Download 1. [20 pts] Find an integrating factor and solve the equation y

22M:034 Engineer Math IV: Differential Equations
Midterm Exam 1 – October 2, 2013
Name
Section number
1. [20 pts] Find an integrating factor and solve the equation
y 0 − 3y = e2t .
Then solve the initial value problem
y 0 − 3y = e2t ,
y(0) = 3.
Solution: We can use the integrating factor µ(t) = e−3t to convert the
original equation into
e−3t y 0 − 3e−3t y = e−3t e2t
or
(e−3t y)0 = e−t .
Then, by integrating both sides of this equation and solving for y we
obtain the general solution of y 0 − 3y = e2t , namely
y = Ce3t − e2t
where C is an arbitrary constant. To satisfy the condition y(0) = 3,
we must choose C = 4. Thus the unique solution of the initial value
problem is
y = 4e3t − e2t .
2. [20 pts] Solve the initial value problem
dy
2x
=
, y(2) = 0
dx
2y + 1
and determine the interval in which the solution exists.
Solution: The differential equation can be written as
(2y + 1)dy = 2xdx.
Integrating the left side with respect to y and the right side with respect
to x gives
y 2 + y = x2 + C,
where C is an arbitrary constant. To satisfy y(2) = 0 we must have
y(2)2 + y(2) = 22 + C,
C = −4.
Hence the solution of the initial value problem is given implicitly by
y 2 + y = x2 − 4.
To obtain the solution explicitly, we solve the above equation for y in
terms of x and we obtain
1
1√ 2
4x − 15 − .
y=±
2
2
This gives two solutions of the original differential equation, only one of
which, however, satisfies the given initial condition. This is the solution
corresponding to the plus sign; thus we obtain
1 √
y = ( 4x2 − 15 − 1)
2
as the solution of the initial value problem. Finally, to determine the
interval in which this solution is valid, we must find the interval containing 2 in which the quantity √under the radical is positive. We find
that the desired interval is x > 215 .
3. [20 pts] At time t = 0 a tank contains Q0 lb of salt dissolved in
100 gal of water. Assume that water containing 41 lb of salt per gallon
is entering the tank at a rate of 3 gal/min, and that the well-stirred
solution is leaving the tank at the same rate. Find an expression for
the amount of salt Q(t) in the tank at any time t.
Solution:
Rate of change
Rate at which
Rate at which
of amount of salt
salt is flowing in
salt is flowing out
in tank in
dQ
dt
lb
min
lb
min
in
=
lb
min
1
×3
4
lb
gal
×
in
−
gal
min
lb
min
Q(t)
×3
100
lb
gal
×
gal
min
The problem yields a separable first-order linear differential equation
with initial condition
3
3
Q(t), Q(0) = Q0 .
Q0 (t) = −
4 100
The general solution of the differential equation is
−3t
Q(t) = Ce 100 + 25,
where C is an arbitrary constant. To satisfy the initial condition, we
must choose C = Q0 − 25. Thus the unique solution of the initial value
problem,
−3t
Q(t) = (Q0 − 25)e 100 + 25,
gives the expression for the amount of salt Q(t) in the tank at any time t.
4. [20 pts] Solve the differential equation
(3x2 y + y 2 )dx + (x3 + 2xy + ey )dy = 0.
Solution: Let M (x, y) = 3x2 y + y 2 and N (x, y) = x3 + 2xy + ey . Then
∂M
∂N
= 3x2 + 2y
and
= 3x2 + 2y,
∂y
∂x
and since these are equal, the equation is exact. Thus there is a potential function ψ = ψ(x, y) such that

∂ψ


(x, y) = 3x2 y + y 2
∂x
∂ψ


(x, y) = x3 + 2xy + ey .
∂y
Our goal is to solve this system of equations for the function ψ(x, y).
The first of these equations can be integrated with respect to x (holding
y constant) to find
ψ(x, y) = x3 y + xy 2 + h(y),
where the function h is an arbitrary differentiable function of y, playing
the role of an arbitrary constant. By differentiating the above equation
with respect to y and comparing the result with the second equation
in the system, we obtain
x3 + 2xy + h0 (y) = x3 + 2xy + ey ,
h0 (y) = ey ,
and as a solution of the above separable equation we can take h(y) = ey
(we do not require the most general one). Finally, solutions of the
original differential equation, (3x2 y + y 2 )dx + (x3 + 2xy + ey )dy = 0,
are given implicitly by
ψ(x, y) = C
or
where C is an arbitrary constant.
x3 y + xy 2 + ey = C,
5. [20 pts] Solve the equation
dy
= y(2 − y).
dx
Also find the equilibrium solutions, classify each one as asymptotically
stable or unstable, and sketch the equilibrium solutions and several
other solutions in the xy-plane.
Solution: To solve the equation,
and we obtain:
Z
dy
dy
y(2 − y)
Z 1
1
( 2 + 2 )dy
y 2−y
Z
Z
dy 1
dy
1
+
2
y
2
2−y
1
1
ln |y| − ln |2 − y|
2
2
ln |y| − ln |2 − y|
y
|
ln |
2−y
y
eln | 2−y |
y
|
|
2−y
y
2−y
y
y
2x
y + Cye
y(1 + Ce2x )
we integrate both sides of
dy
y(2−y)
= dx
Z
=
dx
= x + C1
= x + C1
= x + C1
= 2x + 2C1
= 2x + C2
= e2x+C2
= e2x eC2
= Ce2x
(C2 = 2C1 )
(eC2 > 0)
(C ∈ R)
Ce2x (2 − y)
2Ce2x − Cye2x
2Ce2x
2Ce2x
2Ce2x
y =
.
1 + Ce2x
=
=
=
=
Thus, the solutions are given by
2e2x
y(x) =
.
(c = C1 )
2x
c+e
There are two equilibrium solutions to this equation
y=0
unstable,
y=2
stable.
We sketch the equilibrium solutions and several other solutions on the
next page.