MATH 215/255
Fall 2015
Assignment 2
Due date: October 02, 2015
§1.4, §1.6, Braun’s Book §1.9
1. Consider a pond that initially contains 10 million gal of fresh water. Water containing an undesirable chemical flows into the pond at the rate of 5 million gal/yr, and the
mixture in the pond flows out at the same rate. The concentration γ(t) of chemical
t
in the incoming water varies with time according to the expression γ(t) = 2 + e− 2
g/gal. Construct a mathematical model of this flow process and determine the amount
of chemical in the pond at any time.
Answer.
Let y(t) be the amount of chemical in the pond at time t (with units
million gal and year), then we know that y(0) = 0 and
y 0 (t) = 5 × 106 · γ(t) − 5 × 106 ·
t
y(t)
1
= 5 × 106 (2 + e− 2 ) − y
10 × 106
2
So we get
t
1
y 0 + y = 10 × 106 + 5 × 106 e− 2 ,
2
and y(0) = 0.
t
1
Multiply r(t) on the both sides of y 0 + y = 10 × 106 + 5 × 106 e− 2 , then
2
r(t)y 0 +
t
r(t)
y = [10 × 106 + 5 × 106 e− 2 ]r(t).
2
r(t)
y = [r(t)y]0 =
2
t
r(t)y 0 + r0 (t)y, then r0 = 2r , which implies that r(t) = e 2 . So we get
We are looking for an integrating factor r(t), that is, r(t)y 0 +
t
t
t
t
[e 2 y]0 = [10 × 106 + 5 × 106 e− 2 ]e 2 = 10 × 106 e 2 + 5 × 106 .
So we get
t
e2 y =
Z
t
t
[10 × 106 e 2 + 5 × 106 ] dt = 20 × 106 e 2 + 5 × 106 t + C.
So
t
t
y = 2 × 107 + 5 × 106 te− 2 + Ce− 2 .
Since y(0) = 0, then 0 = 20 × 106 + C, that is, C = −2 × 107 . Hence
t
t
y = 2 × 107 + 5 × 106 te− 2 − 2 × 107 e− 2 .
2. Newton’s law of cooling states that the temperature of an object changes at a rate
proportional to the difference between its temperature and that of its surroundings.
Suppose that the temperature of a cup of coffee obeys Newtons law of cooling. If the
coffee has a temperature of 200◦ F when freshly poured, and 1 min later has cooled to
190◦ F in a room at 70◦ F, determine when the coffee reaches a temperature of 150◦ F.
Answer.
Let y(t) be the temperature of the coffee at any time t, then we know
that y(0) = 200, y(1) = 190. By the Newton’s law, we know that there exists some
constant k such that
y 0 (t)
= k.
70 − y(t)
So we get
y 0 = k(70 − y),
y(0) = 200,
and y(1) = 190.
Since y(t) = 70 is a solution to y 0 = k(70 − y) and y(0) = 200, by the uniqueness
dy
= −k(y − 70),
theorem, then y(t) > 70 for all possible t. Since y 0 = k(70 − y), then
dt
Z
Z
1
1
dy = −k dt. So we get
that is, y−70
dy = − k dt, which implies that
y − 70
ln(y − 70) = −kt + C.
Then y = 70 + e−kt+C = 70 + De−kt . Since y(0) = 200 and y(1) = 190, then
200 = 70 + D,
and
190 = 70 + De−k .
13
> 0. Hence
So we get D = 130 and k = ln 12
13
y = 70 + 130e−(ln 12 )t .
We are looking from time t such that y(t) = 150, then
13
150 = 70 + 130e−(ln 12 )t .
So we get
t=
ln 13
8
≈ 6.07.
ln 13
12
So approximately after 6 minutes, the coffee will reach a temperature of 150◦ F.
3. Find the value b such that (xy 2 + bx2 y) + (x + y)x2 y 0 = 0 is exact, and then find the
general solution to that exact equation.
Answer.
Let M (x, y) = xy 2 + bx2 y and N (x, y) = (x + y)x2 = x3 + x2 y, then
My (x, y) = 2xy + bx2 ,
and Nx (x, y) = 3x2 + 2xy.
Since (xy 2 + bx2 y) + (x + y)x2 y 0 = 0 is exact, then My (x, y) = Nx (x, y), that is,
2xy + bx2 = 3x2 + 2xy, which implies that b = 3. Hence there exists some φ(x, y) such
that
• φx (x, y) = xy 2 + 3x2 y.
• φy (x, y) = x3 + x2 y.
2
• the solution to (xy 2 + 3x2 y) + (x + y)x2 y 0 = 0 is given by φ(x, y) = C.
Since φx (x, y) = xy 2 + 3x2 y, then
Z
1
φ(x, y) = [xy 2 + 3x2 y] dx + f (y) = x2 y 2 + x3 y + f (y).
