Di↵erential Equations

BERNOULLI EQUATIONS

Graham S McDonald

A Tutorial Module for learning how to solve

Bernoulli 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. Integrating factor method
5. Standard integrals
6. Tips on using solutions
Full worked solutions
Section 1: Theory 3

1. Theory
A Bernoulli di↵erential equation can be written in the following
standard form:
dy
+ P (x)y = Q(x)y n ,
dx
where n 6= 1 (the equation is thus nonlinear).
To find the solution, change the dependent variable from y to z, where
z = y 1 n . This gives a di↵erential equation in x and z that is
linear, and can be solved using the integrating factor method.

Note: Dividing the above standard form by y n gives:

1 dy
+ P (x)y 1 n
= Q(x)
y n dx
1 dz
i.e. + P (x)z = Q(x)
(1 n) dx
⇣ ⌘
n dy
where we have used dz
dx = (1 n)y dx .

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Section 2: Exercises 4

2. Exercises
Click on Exercise links for full worked solutions (there are 9
exercises in total)
Exercise 1.
The general form of a Bernoulli equation is
dy
+ P (x)y = Q(x) y n ,
dx
where P and Q are functions of x, and n is a constant. Show that
the transformation to a new dependent variable z = y 1 n reduces
the equation to one that is linear in z (and hence solvable using the
integrating factor method).
Solve the following Bernoulli di↵erential equations:
Exercise 2.
dy 1
y = xy 2
dx x
● Theory ● Answers ● IF method ● Integrals ● Tips
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Section 2: Exercises 5

Exercise 3.
dy y
+ = y2
dx x
Exercise 4.
dy 1
+ y = ex y 4
dx 3
Exercise 5.
dy
x + y = xy 3
dx

Exercise 6.
dy 2
+ y = x2 cos x · y 2
dx x

● Theory ● Answers ● IF method ● Integrals ● Tips

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Section 2: Exercises 6

Exercise 7.
dy (4x + 5)2 3
2 + tan x · y = y
dx cos x
Exercise 8.
dy
x + y = y 2 x2 ln x
dx

Exercise 9.
dy
= y cot x + y 3 cosecx
dx

● Theory ● Answers ● IF method ● Integrals ● Tips

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Section 3: Answers 7

3. Answers
1. HINT: Firstly, divide each term by y n . Then, di↵erentiate z
dy
with respect to x to show that (1 1 n) dx
dz
= y1n dx ,

x2
2. 1
y = 3 + C
x ,

3. 1
y = x(C ln x) ,

4. 1
y3 = ex (C 3x) ,

5. y 2 = 1
2x+Cx2 ,

6. 1
y = x2 (sin x + C) ,

7. 1
y2 = 12 cos x (4x
1
+ 5)3 + C
cos x ,

8. 1
xy = C + x(1 ln x) ,

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Section 3: Answers 8
sin2 x
9. y 2 = 2 cos x+C .

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Section 4: Integrating factor method 9

4. Integrating factor method

Consider an ordinary di↵erential equation (o.d.e.) that we wish to
solve to find out how the variable z depends on the variable x.

If the equation is first order then the highest derivative involved is

a first derivative.

If it is also a linear equation then this means that each term can
involve z either as the derivative dx
dz
OR through a single factor of z .

Any such linear first order o.d.e. can be re-arranged to give the
following standard form:

dz
+ P1 (x)z = Q1 (x)
dx

where P1 (x) and Q1 (x) are functions of x, and in some cases may be
constants.

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Section 4: Integrating factor method 10

A linear first order o.d.e. can be solved using the integrating

factor method.
After writing the equation in standard form, P1 (x) can be identified.
One then multiplies the equation by the following “integrating
factor”:
R
IF= e P1 (x)dx

This factor is defined so that the equation becomes equivalent to:

dx (IF z) = IF Q1 (x),
d

whereby integrating both sides with respect to x, gives:

R
IF z = IF Q1 (x) dx

Finally, division by the integrating factor (IF) gives z explicitly in

terms of x, i.e. gives the solution to the equation.

