Differential Equations
HOMOGENEOUS FUNCTIONS
Graham S McDonald
A Tutorial Module for learning to solve
differential equations that involve
homogeneous functions
● Table of contents
● Begin Tutorial
c 2004 g.s.mcdonald@salford.ac.uk
Table of contents
1.
2.
3.
4.
5.
Theory
Exercises
Answers
Standard integrals
Tips on using solutions
Full worked solutions
Section 1: Theory
3
1. Theory
M (x, y) = 3x2 + xy is a homogeneous function since the sum of
the powers of x and y in each term is the same (i.e. x2 is x to power
2 and xy = x1 y 1 giving total power of 1 + 1 = 2).
The degree of this homogeneous function is 2.
Here, we consider differential equations with the following standard
form:
dy
M (x, y)
=
dx
N (x, y)
where M and N are homogeneous functions of the same degree.
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Section 1: Theory
4
To find the solution, change the dependent variable from y to v, where
y = vx .
The LHS of the equation becomes:
dv
dy
=x
+v
dx
dx
using the product rule for differentiation.
Solve the resulting equation by separating the variables v and x.
Finally, re-express the solution in terms of x and y.
Note. This method also works for equations of the form:
y
dy
=f
.
dx
x
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Section 2: Exercises
5
2. Exercises
Click on Exercise links for full worked solutions (there are 11
exercises in total)
Exercise 1.
Find the general solution of
dy
xy + y 2
=
dx
x2
Exercise 2.
Solve 2xy
dy
= x2 + y 2 given that y = 0 at x = 1
dx
Exercise 3.
Solve
dy
x+y
and find the particular solution when y(1) = 1
=
dx
x
● Theory ● Answers ● Integrals ● Tips
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Section 2: Exercises
6
Exercise 4.
dy
1
Solve x
= x − y and find the particular solution when y(2) =
dx
2
Exercise 5.
dy
x − 2y
Solve
=
and find the particular solution when y(1) = −1
dx
x
Exercise 6.
y 1
dy
x+y
=
, prove that tan−1
= ln x2 + y 2 + A,
dx
x−y
x
2
where A is an arbitrary constant
Given that
Exercise 7.
dy
= x2 + y 2
Find the general solution of 2x2
dx
● Theory ● Answers ● Integrals ● Tips
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Section 2: Exercises
7
Exercise 8.
Find the general solution of (2x − y)
dy
= 2y − x
dx
Note. The key to solving the next three equations is to
recognise that each equation can be written in the form
y
dy
=f
≡ f (v)
dx
x
Exercise 9.
Find the general solution of
y
dy
y
= + tan
dx
x
x
Exercise 10.
Find the general solution of x
y
dy
= y + xe x
dx
● Theory ● Answers ● Integrals ● Tips
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Section 2: Exercises
8
Exercise 11.
Find the general solution of x
p
dy
= y + x2 + y 2
dx
● Theory ● Answers ● Integrals ● Tips
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Section 3: Answers
9
3. Answers
x
1. General solution is y = − ln x+C
,
2. General solution is x = C(x2 − y 2 ) , and particular solution is
x = x2 − y 2 ,
3. General solution is y = x ln (kx) , and particular solution is
y = x + x ln x ,
4. General solution is 1 = Kx(x − 2y) , and particular solution is
2xy − x2 = −2 ,
5. General solution is x2 (x − 3y) = K , and particular solution is
x2 (x − 3y) = 4 ,
6. HINT: Try changing the variables from (x, y) to (x, v), where
y = vx ,
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Section 3: Answers
10
7. General solution is 2x = (x − y)(ln x + C) ,
8. General solution is y − x = K(x + y)3 ,
9. General solution is sin
y
x
= kx ,
10. General solution is y = −x ln(− ln kx) ,
11. General solution is sinh−1
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y
x
= ln x + C .
