Suggested Problems Solution – Sec. 1.6 and p. 78
Math 81, Applied Analysis
Instructor: Dr. Doreen De Leon
1
Section 1.6: 8, 19, 20, 27, 70, 34, 38
y
8. x2 y 0 = xy + x2 e x
Divide both sides by x2 :
y0 =
y
y
+ ex .
x
This equation is homogeneous, so
y
let v =
=⇒ y = xv =⇒ y 0 = xv 0 + v.
x
Plug in:
xv 0 + v = v + ev
xv 0 = ev ← separable
dv
x
= ev
dx
1
−v
e dv = dx
Z
Zx
1
−v
e dv =
dx
x
−e−v = ln |x| + c1 (Note: Remember the absolute value and ”+c.”)
e−v = − ln |x| + c2 (Note: You must multiply both sides by -1 before taking the ln
(can’t take the log of a negative number).)
ln e−v = ln(− ln |x| + c)
−v = ln(c − ln |x|)
v = − ln(c − ln |x|)
y
= − ln(c − ln |x|).
x
So, y = −x ln(c − ln |x|).
19. x2 y 0 + 2xy = 5y 3
Divide both sides by x2 :
2
5
y0 + y = 2 y3.
x
x
This is a Bernoulli equation, with n = 3. So,
Let v = y 1−n = y 1−3 , or v = y −2 .
1
(1)
Then, we apply the chain rule to obtain
v 0 = −2y −3 y 0 .
Solving (1) for y 0 gives
y0 =
5 3 2
y − y.
x2
x
So,
v
0
4
=⇒ v 0 − v
x
5 3 2
= −2y
y − y
x2
x
10 4
= − 2 + y −2
x
x
10 4
=− 2 + v
x
x
10
= − 2 . ← a linear DE
x
−3
Integrating factor:
p(x) = −
R 4
4
=⇒ µ(x) = e − x dx = e−4 ln x = x−4 .
x
Multiply by µ(x) and solve:
x−4 v 0 − 4x−5 v = −10x−6
d
x−4 v = −10x−6
dx
Z
Z
d
−4
x v dx = −10x−6 dx
dx
x−4 v = 2x−5 + c
v = 2x−1 + cx4
Therefore,
y −2 = 2x−1 + cx4 .
20. y 2 y 0 + 2xy 3 = 6x
Divide both sides by y 2 :
y 0 + 2xy = 6xy −2 .
This is a Bernoulli equation with n = −2. So,
Let v = y 1−n = y 1−(−2) , or v = y 3 .
Then, we apply the chain rule to obtain
v 0 = 3y 2 y 0 .
Solving (2) for y 0 gives
y 0 = 6xy −2 − 2xy.
2
(2)
So,
v 0 = 3y 2 6xy −2 − 2xy
= 18x − 6xy 3
= 18x − 6xv
0
=⇒ v + 6xv = 18x. ← a linear DE
Integrating factor:
R
p(x) = 6x =⇒ µ(x) = e
6xdx
2
= e3x .
Multiply by µ(x) and solve:
2
2
2
e3x v 0 + 6xe3x v = 18xe3x
d 3x2 2
e v = 18xe3x
dx
Z
Z
d 3x2 2
e v dx = 18xe3x dx
dx
2
2
e3x v = 3e3x + c
2
v = 3 + ce−3x .
Therefore,
−3x2
3
y = 3 + ce
1
−3x2 3
.
, or y = 3 + ce
27. 3xy 2 y 0 = 3x4 + y 3
Divide both sides by 3xy 2 :
y 0 = x3 y −2 +
1
1
y =⇒ y 0 − y = x3 y −2 .
3x
3x
Again, we have a Bernoulli equation with n = −2. So,
Let v = y 1−n = y 1−(−2) , or v = y 3 .
Then, we apply the chain rule to obtain
v 0 = 3y 2 y 0 .
