 1 and c = 1, we get:
dy
= −2 + y.
dt
2
(2)
Problem 2
A spherical raindrop evaporates at a rate proportional to its surface area. Write a differential equation for
the volume of the raindrop as a function of time.
2.1
Solution
We know that after a time ∆t, the raindrop will lose a volume ∆V , which is proportional to the surface of
the raindrop 4πr2 (r is radius of the sphere). Thus, we can write:
∆V = −a &middot; 4πr2 &middot; (∆t)
(3)
with a &gt; 0. The minus sign is there because the volume is decreasing. The volume of a sphere is given by:
V =
4 3
πr .
3
(4)
Therefore, we have:
r=
3
V
4π
1
1/3
(5)
We finally get (∆t and ∆V become infinitesimal):
dV
= −kV 2/3 .
dt
(6)
with k &gt; 0.
3
Problem 3
For small, slowly falling objects, the assumption that the drag force is proportional to the velocity is a good
one. For larger, more rapidly falling objects, it is more accurate to assume that the drag force is proportional
to the square of the velocity.
1. Write a differential equation for the velocity of a falling object of mass m if the drag force is proportional
to the square of the velocity.
2. Determine the limiting velocity after a long time.
3.1
Solution
1. When the drag force was proportional to velocity (v), the equation was:
m
dv
= mg − γv.
dt
(7)
Now the drag force is proportional to the square of the velocity (v 2 ):
m
dv
= mg − γv 2 .
dt
(8)
You cannot use the general formula to find the solution because equation (8) is first order but nonlinear:
you cannot write equation (8) as a linear combination of v and dv
dt .
2. When the limiting velocity is reached, the system is said to have reached the equilibrium. When the
equilibrium is reached the velocity is constant and thus, we have dv
dt = 0. Therefore, we get:
r
mg
(9)
v(t) =
γ
4
Problem 4
Solve each of the following initial value problems and plot the solutions for several values of y0 . Then,
describe in a few words how the solution resemble, and differ from each, other.
1.
dy
dt
= −y + 5, y(0) = y0
2.
dy
dt
= −2y + 5, y(0) = y0
3.
dy
dt
= −2y + 10, y(0) = y0
4.1
Solution
To solve this problem, we just use the general solution.
1. y = 5 + (y0 − 5)e−t
2. y = 5 + (y0 − 5)e−2t
3. y =
5
2
+ (y0 − 5)e−2t
The equilibrium solution is 5 for (1) and (2). The equilibrium is
for (2) and (3) than it is for (1).
2
5
2
for (3). The equilibrium is reached faster
5
Problem 5
Here is an alternative way to solve the equation:
dy
= ay − b.
dt
(10)
1. Solve the simpler equation:
dy
= ay.
dt
(11)
Call the solution y1 (t),
2. Observe that the only difference between equation (10) and equation (11) is the constant −b in equation (10). Therefore it may seem reasonable to assume that the solutions of these two equations also
differ only by a constant. Test thus assumption by trying to find a constant k such that y(t) = y1 (t)+k
is a solution of equation (10).
5.1
Solution
1. We have successively:
dy
= ay,
dt
1 dy
= a,
y dt
d
(ln y) = a,
dt
Z
Z
d
(ln y) = dt a,
dt
dt
(12)
(13)
(14)
(15)
ln y = at + c,
(16)
y = ec eat .
(17)
at
Therefore, y1 = Ce .
2. Now, we want to find k such y1 (t) + k is a solution of equation (10). Thus, we want:
d
Ceat + k = a Ceat + k − b.
dt
(18)
aCeat = aCeat + ak − b,
(19)
We have:
k=
The solution of equation (10) is y =
6
b
a
b
.
a
(20)
+ Ceat , which is what we expected.
Problem 6
A falling object satisfies the initial value problem:
v
dv
= 9.8 − ,
dt
5
v(0) = 0.
1. Find the time that must elapse for the object to reach 98% of its limiting velocity.
2. How far does the object fall in the time found in part (1).
3
(21)
(22)
6.1
Solution
1. The general solution of the equation is:
t
v(t) = 49(1 − e− 5 ).
(23)
We want:
v(T ) = 0.98v(t∞ ),
= 0.98 &middot; 49.