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Section 4: Standard integrals
11
4. Standard integrals
f (x)
n
x
1
x
x
e
sin x
cos x
tan x
cosec x
sec x
sec2 x
cot x
sin2 x
cos2 x
R
f (x)dx
xn+1
n+1
(n 6= −1)
ln |x|
ex
− cos x
sin x
− ln
|cos x|
ln tan x2 ln |sec x + tan x|
tan x
ln |sin x|
x
sin 2x
2 −
4
x
sin 2x
2 +
4
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f (x)
n
0
[g (x)] g (x)
g 0 (x)
g(x)
x
a
sinh x
cosh x
tanh x
cosech x
sech x
sech2 x
coth x
sinh2 x
cosh2 x
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f (x)dx
[g(x)]n+1
n+1
(n 6= −1)
ln |g (x)|
ax
(a > 0)
ln a
cosh x
sinh x
ln cosh x ln tanh x2 2 tan−1 ex
tanh x
ln |sinh x|
sinh 2x
− x2
4
sinh 2x
+ x2
4
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Section 4: Standard integrals
f (x)
1
a2 +x2
√ 1
a2 −x2
√
a2 − x2
12
R
f (x) dx
f (x)
1
a
tan−1
1
a2 −x2
R
(a > 0)
1
x2 −a2
f (x) dx
a+x 1
2a ln a−x (0 < |x| < a)
x−a 1
ln
2a
x+a (|x| > a > 0)
sin−1
x
a
√ 1
a2 +x2
√
2
2
ln x+ aa +x (a > 0)
(−a < x < a)
√ 1
x2 −a2
√
2
2
ln x+ xa −a (x > a > 0)
a2
2
−1
sin
√
+x
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x
a
x
a
a2 −x2
a2
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√
i √
a2 +x2
a2
2
x2 −a2
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a2
2
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h
h
sinh−1
− cosh−1
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x
a
i
√
x a2 +x2
2
a
i
√
2
2
+ x xa2−a
+
x
a
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Section 5: Tips on using solutions
13
5. Tips on using solutions
● When looking at the THEORY, ANSWERS, INTEGRALS or
TIPS pages, use the Back button (at the bottom of the page) to
return to the exercises.
● Use the solutions intelligently. For example, they can help you get
started on an exercise, or they can allow you to check whether your
intermediate results are correct.
● Try to make less use of the full solutions as you work your way
through the Tutorial.
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Solutions to exercises
14
Full worked solutions
Exercise 1.
RHS = quotient of homogeneous functions of same degree (= 2)
d
xvx + v 2 x2
Set y = vx : i.e.
(vx) =
dx
x2
dv
i.e.
x
+ v = v + v2
dx
dv
Separate variables
x
= v 2 (subtract v from both sides)
dx
Z
Z
dv
dx
and integrate :
=
v2
x
1
= ln x + C
i.e.
−
v
x
= ln x + C
Re-express in terms of x,y : −
y
−x
.
i.e.
y =
ln x + C
Return to Exercise 1
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Solutions to exercises
15
Exercise 2.
Standard form:
dy
x2 + y 2
=
dx
2xy
i.e. quotient of homogeneous functions
that have the same degree
Set y = xv:
d
x2 + x2 v 2
(xv) =
dx
2x · xv
i.e.
x
dx
x2 (1 + v 2 )
dv
+
v=
dx dx
2x2 v
i.e.
x
1 + v2
dv
+v =
dx
2v
x
dv
1 + v2
v(2v)
=
−
dx
2v
(2v)
Separate variables
(x, v) and integrate:
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Solutions to exercises
16
i.e.
x
dv
1 − v2
=
dx
2v
Z
Z
2v
dx
dv
=
1 − v2
x
Z
Z
−2v
dx
−
dv =
1 − v2
x
i.e.
d
Note:
(1 − v 2 ) = −2v i.e.
dv
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i.e.
− ln(1 − v 2 ) = ln x + ln C
i.e.
ln[(1 − v 2 )−1 ] = ln(Cx)
i.e.
1
= Cx
1 − v2
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Solutions to exercises
17
Re-express in terms of x and y:
i.e.
i.e.
i.e.
Particular solution:
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x=1
y=0
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1
2 = Cx
1 − xy 2
x2
= Cx
2
x − y2
x
= x2 − y 2 .
C
gives
− C1 = 1 − 0
i.e.
C=1
gives
x2 − y 2 = x .
Return to Exercise 2
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Solutions to exercises
Exercise 3.