Since y 0 = x3 y −2 +
1
y, we have
3x
1
3 −2
v = 3y x y + y
3x
1
= 3x3 + y 3
x
1
3
= 3x + v
x
1
0
3
=⇒ v − v = 3x . ← a linear DE
x
0
2
3
Integrating factor:
p(x) = −
R 1
1
=⇒ µ(x) = e − x dx = e− ln x = x−1 .
x
Multiply by µ(x) and solve:
x−1 v 0 − x−2 v = 3x2
d
x−1 v = 3x2
dx
Z
Z
d
−1
x v dx = 3x2 dx
dx
x−1 v = x3 + c
v = x4 + cx.
Therefore,
y 3 = x4 + cx, or y = x4 + cx
13
.
70. As in the text discussion, suppose that an airplane maintains a heading toward an
airport at the origin. If v0 = 500 mi/h and w = 50 mi/h (with the wind blowing
due north), and the plane begins at the point (200, 150), show that its trajectory is
described by
p
1
y + x2 + y 2 = 2(200x9 ) 10 .
The equation for the trajectory is
y 2 21
y
dy
= −k 1+
.
dx
x
x
This is homogeneous, so
let v =
y
=⇒ y = xv =⇒ y 0 = xv 0 + v.
x
Plug in:
1
xv 0 + v = v − k(1 + v 2 ) 2
1
xv 0 = −k(1 + v 2 ) 2
Z
Z
dv
k
− dx
1 =
x
(1 + v 2 ) 2
√
ln v + 1 + v 2 = −k ln |x| + c1 (Recall the need for absolute values until exponentiation.)
√
2
eln|v+ 1+v |
√
v + 1 + v2
r
y 2
y
+ 1+
x
x
r
y 2
y+x 1+
x
p
y + x2 + y 2
= e−k ln |x|+c1
= cx−k
= cx−k
= cx1−k
= cx1−k .
4
Since k =
w
50
1
=
= , we obtain
v0
500
10
p
p
1
9
y + x2 + y 2 = cx1− 10 =⇒ y + x2 + y 2 = cx 10 .
Solve for c: y(200) = 150 so
150 +
√
9
1502 + 2002 = c(200) 10
9
400 = c(200) 10
1
400
10 .
c=
9 = 2(200)
200 10
So,
y+
p
p
1
1
9
x2 + y 2 = 2(200) 10 x 10 , or y + x2 + y 2 = 2(200x9 ) 10 .
34. (2xy 2 + 3x2 )dx + (2x2 y + 4y 3 )dy = 0
M (x, y) = 2xy 2 + 3x2 , N (x, y) = 2x2 y + 4y 3 .
Check if exact:
∂M
∂N
∂M
= 4xy,
= 4xy =
=⇒ exact.
∂y
∂x
∂y
Next, find F (x, y):
Z
F (x, y) =
M (x, y)dx
Z
=
(2xy 2 + 3x2 )dx
= x2 y 2 + x3 + g(y).
Find g(y):
∂F (x, y)
= N (x, y) =⇒ 2x2 y + g 0 (y) = 2x2 y + 4y 3
∂y
Z
0
3
g (y) = 4y =⇒ g(y) = 4y 3 dy = y 4 .
So, F (x, y) = x2 y 2 + x3 + y 4 .
Solution: F (x, y) = c, so
x2 y 2 + x3 + x4 = c.
38. (x + tan−1 y)dx +
x+y
dy = 0
1 + y2
M (x, y) = x + tan−1 y, N (x, y) =
x+y
.
1 + y2
Check if exact:
∂M
1
∂N
1
∂M
=
,
=
=
=⇒ exact.
2
2
∂y
1 + y ∂x
1+y
∂y
5
Next, find F (x, y):
Z
F (x, y) =
N (x, y)dy
Z
=
x+y
dy
1 + y2
= x tan−1 y +
1
ln(1 + y 2 ) + h(x).
2
Find h(x):
∂F (x, y)
= M (x, y) =⇒ tan−1 y + h0 (x) = x + tan−1 y
∂x
Z
1
0
h (x) = x =⇒ h(x) = xdx = x2 .
2
So, F (x, y) = x tan−1 y +
1
1
ln(1 + y 2 ) + x2 . Solution: F (x, y) = c, so
2
2
x tan−1 y +
2
1
1
ln(1 + y 2 ) + x2 = c.
