# Persamaan Diferensial Bernoulli by deyedeex

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```									Diﬀerential Equations BERNOULLI EQUATIONS
Graham S McDonald A Tutorial Module for learning how to solve Bernoulli diﬀerential equations

q Table of contents q Begin Tutorial

c 2004 g.s.mcdonald@salford.ac.uk

Table of contents
1. 2. 3. 4. 5. 6. Theory Exercises Answers Integrating factor method Standard integrals Tips on using solutions Full worked solutions

Section 1: Theory

3

1. Theory
A Bernoulli diﬀerential equation can be written in the following standard form: dy + P (x)y = Q(x)y n , dx where n = 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 diﬀerential 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 y n dx dz 1 i.e. + P (x)z (1 − n) dx where we have used
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dz dx

= Q(x) = Q(x)

dy = (1 − n)y −n 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 diﬀerential equations: Exercise 2. dy 1 − y = xy 2 dx x q Theory q Answers q IF method q Integrals q 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

q Theory q Answers q IF method q Integrals q Tips
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Section 2: Exercises

6

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

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

q Theory q Answers q IF method q Integrals q Tips
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Section 3: Answers

7

3. Answers
1. HINT: Firstly, divide each term by y n . Then, diﬀerentiate z dy dz 1 with respect to x to show that (1−n) dx = y1 dx , n 2. 3. 4.
1 y 1 y 1 y3

= −x + 3

2

C x

,

= x(C − ln x) , = ex (C − 3x) ,
1 2x+Cx2

5. y 2 = 6. 7. 8.
1 y 1 y2 1 xy

,

= x2 (sin x + C) , =
1 12 cos x (4x

+ 5)3 +

C cos x

,

= C + x(1 − ln x) ,
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Section 3: Answers

8

9. y 2 =

sin x 2 cos x+C

2

.

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Section 4: Integrating factor method

9

4. Integrating factor method
Consider an ordinary diﬀerential 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 dz involve z either as the derivative dx 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”: IF= e
P1 (x)dx

This factor is deﬁned so that the equation becomes equivalent to:
d dx (IF z)

= IF Q1 (x),

whereby integrating both sides with respect to x, gives: 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
f (x) x
n 1 x x

f (x)dx
xn+1 n+1

f (x) [g (x)] g (x) a sinh x cosh x tanh x cosech x sech x sech2 x coth x sinh2 x cosh2 x
g (x) g(x) x n

f (x)dx
[g(x)]n+1 n+1

(n = −1)

(n = −1)

e sin x cos x tan x cosec x sec x sec2 x cot x sin2 x cos2 x

ln |x| ex − cos x sin x − ln |cos x| ln tan x 2 ln |sec x + tan x| tan x ln |sin x| x sin 2x 2 − 4 x sin 2x 2 + 4

ln |g (x)| ax (a > 0) ln a cosh x sinh x ln cosh x ln tanh x 2 2 tan−1 ex tanh x ln |sinh x| sinh 2x −x 4 2 sinh 2x +x 4 2

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Section 5: Standard integrals

12

f (x)
1 a2 +x2 1 a

f (x) dx tan−1
x a

f (x)
1 a2 −x2 1 x2 −a2

f (x) dx
1 2a 1 2a

ln ln

a+x a−x x−a x+a

(0 < |x| < a) (|x| > a > 0)

(a > 0)

√ 1 a2 −x2

sin−1

x a

√ 1 a2 +x2 √ 1 x2 −a2

ln ln

√ x+ a2 +x2 a √ x+ x2 −a2 a

(a > 0) (x > a > 0)
√ x a2 +x2 a2 √ 2 2 + x x2−a a

(−a < x < a) √

a2 − x2

a2 2

sin−1 +x
√

x a

√ √

a2 +x2

a2 2 a2 2

sinh−1 − cosh−1

x a x a

+

a2 −x2 a2

x2 −a2

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Section 6: Tips on using solutions

13

6. Tips on using solutions
q 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. q 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. q 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 y n dx

DIVIDE by y n : SET z = y 1−n : i.e. i.e. SUBSTITUTE i.e. where

+ P (x)y 1−n = Q(x)

dz dx

dy = (1 − n)y (1−n−1) dx

1 dz (1−n) dx 1 dz (1−n) dx dz dx

=

1 dy y n dx

+ P (x)z = Q(x) linear in z

+ P1 (x)z = Q1 (x)

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

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Solutions to exercises

16

Integrating factor, ∴ i.e. i.e. i.e.
1 Use z = y :

IF = e x

1 x dx

= eln x = x

dz + z = −x2 dx d [x · z] = −x2 dx x2 dx x3 +C 3
3

xz = − xz = −
x y

= −x + C 3

i.e.

