Di↵erential Equations

EXACT EQUATIONS

Graham S McDonald

A Tutorial Module for learning the technique

of solving exact di↵erential equations

● Table of contents
● Begin Tutorial

c 2004 g.s.mcdonald@salford.ac.uk
 Table of contents
1. Theory
2. Exercises
3. Answers
4. Standard integrals
5. Tips on using solutions
Full worked solutions
Section 1: Theory 3

1. Theory
We consider here the following standard form of ordinary di↵erential
equation (o.d.e.):

P (x, y)dx + Q(x, y)dy = 0

@Q
If @P
@y = @x then the o.de. is said to be exact.

This means that a function u(x, y) exists such that:

@u @u
du = dx + dy
@x @y
= P dx + Q dy = 0 .

One solves @u
@x = P and @u
@y = Q to find u(x, y).

Then du = 0 gives u(x, y) = C, where C is a constant.

This last equation gives the general solution of P dx + Q dy = 0.

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Section 2: Exercises 4

2. Exercises
Click on Exercise links for full worked solutions (there are 11
exercises in total)
Show that each of the following di↵erential equations is exact and
use that property to find the general solution:

Exercise 1.
1 y
dy dx = 0
x x2

Exercise 2.
dy
2xy + y 2 2x = 0
dx

Exercise 3.
2(y + 1)ex dx + 2(ex 2y)dy = 0

● Theory ● Answers ● Integrals ● Tips

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Section 2: Exercises 5

Exercise 4.
(2xy + 6x)dx + (x2 + 4y 3 )dy = 0

Exercise 5.
dy
(8y x y)
2
+x xy 2 = 0
dx

Exercise 6.
(e4x + 2xy 2 )dx + (cos y + 2x2 y)dy = 0

Exercise 7.
(3x2 + y cos x)dx + (sin x 4y 3 )dy = 0

● Theory ● Answers ● Integrals ● Tips

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Section 2: Exercises 6

Exercise 8.
x2
x tan 1
y · dx + · dy = 0
2(1 + y )
2

Exercise 9.
(2x + x2 y 3 )dx + (x3 y 2 + 4y 3 )dy = 0

Exercise 10.
dy
(2x3
3x y + y )
2
= 2x3
3
6x2 y + 3xy 2
dx

Exercise 11.
(y 2 cos x sin x)dx + (2y sin x + 2)dy = 0

● Theory ● Answers ● Integrals ● Tips

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Section 3: Answers 7

3. Answers

1. y = Ax ,

2. y 2 x x2 = A ,

3. (y + 1)ex y2 = A ,

4. x2 y + 3x2 + y 4 = A ,

5. 2 x (1
1 2
y 2 ) + 4y 2 = A ,

6. 1 4x
4e + x2 y 2 + sin y = A ,

7. x3 + y sin x y4 = A ,
x2
8. 2 tan 1
y =A,
x3 y 3
9. x +2
3 + y4 = A ,

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Section 3: Answers 8
x4 y4
10. 2 2x y +
3 3
2
2 2
x y 4 =A,

11. y 2 sin x + cos x + 2y = A.

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Section 4: Standard integrals 9

4. Standard integrals
R R
f (x) f (x)dx f (x) f (x)dx
xn+1 n [g(x)]n+1
xn
n+1 (n 6= 1) [g (x)] g (x) 0
n+1 (n 6= 1)
g 0 (x)
1
x ln |x| g(x) ln |g (x)|
ax
ex
ex a x
ln a (a > 0)
sin x cos x sinh x cosh x
cos x sin x cosh x sinh x
tan x ln |cos x| tanh x ln cosh x
cosec x ln tan x2 cosech x ln tanh x2
sec x ln |sec x + tan x| sech x 2 tan 1 ex
sec2 x tan x sech2 x tanh x
cot x ln |sin x| coth x ln |sinh x|
sin2 x x
2
sin 2x
4 sinh2 x sinh 2x
4
x
2
cos2 x x
2 + sin 2x
4 cosh2 x sinh 2x
4 + x
2