Set y = xv:
x
18
dv
+v
dx
=
=
i.e. x
dv
dx
=
x + xv
x
x
(1 + v) = 1 + v
x
1
Separate variables and integrate:
Z
Z
dv
i.e.
i.e.
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v
v
dx
x
= ln x + ln k
= ln (kx)
=
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(ln k = constant)
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Solutions to exercises
19
Re-express in terms of x and y:
y
= ln (kx)
x
i.e. y = x ln (kx) .
Particular solution with y = 1 when x = 1:
i.e.
i.e.
i.e.
1 = ln (k)
k = e1 = e
y = x ln (ex)
= x[ln e + ln x]
= x[1 + ln x]
y = x + x ln x .
Return to Exercise 3
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Solutions to exercises
20
Exercise 4.
dy
dx
=
x−y
x
dv
i.e. x dx
+v =1−v
: Set y = vx:
i.e.
i.e. i.e.
dv
x dx
= 1 − 2v
1
2
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R
dv
1−2v
=
R
dx
x
ln(1 − 2v) = ln x + ln k
h
i
1
ln (1 − 2v)− 2 − ln x = ln k
i.e.
i.e.
ln
1
1
(1−2v) 2 x
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= ln k
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Solutions to exercises
21
i.e.
Re-express in x, y:
i.e.
(square both sides)
i.e.
1
2
i.e.
1 = kx(1 − 2v)
12
1 = kx 1 − 2y
x
12
1 = kx x−2y
x
1 = K x2 x−2y
, (k 2 = K)
x
1 = K x(x − 2y)
1 = K · 2 · (2 − 2
Particular solution:
y(2) =
1
2
x=2
y = 12
gives
1
2
) = K · 2 · 1, i.e. K =
1
2
2 = x2 − 2xy.
Return to Exercise 4
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Solutions to exercises
Exercise 5.
Set y = xv:
22
x
dv
+v
dx
dv
i.e. x
dx
=
x − 2xv
x
1 − 2v
=
1 − 3v
=
Separate variables and integrate:
Z
Z
dv
dx
=
1 − 3v
x
1
i.e.
ln (1 − 3v) = ln x + ln k
(ln k = constant)
(−3)
i.e.
ln (1 − 3v) = −3 ln x − 3 ln k
i.e ln (1 − 3v) + ln x3 = −3 ln k
i.e
ln [x3 (1 − 3v)] = −3 ln k
i.e
x3 (1 − 3v) = K
(K = constant)
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Solutions to exercises
23
Re-express in terms of x and y:
3y
x3 1 −
x
x
−
3y
i.e. x3
x
= K
= K
i.e. x2 (x − 3y)
= K .
Particular solution with y(1) = −1:
∴
1(1 + 3) = K
x (x − 3y) = 4 .
i.e. K = 4
2
Return to Exercise 5
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Solutions to exercises
24
Exercise 6.
Already in standard form, with quotient of two first degree
homogeneous functions.
Set y = xv:
x
dv
+v
dx
=
x + vx
x − vx
dv
dx
=
x(1 + v)
−v
x(1 − v)
=
1 + v − v(1 − v)
1−v
=
1 + v2
1−v
i.e.
x
i.e.
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dv
dx
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Solutions to exercises
25
Separate variables and integrate:
Z
Z
dx
1−v
dv =
2
1+v
x
Z
Z
Z
dv
1
2v
dx
i.e.
−
=
1 + v2
2
1 + v2
x
1
i.e. tan−1 v − ln(1 + v 2 ) = ln x + A
2
Re-express in terms of x and y:
1
y2
−1 y
tan
−
ln 1 + 2 = ln x + A
x
2
x
2
y
1
x
1
+
y2
i.e. tan−1
=
ln
+ ln x2 + A
2
x
2
x
2
2
1
x + y2
2
=
ln
·
6
x
+A
2
6 x2
Return to Exercise 6
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Solutions to exercises
26
Exercise 7.
dy
x2 + y 2
=
dx
2x2
Set y = xv:
x
dv
+v
dx
=
=
i.e. x
dv
dx
=
=
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x2 + x2 v 2
2x2
1 + v2
2
1 + v2
2v
−
2
2
1 + v 2 − 2v
2
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Solutions to exercises
27
Separate variables and integrate:
Z
dv
1 − 2v + v 2
Z
dv
i.e.