2
2
p. 78: 1, 3, 10, 17, 35
1. x3 + 3y − xy 0 = 0
Rewrite in a “standard” form:
xy 0 − 3y = x3
3
y 0 − y = x2 .
x
This is a linear equation.
Integrating factor:
p(x) = −
R 3
3
=⇒ µ(x) = e − x dx = e−3 ln x = x−3 .
x
Multiply by µ(x) and solve:
3
x−3 y 0 − x−3 y = x2 (x−3 )
x
−3 0
x y − 3x−4 y = x−1
d −3
1
(x y) =
dx
Z
Zx
d −3
1
(x y)dx =
dx
dx
x
x−3 y = ln |x| + c
y = (ln |x| + c)x3 .
6
3. xy + y 2 − x2 y 0 = 0
Rewrite in a “standard” form:
x2 y 0 − xy = y 2
y
y2
y0 − = 2 .
x
x
This equation is homogeneous (as well as Bernoulli). I will solve it as a homogeneous
equation, leaving the Bernoulli version for you.
dy
dv
y
=⇒ y = xv =⇒
= x + v.
Let v =
x
dx
dx
Plugging in gives
dv
x + v − v = v2
dx
dv
x
= v2.
dx
This is a separable equation.
1
1
dv = dx
2
Zx
Z v
1
1
dv =
dx
2
v
x
1
− = ln |x| + c
v
1
v=−
ln |x| + c
1
y
=−
.
=⇒
x
ln |x| + c
So,
x
y=−
.
ln |x| + c
10. y 0 = 1 + x2 + y 2 + x2 y 2
At first glance, this looks a bit scary. However, if you notice that the right-hand side
can be factored to give
y 0 = (1 + x2 )(1 + y 2 ),
you can see that this equation is separable. So,
dy
= (1 + x2 )(1 + y 2 )
dx
1
dy = (1 + x2 )dx
2
1+y
Z
Z
1
dy = (1 + x2 )dx
1 + y2
1
tan−1 y = x + x3 + c
3
1
y = tan x + x3 + c .
3
7
17. ex + yexy + (ey + xeyx )y 0 = 0
The only possibility for solving this one is if the equation is exact.
M (x, y) = ex + yexy , N (x, y) = ey + xeyx .
Check if exact:
∂N
∂M
∂M
= exy + xyexy ,
= eyx + xyeyx =
=⇒ exact.
∂y
∂x
∂y
Next, find F (x, y):
Z
F (x, y) =
M (x, y)dx
Z
=
(ex + yexy )dx
= ex + exy + g(y).
Find g(y):
∂F (x, y)
= N (x, y) =⇒ xexy + g 0 (y) = ey + xeyx
∂y
Z
0
y
g (y) = e =⇒ g(y) = ey dy = ey .
So, F (x, y) = ex + exy + ey . Solution: F (x, y) = c, so
ex + exy + ey = c.
35.
2xy + 2x
dy
=
dx
x2 + 1
This equation is separable and linear! Check it out.
• Separable: Rewrite the equation as
dy
2x
= 2
(y + 1).
dx
x +1
Then,
1
2x
dy = 2
dx
y+1
x +1
Z
Z
1
2x
dy =
dx
2
y+1
x +1
ln |y + 1| = ln(x2 + 1) + c1
2
eln |y+1| = eln(x +1)+c1
y + 1 = c(x2 + 1)
y = c(x2 + 1) − 2.
8
• Linear: Rewrite the equation as
y0 −
2x
2x
y
=
.
x2 + 1
x2 + 1
Integrating factor:
p(x) = −
R
2x
1
2
(− 22x )dx
x +1
=⇒
µ(x)
=
e
.
= e− ln(x +1) = 2
2
x +1
x +1
Multiply by µ(x) and solve:
y0
2x
2x
− 2
y= 2
2
2
x + 1 (x + 1)
(x + 1)2
d
y
2x
= 2
2
dx x + 1
(x + 1)2
Z
Z
2x
d
y
dx =
dx
2
2
dx x + 1
(x + 1)2
1
y
=− 2
+c
2
x +1
x +1
y = c(x2 + 1) − 1.
9