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Section 4: Standard integrals 10

R R
f (x) f (x) dx f (x) f (x) dx
1
a2 +x2
1
a tan 1 x
a a2
1
x2
1
2a ln a+x
a x (0 < |x| < a)

(a > 0) x2
1
a2
1
2a ln x a
x+a (|x| > a > 0)

p
x+ a2 +x2
p 1
a2 x2
sin 1 x
a
p 1
a2 +x2
ln a (a > 0)
p
x+ x2 a2
( a < x < a) p 1
x2 a2
ln a (x > a > 0)

p 2 ⇥ p 2
h p i
2 2
a2 x2 a
2 sin 1 x
a a2 +x2 a
2 sinh 1 xa + x aa2+x
p i p 2
h p i
2 2 2 2
+ x aa2 x x2 a2 a
2 cosh 1 xa + x xa2 a

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Section 5: Tips on using solutions 11

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 12

Full worked solutions

Exercise 1.
Standard form: P (x, y)dx + Q(x, y)dy = 0

y
i.e. P (x, y) = x2 and Q(x, y) = 1
x

@Q
Equation is exact if @P
@y = @x

Check: @P
@y = 1
x2 = @Q
@x ) o.d.e. is exact.

Since equation exact, u(x, y) exists such that

@u @u
du = dx + dy
@x @y

= P dx + Q dy = 0
and equation has solution u = C, C = constant.
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Solutions to exercises 13
y
@u
@x =P gives i) @u
@x = x2
@u
@y =Q gives ii) @u
@y = 1
x

Integrate i) partially with respect to x,

y
u = + (y),
x

where (y) is an arbitrary function of y.

Di↵erentiate with respect to y,

@u 1 @ 1 d
= + = +
@y x @y x dy
(since = (y) only)

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Solutions to exercises 14

Compare with equation ii)

1 d 1
+ =
x dy x
d
i.e. = 0
dy
i.e. = C 0 , C 0 = constant
y
and u = + C 0.
x

du = 0 implies u = C, C = constant

) y
x =A , A=C C0

= constant.
Return to Exercise 1

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Solutions to exercises 15

Exercise 2.

Standard form: (y 2 2x)dx + 2xy dy = 0

@Q
Exact if @P
@y = @x , where P (x, y) = y 2 2x
Q(x, y) = 2xy

@Q
@P
@y = 2y = @x i.e. o.d.e. is exact.
) u(x, y) exists such that du = @u
@x dx + @u
@y dy

= P dx + Q dy = 0,

giving i) @u
@x = y2 2x , ii) @u
@y = 2xy.

Integrate i): u = xy 2 x2 + (y) , is arbitrary function.

Di↵erentiate and compare with ii):

d
@u
@y = 2xy + dy = 2xy
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Solutions to exercises 16

) d
dy =0 and = C0 (constant)
) u = xy 2 x2 + C 0
du = 0 implies u = C, ) xy 2 x2 = A , where A = C C 0.

Return to Exercise 2

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Solutions to exercises 17

Exercise 3.

P (x, y) dx + Q(x, y) dy = 0 where P (x, y) = 2(y + 1)ex

Q(x, y) = 2(ex 2y)

@P
@y = 2ex = @Q
@x , ) o.d.e. is exact.

) u(x, y) exists such that du = @u

@x dx + @u
@y dy
= P dx + Q dy = 0,
Giving i) @u
@x = 2(y + 1)ex , ii) @u
@y = 2(ex 2y) .

Integrate i): u = 2(y + 1)ex + (y)

d
Di↵erentiate: @u
@y = 2ex + dy = 2(ex 2y) , using ii)

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Solutions to exercises 18
d
R R
i.e. dy = 4y i.e. d = 4 y dy i.e. = 2y 2 + C 0

) u = 2(y + 1)ex 2y 2 + C 0

du = 0 gives u = C,
) (y + 1)ex y2 = A , where A = (C C 0 )/2 .

Return to Exercise 3

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Solutions to exercises 19

Exercise 4.