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Section 5: Standard integrals 11

5. 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 5: Standard integrals 12

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 6: Tips on using solutions 13

6. Tips on using solutions

● When looking at the THEORY, ANSWERS, IF METHOD,

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.
dy
+ P (x)y = Q(x)y n
dx
1 dy
DIVIDE by y n : y n dx + P (x)y 1 n
= Q(x)

n 1) dy
SET z = y 1 n
: i.e. dz
dx = (1 n)y (1 dx

1 dy
i.e. 1dz
(1 n) dx = y n dx

SUBSTITUTE 1dz
(1 n) dx + P (x)z = Q(x)

i.e. dz
dx + P1 (x)z = Q1 (x) linear in z

where P1 (x) = (1 n)P (x)

Q1 (x) = (1 n)Q(x) .
Return to Exercise 1
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Solutions to exercises 15

Exercise 2.
dy
This is of the form + P (x)y = Q(x)y n where
dx
1
where P (x) =
x
Q(x) = x
and n = 2
1 dy 1 1
DIVIDE by y n : i.e. 2
y =x
y dx x
dz 2 dy 1 dy
SET z = y 1 n
= y : i.e.
1
= y =
dx dx y 2 dx
dz 1
) z=x
dx x
dz 1
i.e. + z= x
dx x

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Solutions to exercises 16
R 1
Integrating factor, IF = e x dx = eln x = x
dz
) x + z = x2
dx
d
i.e. [x · z] = x2
dx
Z
i.e. xz = x2 dx

x3
i.e. xz = +C
3
x3
Use z = y:
1 x
y = 3 +C

1 x2 C
i.e. = + .
y 3 x
Return to Exercise 2

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Solutions to exercises 17

Exercise 3.
dy
This is of the form + P (x)y = Q(x)y n
dx
where P (x) = x1 ,

Q(x) = 1,

and n=2

1 dy
DIVIDE by y n : i.e. y 2 dx + x1 y 1
=1

2 dy 1 dy
SET z = y 1 n
=y 1
: i.e. dz
dx = 1·y dx = y 2 dx

) dz
dx + x1 z = 1

i.e. dz
dx
1
xz = 1

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Solutions to exercises 18
R dx 1 1
Integrating factor, IF = e x =e ln x
=e
ln x
=
x

) 1 dz
x dx
1
x2 z = 1
x

⇥1 ⇤
i.e. d
dx x ·z = 1
x

R
i.e. 1
x ·z = dx
x

i.e. z
x = ln x + C

Use z = y1 : 1
yx =C ln x

i.e. 1
y = x(C ln x) .
Return to Exercise 3

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Solutions to exercises 19

Exercise 4.
dy
This of the form + P (x)y = Q(x)y n
dx
1
where P (x) =
3
Q(x) = ex
and n = 4
1 dy 1 3
DIVIDE by y n : i.e. + y = ex
y 4 dx 3
dz 4 dy 3 dy
SET z = y 1 n
=y 3
: i.e. = 3y =
dx dx y 4 dx
1 dz 1
) + z = ex
3 dx 3
dz
i.e. z= 3ex
dx

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Solutions to exercises 20
R
Integrating factor, IF = e dx
=e x

x dz
) e e x
z= 3e x
· ex
dx
d
i.e. [e · z] = 3
x
dx
Z
i.e. e x·z = 3 dx

i.e. e x
·z = 3x + C

Use z = y3 :
1
e x
· 1
y3 = 3x + C
1
i.e. 3
= ex
(C 3x) .
y
Return to Exercise 4

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Solutions to exercises 21

Exercise 5.
dy y 1
Bernoulli equation: + = y 3 with P (x) = , Q(x) = 1, n = 3
dx x x

1 dy
DIVIDE by y n i.e. y 3 : y 3 dx + x1 y 2
=1

3 dy
SET z = y 1 n
i.e. z=y 2
: dz
dx = 2y dx

1 dy
i.e. 1 dz
2 dx = y 3 dx

) 1 dz
2 dx + x1 z = 1

i.e. dz
dx
2
xz = 2

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Solutions to exercises 22
R dx 2 1
Integrating factor, IF = e 2 x =e 2 ln x
=e ln x
= 2
x

) 1 dz
x2 dx
2
x3 z = 2
x2

⇥ ⇤
i.e. d
dx
1
x2 z = 2
x2

i.e. 1
x2 z = ( 2) · ( 1) x1 + C

i.e. z = 2x + Cx2

Use z = y2 :
1
y2 = 1
2x+Cx2 .
Return to Exercise 5

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Solutions to exercises 23

Exercise 6.
dy
This is of the form + P (x)y = Q(x)y n where
dx
2
where P (x) =
x
Q(x) = x2 cos x
and n = 2
1 dy 2 1
DIVIDE by y : n
i.e. 2
+ y = x2
cos x
y dx x
dz dy 1 dy
SET z = y 1 n = y 1 : i.e. = 1·y 2 =
dx dx y 2 dx
dz 2
) + z = x2 cos x
dx x
dz 2
i.e. z = x2 cos x
dx x