2
Take the partial derivative with respect to y, then
φy (x, y) = x2 y + x3 + f 0 (y).
Since φy (x, y) = x3 + x2 y, then x3 + x2 y = x2 y + x3 + f 0 (y), that is, f 0 (y) = 0, which
implies that we can take f (y) = 0. So
1
φ(x, y) = x2 y 2 + x3 y.
2
So the solution is:
1 2 2
x y + x3 y = C.
2
4. Find the general solution to
Answer.
Let M (x, y) =
y
x
y
+ 6x + (ln x + 2)y 0 = 0 for x > 0.
x
+ 6x and N (x, y) = ln x + 2, then
My (x, y) =
1
,
x
and Nx (x, y) =
1
.
x
y
1
, which implies that
+ 6x + (ln x + 2)y 0 = 0 is exact.
x
x
So there exists some function φ(x, y) such that
So My (x, y) = Nx (x, y) =
• φx (x, y) =
y
x
+ 6x.
• φy (x, y) = ln x + 2.
y
+ 6x + (ln x + 2)y 0 = 0 is φ(x, y) = C.
• the solution to
x
y
Since φx (x, y) = + 6x, then
x
Z h
i
y
φ(x, y) =
+ 6x dx + f (y) = y ln x + 3x2 + f (y).
x
Take the partial derivative with respect to y, then
φy (x, y) = ln x + f 0 (y).
Since φy (x, y) = ln x + 2, then ln x + 2 = ln x + f 0 (y), that is, f 0 (y) = 2. So we can
take f (y) = 2y. Hence
φ(x, y) = y ln x + 3x2 + 2y.
Hence the solution is:
y ln x + 3x2 + 2y = C.
3
5. Verify r(x, y) =
1
is an integrating factor for x2 y 3 + x(1 + y 2 )y 0 = 0 and find the
xy 3
general solution.
Answer.
Multiply
2 3
1
2 y 3 + x(1 + y 2 )y 0 = 0, then x y +
on
the
both
sides
of
x
xy 3
xy 3
x(1 + y 2 ) 0
y = 0, that is, we have
xy 3
x+
1 + y2 0
y = 0.
y3
1 + y2
, then My (x, y) = 0 and Nx (x, y) = 0. So
y3
1 + y2 0
y = 0 is exact. Hence we know that
My (x, y) = Nx (x, y) = 0, that is, x +
y3
1
r(x, y) = 3 is an integrating factor for x2 y 3 + x(1 + y 2 )y 0 = 0.
xy
Let M (x, y) = x and N (x, y) =
Since x +
1 + y2 0
y = 0 is exact, then there exists some function φ(x, y) such that
y3
• φx (x, y) = x.
1 + y2
1
1
= 3+ .
3
y
y
y
• the solution to the differential equation is φ(x, y) = C.
• φy (x, y) =
Since φx (x, y) = x, then
Z
φ(x, y) =
x dx + f (y) =
x2
+ f (y).
2
Take the partial derivative with respect to y, then
φy (x, y) = f 0 (y).
Since φy (x, y) =
1
1
+ , then
3
y
y
f 0 (y) =
So
Z f (y) =
1
1
+
y3 y
1
1
+ .
3
y
y
dy = −
1
+ ln |y|.
2y 2
So we get
φ(x, y) =
Hence the solution is:
x2
1
− 2 + ln |y|.
2
2y
x2
1
− 2 + ln |y| = C.
2
2y
4
6. Use the integrating factor r(x) = x to solve the differential equation (3xy + y 2 ) + (x2 +
xy)y 0 = 0.
Answer.
Since r(x) = x is an integrating factor of (3xy + y 2 ) + (x2 + xy)y 0 = 0,
then x(3xy + y 2 ) + x(x2 + xy)y 0 = 0 is exact. Then there exists some function φ(x, y)
such that
• φx (x, y) = x(3xy + y 2 ) = 3x2 y + xy 2 .
• φy (x, y) = x(x2 + xy) = x3 + x2 y.
• the solution to the differential equation is φ(x, y) = C.
Since φx (x, y) = 3x2 y + xy 2 , then
Z
1
φ(x, y) = [3x2 y + xy 2 ] dx + f (y) = x3 y + x2 y 2 + f (y).
2
Take the partial derivative with respect to y, then
φy (x, y) = x3 + x2 y + f 0 (y).
Since φy (x, y) = x3 + x2 y, then x3 + x2 y = x2 + x2 y + f 0 (y), that is, f 0 (y) = 0. So we
can take f (y) = 0. Hence
1
φ(x, y) = x3 y + x2 y 2 .
2
So the solution to the differential equation is:
1
x3 y + x2 y 2 = C.