(24)
Therefore, we get:
T
1 − e− 5 = 0.98,
0.02 = e
− T5
,
T
ln 0.02 = − ,
5
T ≈ 19.56s
2. To find the distance travelled by the object in 19.56s, we need to integrate the speed:
Z 19.56
t
x=
dt 49(1 − e− 5 ),
(25)
(26)
(27)
(28)
(29)
0
19.56
x = 49 t + 5e−t 5 0
,
(30)
= 718.34
7
Problem 7
The half-life of a radioactive material is the time required for an amount of this material to decay to one-half
its original value. Show that for any radioactive material that decays according to the equation dQ
dt = −rQ,
the half-life τ and the decay rate r satisfy the equation rτ = ln 2.
7.1
Solution
We know that the solution of:
dQ
= −rQ,
dt
is Q(t) = Q0 e−rt . After one half-time, Q(τ ) = Q20 . Thus, we have:
Q0
= Q0 e−rτ ,
2
2 = erτ ,
rτ = ln 2.
8
(31)
(32)
(33)
(34)
Problem 8
Consider an electric circuit containing a capacitor, resistor, and battery. The charge Q(t) on the capacitor
satisfies the equation:
dQ Q
+
= V,
(35)
R
dt
C
where R is the resistance, C is the capacitance, and V is the constant voltage supplied by the battery.
1. If Q(0) = 0, find Q(t) at any time t, sketch the graph of Q versus t.
2. Find the limiting value QL that Q(t) approaches after a long time.
3. Suppose that Q(t1 ) = QL and that at time t = t1 the battery is removed and the circuit closed again.
Find Q(t) for t &gt; t1 and sketch its graph.
4
8.1
Solution
1. Let’s rewrite equation (35):
dQ
Q
V
=−
+ ,
dt
RC
R
with Q(0) = 0. We know that the solution is:
t
(36)
Q(t) = V C + ke− RC .
(37)
0 = V C + k.
(38)
t
Q(t) = V C 1 − e− RC .
(39)
Using the initial condition, we find:
Thus, we finally have:
t
2. When t → ∞, e− RC → 0. Thus, the limiting value QL is V C.
3. We are solving an equation similar to equation (35) but the initial condition is different. Since there
is no battery, equation (35) becomes:
dQ Q
+
= 0,
(40)
R
dt
C
with Q(t1 ) = V C. The general solution is:
t
Q(t) = ke− RC .
(41)
Using the initial condition, we find:
t1
V C = ke− RC .
(42)
Thus, we finally get:
Q(t) = V Ce−
9
t−t1
RC
.
Problem 9
Determine the order of the given differential equation and whether the equation is linear or nonlinear.
2
1. t2 ddt2y + t dy
dt + 2y = sin t
d3 y
dt3
d2 y
dt2
2.
d4 y
dt4
+
3.
d2 y
dt2
+ sin(t + y) = sin t
9.1
+
+
dy
dt
+y =1
Solution
1. Second order because of
d2 y
dt2 ,
linear.
2. Fourth order because of
d4 y
dt4 ,
linear.
3. Second order because of
10
2
d y
dt2 ,
nonlinear because of sin(t + y).
Problem 10
Verify that each given function is a solution of the differential equation:
1. y00 − y = 0; y(t) = et .
2. ty0 − y = t2 ; y = 3t + t2 .
3. y0000 + 4y000 + 3y = t; y(t) = 3t .
5
(43)
10.1
Solution
1.
dy
= et ,
dt
d2 y
= et .
dt2
2.
(44)
(45)
et − et = 0.
(46)
dy
= 3 + 2t.
dt
(47)
t(3 + 2t) − (3t + t2 ) = t2 .
(48)
3.
1
dy
= ,
dt
3
d2 y
= 0,
dt2
3
d y
= 0,
dt3
d4 y
= 0.
dt4
3
11
t
= t.
3
(49)
(50)
(51)
(52)
(53)
Problem 11
Determine the order of the given partial differential equation and state whether the equation is linear or
nonlinear. Partial derivatives are denoted by subscripts.
1. uxx + uyy + uzz = 0
2. uxx + uyy + uux + uuy + u = 0
3. uxxxx + 2uxxyy + uyyyy = 0
11.1
Solution
1. Second order, linear.
2. Second order, nonlinear because of uux and uuy .
3. Fourth order.
6
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