(1 − v)2
=
=
Z
1
2
Z
1
2
dx
x
dx
x
[Note: 1 − v is a linear function of v, therefore use standard integral
and divide by coefficient of v. In other words,
w =1−v
= −1
dw
dv
and
R
dv
(1−v)2
=
1
(−1)
Z
dw
i.e. −
w2
1
i.e. − −
w
1
i.e.
1−v
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R
dw
w2 .]
Z
1
dx
2
x
1
ln x + C
2
1
ln x + C
2
=
=
=
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Solutions to exercises
28
Re-express in terms of x and y:
i.e.
i.e.
1
1
ln x + C
=
1 − xy
2
1
x
=
ln x + C
x−y
2
2x = (x − y)(ln x + C 0 ),
(C 0 = 2C).
Return to Exercise 7
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Solutions to exercises
29
Exercise 8.
dy
dx
=
2y−x
2x−y .
dv
x dx
+v =
Set y = vx,
dv
∴ x dx
=
Partial fractions:
2v−1−v(2−v)
2−v
=
v 2 −1
2−v
2−v
v 2 −1
+
B
v+1
=
i.e.
A
v−1
;
=
2v−1
2−v
R
2−v
v 2 −1 dv
=
R
dx
x
A(v+1)+B(v−1)
v 2 −1
A + B = −1
A−B =2
2A = 1
i.e. A = 12 , B = − 32
i.e.
1
2
R
1
v−1
i.e.
1
2
ln(v − 1) −
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−
3
v+1 dv
3
2
=
R
dx
x
ln(v + 1) = ln x + ln k
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Solutions to exercises
1
(v−1) 2
i.e.
ln
i.e.
v−1
(v+1)3 x2
3
(v+1) 2
x
30
= ln k
= k2
y
x
Re-express in x, y:
y
x
−1
= k2
3
2
+1 x
y−x
x
y+x 3
x
i.e.
= k2
x2
i.e. y − x = K(y + x)3 .
Return to Exercise 8
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Solutions to exercises
31
Exercise 9.
RHS is only a function of v = xy , so substitute and separate variables.
Set y = xv:
x
i.e.
dv
+v
dx
dv
x
dx
Separate variables and integrate:
Z
Z
dv
=
tan v
Z
Z
cos v
{ Note:
dx ≡
sin v
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= v + tan v
=
tan v
dx
x
f 0 (v)
dv = ln[f (v)] + C }
f (v)
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Solutions to exercises
32
i.e.
ln[sin v]
sin v
i.e. ln
x
sin v
i.e.
x
i.e. sin v
= ln x + ln k
=
(ln k = constant)
ln k
= k
= kx
Re-express in terms of x and y:
sin
y
x
= kx.
Return to Exercise 9
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Solutions to exercises
33
Exercise 10.
y
y
dy
=
+ e( x )
dx
x
i.e. RHS is function of v = xy , only.
Set y = vx:
dv
+v
dx
dv
i.e.
x
dx
Z
x
e−v dv
i.e.
i.e.
i.e.
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− e−v
e−v
II
= v + ev
= ev
Z
dx
=
x
= ln x + ln k
= ln(kx)
= − ln(kx)
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Solutions to exercises
34
Re-express in terms of x, y:
y
e− x
y
i.e.
−
x
i.e.
y
= − ln(kx)
=
ln[− ln(kx)]
= −x ln[− ln(kx)].
Return to Exercise 10
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Solutions to exercises
35
Exercise 11.
dy
dx
=
=
y
1p 2
+
x + y2
x r
x
y 2
y
+ 1+
x
x
[Note RHS is a function of only v = xy , so substitute and separate
the variables]
i.e. Set y = xv:
x
i.e.
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dv
+v
dx
dv
x
dx
II
p
= v + 1 + v2
p
=
1 + v2
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Solutions to exercises
36
Separate variables and integrate:
Z
dv
√
1 + v2
Z
dv
√
{ Standard integral:
1 + v2
i.e.
sinh−1 (v)
Z
=
dx
x
=
sinh−1 (v) + C }
=
ln x + A
Re-express in terms of x and y
y
sinh−1
= ln x + A .
x
Return to Exercise 11
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