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Solutions to exercises 24
R 2
R dx 2 1
Integrating factor, IF = e x dx =e 2 x =e 2 ln x
=eln x
= 2
x
1 dz 2 x2
) 2 3
z = 2 cos x
x dx x x

d 1
i.e. 2
· z = cos x
dx x
Z
1
i.e. 2
· z = cos x dx
x
1
i.e. 2
· z = sin x + C
x

Use z = y1 : 1
x2 y = sin x + C
1
i.e. = x2 (sin x + C) .
y
Return to Exercise 6

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Solutions to exercises 25

Exercise 7.
Divide by 2 to get standard form:

dy 1 (4x + 5)2 3
+ tan x · y = y
dx 2 2 cos x
dy
This is of the form + P (x)y = Q(x)y n
dx

1
where P (x) = tan x
2

(4x + 5)2
Q(x) =
2 cos x

and n = 3

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Solutions to exercises 26
1 dy 1 (4x + 5)2
DIVIDE by y n : i.e. + tan x · y 2
=
y dx 2
3 2 cos x

dz 3 dy 2 dy
SET z = y 1 n
=y 2
: i.e. = 2y =
dx dx y 3 dx

1 dz 1 (4x + 5)2
) + tan x · z =
2 dx 2 2 cos x

dz (4x + 5)2
i.e. tan x · z =
dx cos x

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Solutions to exercises 27
R R
 R
sin x f 0 (x)
dx
Integrating factor, IF = e tan x·dx
=e cos x dx
⌘e f (x)

= eln cos x = cos x

dz (4x+5)2
) cos x cos x tan x · z = cos x
dx cos x
dz
i.e. cos x sin x · z = (4x + 5)2
dx
d
i.e. [cos x · z] = (4x + 5)2
dx
Z
i.e. cos x · z = (4x + 5)2 dx
✓ ◆
1 1
i.e. cos x · z = · (4x + 5)3 + C
4 3
Use z = y2 :
1 cos x
y2 = 12 (4x
1
+ 5)3 + C
1 1 C
i.e. = (4x + 5) +
3
.
y 2 12 cos x cos x
Return to Exercise 7
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Solutions to exercises 28

Exercise 8.

dy
Standard form: dx + 1
x y = (x ln x)y 2

i.e. P (x) = x1 , Q(x) = x ln x , n = 2

1 dy
DIVIDE by y 2 : y 2 dx + 1
x y 1
= x ln x

2 dy 1 dy
SET z = y 1
: dz
dx = y dx = y 2 dx

) dz
dx + 1
x z = x ln x

i.e. dz
dx
1
x ·z = x ln x

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Solutions to exercises 29
R dx 1 1
Integrating factor: IF = e x =e ln x
=e ln x
=
x
) 1 dz
x dx
1
x2 z = ln x
⇥1 ⇤
i.e. d
dx xz = ln x
R
i.e. 1
xz = ln x dx + C 0
R R
[ Use integration by parts: dv
u dx dx = uv v du
dx dx,

with u = ln x , dv
dx =1]
⇥ R ⇤
i.e. 1
xz = x ln x x· 1
x dx +C

Use z = y1 : 1
xy = x(1 ln x) + C .
Return to Exercise 8

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Solutions to exercises 30

Exercise 9.

dy
Standard form: dx (cot x) · y = (cosec x) y 3

1 dy
DIVIDE by y 3 : y 3 dx (cot x) · y 2
= cosec x

3 dy 1 dy
SET z = y 2
: dz
dx = 2y dx = 2· y 3 dx

) 1 dz
2 dx cot x · z = cosec x

i.e. dz
dx + 2 cot x · z = 2 cosec x

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Solutions to exercises 31
R R f 0 (x)
cos x 2 dx
Integrating factor: IF = e2 sin x dx ⌘e f (x) = e2 ln(sin x) = sin2 x.

) sin2 x · dz
dx + 2 sin x · cos x · z = 2 sin x

⇥ 2 ⇤
i.e. d
dx sin x · z = 2 sin x

i.e. z sin2 x = ( 2) · ( cos x) + C

sin2 x
Use z = y2 :
1
y2 = 2 cos x + C

sin2 x
i.e. y =
2
2 cos x+C .
Return to Exercise 9

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