MATH 251 Section 004 Home Work 03
Sketch of Solutions
Problem 1. An object with mass m = 0.5 kg is thrown upwards with initial velocity v0 = 20 meters
from the roof of a building which is 30 meters high. Assume there is a force due to air resistance that
is proportional to the velocity v of the object with a positive constant of proportionality (coefficient
1
of drag) k = 10
. You may use g = 10 meters per second squared as the gravitational constant.
Calculate the velocity of the object as a function of time. If the object is allowed to continue
indefinitely, what would be its velocity eventually.
Solution. The following initial value problem describes the motion of the object (assuming the
downward direction to be positive and also note that drag is proportional to the velocity, not the
square of the velocity):
0.5
1
dv
= 0.5 · 10 − v
dt
10
v(0) = −20
Solving the above IVP yeilds:
1
v(t) = 50 − 70e− 5 t
Note that the equation is separable, so the terminal velocity is the asymptotically stable equilibrium
solution v(t) = 50
lim v(t) = 50
t→∞
Problem 2.
(a) Consider the following autonomous differential equation:
dy
= (y − 1)2 (y 2 − 4)
dt
Find all the equilibrium solutions and classify them according to their stability.
(b) Calculate limt→∞ y(t), where y(t) is the solution to the following initial value problem:
dy
= y 3 − y,
dt
y(0) =
9
10
(c) The velocity, given in meters per second, of a certain particle is given by the initial value
problem
dv
1
= 1000 − v 2 ,
v(0) = 5
dt
40
Approximately how fast will the particle be moving after a very long time?
1
Solution.
(a)
dy
= (y − 1)2 (y − 2)(y + 2)
dt
So the equilibrium solutions are as follows:
y = −2
Stable.
y=1
Semi-stable.
y=2
Unstable.
(b)
dy
= y(y − 1)(y + 1)
dt
So the equilibrium solutions are as follows:
y = −1
Unstable.
y=0
Stable.
y=1
Unstable.
So, if y0 =
9
10 ,
lim y(t) = 0
t→∞
(c)
1 2
v = 0 ⇒ v = 200
40
So, v(t) = 200 is an equilibrium solution and in fact stable (check!). So if v(0) = 5,
1000 −
lim v(t) = 200
t→∞
Problem 3. Solve the following initial value problems:
(a)
(b)
dy
= 0,
dx
π
π
y( ) =
2
2
dy
− sin(x) sin(2y) + y cos(x) + 2 cos(x) cos(2y) + sin(x)
= 0,
dx
cos(x + y) + 2x + cos(x + y) + 4y
π
y( ) = π
2
Solution.
(a) Note that
∂M
∂y
= − sin(x + y) =
∂N
∂x .
So the equation is exact.
2
Let ψ(x) = C be the general solution. So ∂ψ
∂x = cos(x + y) + 2x ⇒ ψ = sin(x + y) + x + C1 (y).
∂ψ
The other partial ∂y = cos(x + y) + 4y ⇒ ψ(x) = sin(x + y) + 2y 2 + C2 (x).
Comparing the two expressions, we obtain the implicit general solution: sin(x+y)+x2 +2y 2 =
C.
Solution to IVP: sin(x + y) + x2 + 2y 2 =
3π 2
4 .
2
(b) Note that
∂M
∂y
= 2 sin(x) cos(2y) + cos(x) =
∂N
∂x .
So the equation is exact.
∂ψ
∂x
Let ψ(x) = C be the general solution. So
= − sin(x) sin(2y)+y cos(x) ⇒ ψ = cos(x) sin(2y)+
=
2
cos(x)
cos(2y) + sin(x) ⇒ ψ(x) = cos(x) sin(2y) +
y sin(x) + C1 (y). The other partial ∂ψ
∂y
y sin(x) + C2 (x).
Comparing the two, we obtain the implicit general solution: cos(x) sin(2y) + y sin(x) = C.
Solution to the IVP: cos(x) sin(2y) + y sin(x) = π.
Problem 4. Consider the following differential equation:
2x + αy + 2xex
2
+ 2y + x
dy
=0
dx
Determine the value of α which makes this equation exact.
Solution.
Solve :
∂M
∂y
=
∂N
∂x
to get α = 1.
Problem 5. Consider the following direction field of a first order ODE:
3
2
1
0
-1
0
2
4
6
Are the following statements TRUE or FALSE:
(a) This is the direction field of a separable equation.
(b) This is the direction field of an autonomous equation.
(c) This is the direction field of an exact equation.
(d) This is the direction field of an ODE of the form y 0 = f (t).
3
8
Find out the equilibrium solutions and classify them.
Solution. All but (d) are TRUE.
y(t) = 0 and y(t) = 2 are asymptotically stable equilibrium solutions. y(t) = 1 is an unstable
equilibrium solution.
4