P (x, y)dx + Q(x, y)dy = 0 where P (x, y) = 2xy + 6x ,

Q(x, y) = x2 + 4y 3

@P
@y = 2x = @Q
@x , ) o.d.e. is exact.

) u(x, y) exists such that du = @u

@x dx + @u
@y dy
= P dx + Q dy = 0,
Giving i) @u
@x = 2xy + 6x , ii) @u
@y = x2 + 4y 3 .

Integrate i): u = x2 y + 3x2 + (y)

d
Di↵erentiate: @u
@y = x2 + dy = x2 + 4y 3 , using ii)

d
R R
i.e. dy = 4y 3
i.e. d =4 y 3 dy i.e. = y4 + C 0

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Solutions to exercises 20

) u = x2 y + 3x2 + y 4 + C 0

du = 0 gives u = C,
) x2 y + 3x2 + y 4 = A , where A = C C0 .

Return to Exercise 4

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Solutions to exercises 21

Exercise 5.
(x xy 2 )dx + (8y x2 y)dy = 0

P (x, y) = x xy 2
Q(x, y) = 8y x2 y . @P
@y = 2xy = @Q
@x , ) o.d.e. is exact.

) u(x, y) exists where du = @u

@x dx + @u
@y dy

= P dx + Q dy = 0

Giving i) @u
@x =x xy 2 ; ii) @u
@y = 8y x2 y.

Integrate i): u = 12 x2 (1 y 2 ) + (y)

d
Di↵erentiate: @u
@y = 1 2
2x · 2y + dy = 8y x2 y , using ii)

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Solutions to exercises 22
R R
) d
dy = 8y i.e. d =8 ydy

i.e. (y) = 4y 2 + C 0 and u = 21 x2 (1 y 2 ) + 4y 2 + C 0

du = 0 gives u = C, ) 2 x (1
1 2
y 2 ) + 4y 2 = A , A = C C0 .

Return to Exercise 5

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Solutions to exercises 23

Exercise 6.

P (x, y)dx + Q(x, y)dy = 0 where P (x, y) = e4x + 2xy 2 ,

Q(x, y) = cos y + 2x2 y

@P
@y = 4xy = @Q
@x , ) o.d.e. is exact.

) u(x, y) exists such that du = @u

@x dx + @u
@y dy
= P dx + Q dy = 0,
Giving i) @u
@x = e4x + 2xy 2 , ii) @u
@y = cos y + 2x2 y .

Integrate i): u = 14 e4x + x2 y 2 + (y)

d
Di↵erentiate: @u
@y = 2x2 y + dy = cos y + 2x2 y , using ii)

d
R R
i.e. dy = cos y i.e. d = cos y dy i.e. = sin y + C 0

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Solutions to exercises 24

) u = 41 e4x + x2 y 2 + sin y + C 0

du = 0 gives u = C,
) 1 4x
4e + x2 y 2 + sin y = A , where A = C C0 .

Return to Exercise 6

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Solutions to exercises 25

Exercise 7.

P (x, y) dx + Q(x, y) dy = 0 where P (x, y) = 3x2 + y cos x

Q(x, y) = sin x 4y 3

@P
@y = cos x = @Q
@x , ) o.d.e. is exact.

) u(x, y) exists such that du = @u

@x dx + @u
@y dy

= P dx + Q dy = 0

Giving i) @u
@x = 3x2 + y cos x, ii) @u
@y = sin x 4y 3 .

Integrate i): u = x3 + y sin x + (y)

d
Di↵erentiate: @u
@y = sin x + dy = sin x 4y 3 , using ii)

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Solutions to exercises 26
R R
) d
dy = 4y 3
i.e. d = 4 y 3 dy

i.e. = y4 + C 0 and u = x3 + y sin x y4 + C 0

du = 0 gives u = C, ) x3 + y sin x y4 = A , A = C C0 .

Return to Exercise 7

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Solutions to exercises 27

Exercise 8.