2
7. Find an integrating factor r(x) for (4xy + 3y 2 − x) + x(x + 2y)y 0 = 0.
Let M (x, y) = 4xy + 3y 2 − x and N (x, y) = x(x + 2y) = x2 + 2xy, then
Answer.
My (x, y) = 4x + 6y,
and Nx (x, y) = 2x + 2y.
So My (x, y) 6= Nx (x, y), that is, (4xy+3y 2 −x)+x(x+2y)y 0 = 0 is not exact. Multiply
r(x) on the both sides of (4xy + 3y 2 − x) + x(x + 2y)y 0 = 0, then
r(x)(4xy + 3y 2 − x) + r(x)x(x + 2y)y 0 = 0.
We are looking for an integrating factor r(x), that is, r(x)(4xy + 3y 2 − x) + r(x)x(x +
2y)y 0 = 0 is exact, then
∂
∂
∂
[r(x)(4xy + 3y 2 − x)] =
[r(x)x(x + 2y)] =
[r(x)(x2 + 2xy)].
∂y
∂x
∂x
So we get
r(x)(4x + 6y) = r0 (x)(x2 + 2xy) + r(x)(2x + 2y).
Then r(x)(4x+6y −2x−2y) = r0 (x)(x2 +2xy), that is, r(x)(2x+4y) = r0 (x)x(x+2y).
2
So we get 2r = xr0 , that is, r0 = r, which implies that
x
R
r(x) = e
2
x
5
dx
= e2 ln x = x2 .
8. Solve y 0 = −
x2 + y 2 + x
and y(2) = −1.
xy
Answer.
Rewrite the equation, we get (x2 + y 2 + x) + xyy 0 = 0. Let M (x, y) =
x2 + y 2 + x and N (x, y) = xy, then My (x, y) = 2y and Nx (x, y) = y. So My (x, y) 6=
Nx (x, y), that is, (x2 + y 2 + x) + xyy 0 = 0 is not exact. Multiply r(x) on the both
sides of (x2 + y 2 + x) + xyy 0 = 0, then
r(x)(x2 + y 2 + x) + r(x)xyy 0 = 0.
We are looking for an integrating factor r(x), that is, r(x)(x2 + y 2 + x) + r(x)xyy 0 = 0
is exact, then
∂
∂
[r(x)(x2 + y 2 + x)] =
[r(x)xy].
∂y
∂x
So we have
r(x) · 2y = r0 (x)xy + r(x)y.
So we have r(x)y = r0 (x)xy, that is, r = xr0 . Hence r0 =
R
r(x) = e
1
x
dx
r
, which implies that
x
= eln x = x.
Hence we know that x(x2 + y 2 + x) + x2 yy 0 = 0 is exact. Then there exists some
function φ(x, y) such that
• φx (x, y) = x(x2 + y 2 + x) = x3 + xy 2 + x2 .
• φy (x, y) = x2 y.
• the solution to the differential equation is φ(x, y) = C.
Since φx (x, y) = x3 + xy 2 + x2 , then
Z
x4 1 2 2 x3
+ x y +
+ f (y).
φ(x, y) = [x3 + xy 2 + x2 ] dx + f (y) =
4
2
3
Take the partial derivative with respect to y, then
φy (x, y) = x2 y + f 0 (y).
Since φy (x, y) = x2 y, then x2 y = x2 y + f 0 (y), that is, f 0 (y) = 0. So we can take
f (y) = 0. Hence
x4 1 2 2 x3
φ(x, y) =
+ x y + .
4
2
3
So the solution to the differential equation is:
x4 1 2 2 x3
+ x y +
= C.
4
2
3
Since y(2) = −1, then
problem is:
16
4
+
4
2
+
8
3
= C, that is, C =
x4 1 2 2 x3
26
+ x y +
= .
4
2
3
3
6
26
3 .
Hence the solution to our
9. Consider the differential equation y 0 = y(y − 1)(y − 2).
(a) Draw the phase diagram, and classify the critical points stable or unstable.
Answer.
Solve y(y − 1)(y − 2) = 0, then y = 0, or y = 1 or y = 2. So we
have three critical points 0, 1 and 2. It’s easy to see that phase diagram is the
following:
2
1
0
So we know that 0 and 2 are unstable, and 1 is stable.
(b) Find lim y(x) for the solution with the initial condition y(0) = 1.5.
x→∞
Answer.
Since 1.5 ∈ (1, 2), then y 0 (x) < 0. So limx→∞ y(x) = 1.
10. For the differential equation y 0 = y 2 (1 − y 2 ), draw the phase diagram, and classify the
critical points stable or unstable.
Answer.
Solve y 2 (1 − y 2 ) = 0, then y = 0, y = 1 or y = −1. So we have three
critical points 0, 1 and −1. It’s easy to see that phase diagram is the following:
1
0
−1
So we know that −1 and 0 are unstable, and 1 is stable.
7