P (x, y) dx + Q(x, y) dy = 0 where P (x, y) = x tan 1 y

x2
Q(x, y) = 2(1+y2 )

@P
@y = x
1+y 2 = @Q
@x , ) o.d.e. is exact.

) u(x, y) exists such that du = @u

@x dx + @u
@y dy
= P dx + Q dy = 0,
x2
Giving i) @u
@x = x tan 1
y, ii) @u
@y = 2(1+y 2 ) .

x2
Integrate i): u= 2 tan 1
y + (y)

x2 d x2
Di↵erentiate: @u
@y = 1
2 (1+y 2 ) + dy = 2(1+y 2 ) , using ii)

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Solutions to exercises 28

) d
dy =0 i.e. (y) = C 0

x2
and u = 2 tan 1
y + C0

du = 0 implies u = C , C = constant
x2
) 2 tan 1
y = A, A = C C0 .

Return to Exercise 8

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Solutions to exercises 29

Exercise 9.

P (x, y)dx + Q(x, y)dy = 0 where P (x, y) = 2x + x2 y 3 ,

Q(x, y) = x3 y 2 + 4y 3

@P
@y = 3x2 y 2 = @Q
@x , ) o.d.e. is exact.

) u(x, y) exists such that du = @u

@x dx + @u
@y dy
= P dx + Q dy = 0,
Giving i) @u
@x = 2x + x2 y 3 , ii) @u
@y = x3 y 2 + 4y 3 .

x3 y 3
Integrate i): u=x + 2
3 + (y)

d
Di↵erentiate: @u
@y = x3 y 2 + dy = x3 y 2 + 4y 3 , using ii)

d
R R
i.e. dy = 4y 3
i.e. d =4 y 3 dy i.e. = y4 + C 0

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Solutions to exercises 30
x3 y 3
) u=x + 2
3 + y4 + C 0

du = 0 gives u = C,
x3 y 3
) x +
2
3 + y 4 = A , where A = C C0 .

Return to Exercise 9

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Solutions to exercises 31

Exercise 10.

(2x3 6x2 y + 3xy 2 )dx + ( 2x3 + 3x2 y y 3 )dy = 0

@P
@y = 6x2 + 6xy = @Q
@x , ) o.d.e. is exact.

) u(x, y) exists such that du = @u

@x dx + @u
@y dy

= P dx + Q dy = 0

Giving i) @u
@x = 2x3 6x2 y + 3xy 2 , ii) @u
@y = 2x3 + 3x2 y y3 .

x4
Integrate i): u= 2 2x3 y + 32 x2 y 2 + (y)

Di↵erentiate: @u
@y = 2x3 + 3x2 y + ddy = 2x3 + 3x2 y y 3 , using ii)

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Solutions to exercises 32
R R
) d
dy = y 3
i.e. d = y 3 dy

i.e. (y) = 1 4
4y + C0

x4 y4
and u(x, y) = 2 2x y +
3 3 2 2
2x y 4 + C0

du = 0 gives u = C,

x4 y4
) 2 2x y +
3 3 2 2
2x y 4 =A, A=C C0 .

Return to Exercise 10

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Solutions to exercises 33

Exercise 11.

P (x, y)dx + Q(x, y)dy = 0 where P (x, y) = y 2 cos x sin x ,

Q(x, y) = 2y sin x + 2

@P
@y = 2y cos x = @Q
@x , ) o.d.e. is exact.

) u(x, y) exists such that du = @u

@x dx + @u
@y dy
= P dx + Q dy = 0,
Giving i) @u
@x = y 2 cos x sin x , ii) @u
@y = 2y sin x + 2 .

Integrate i): u = y 2 sin x + cos x + (y)

d
Di↵erentiate: @u
@y = 2y sin x + dy = 2y sin x + 2 , using ii)

d
R R
i.e. dy =2 i.e. d =2 dy i.e. = 2y + C 0

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Solutions to exercises 34

) u = y 2 sin x + cos x + 2y + C 0

du = 0 gives u = C,

) y 2 sin x + cos x + 2y = A , where A = C C0 .

Return to Exercise 11

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