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

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Solutions to exercises

17

Exercise 3. This is of the form
1 where P (x) = x ,

dy + P (x)y = Q(x)y n dx

Q(x) = 1, and n=2 i.e. i.e. ∴ i.e.
1 dy y 2 dx dz dx 1 + x y −1 = 1

DIVIDE by y n : SET z = y 1−n = y −1 :

dy dy = −1 · y −2 dx = − y12 dx

dz 1 − dx + x z = 1 dz dx 1 − x z = −1

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Solutions to exercises

18
dx x

Integrating factor, IF = e−

= e− ln x = eln x

−1

=

1 x

∴ i.e. i.e. i.e.
1 Use z = y :

1 dz x dx d dx 1 x z x

−
1 x

1 x2 z

1 = −x

1 · z = −x dx x

·z =−

= − ln x + C = C − ln x

1 yx

i.e.

1 y

= x(C − ln x) . Return to Exercise 3

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Solutions to exercises

19

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

dz − z = −3ex dx
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Solutions to exercises

20

Integrating factor, ∴ i.e. i.e. i.e. Use z =
1 y3 :

IF = e− e−x

dx

= e−x

dz − e−x z = −3e−x · ex dx

d −x [e · z] = −3 dx e−x · z = −3 dx

e−x · z = −3x + C e−x ·
1 y3

= −3x + C

i.e.

1 = ex (C − 3x) . y3 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 DIVIDE by y n i.e. y 3 : SET z = y 1−n i.e. z = y −2 : i.e. ∴ i.e.
1 dy y 3 dx 1 + x y −2 = 1

dz dx

dy = −2y −3 dx 1 dy y 3 dx

dz − 1 dx = 2

dz 1 − 1 dx + x z = 1 2 dz dx 2 − x z = −2

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Solutions to exercises

22

Integrating factor, ∴ i.e. i.e. i.e. Use z =
1 y2 : 1 dz x2 dx d dx 1 x2 z

IF = e−2 −
2 x3 z

dx x

= e−2 ln x = eln x

−2

=

1 x2

2 = − x2

1 x2 z

2 = − x2

1 = (−2) · (−1) x + C

z = 2x + Cx2 y2 =
1 2x+Cx2

. Return to Exercise 5

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Solutions to exercises

23

Exercise 6. This is of the form dy + 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 DIVIDE by y n : i.e. + y −1 = −x2 cos x 2 dx y x dz 1 dy dy SET z = y 1−n = y −1 : i.e. = −1 · y −2 =− 2 dx dx y dx 2 dz ∴ − + z = −x2 cos x dx x dz 2 i.e. − z = x2 cos x dx x

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Solutions to exercises

24

Integrating factor,

IF = e ∴ i.e. i.e. i.e.

2 −x dx

= e−2

dx x

= e−2 ln x = eln x

−2

=

1 x2

1 dz 2 x2 − 3 z = 2 cos x 2 dx x x x d 1 · z = cos x dx x2 1 · z = cos x dx x2 1 · z = sin x + C x2
1 x2 y

1 Use z = y :

= sin x + C

i.e.

1 = 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 where P (x) = 1 tan x 2 (4x + 5)2 2 cos x 3

Q(x) and n

= =

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Solutions to exercises

26

DIVIDE by y n :

i.e.

1 dy 1 (4x + 5)2 + tan x · y −2 = 3 dx y 2 2 cos x

SET z = y 1−n = y −2 :

i.e.

dz dy 2 dy = −2y −3 =− 3 dx dx y dx

∴

−

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

i.e.

(4x + 5)2 dz − tan x · z = dx cos x

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Solutions to exercises

27

Integrating factor,

IF = e

− tan x·dx

=e

sin − cos x dx x

≡e

f (x) f (x)

dx

= eln cos x = cos x ∴ i.e. i.e. i.e. i.e. Use z =
1 y2 :

cos x

dz (4x+5)2 −cos x tan x · z = cos x dx cos x dz cos x − sin x · z = (4x + 5)2 dx d [cos x · z] = (4x + 5)2 dx cos x · z = cos x · z =
cos x y2

(4x + 5)2 dx 1 4 1 · (4x + 5)3 + C 3

=

1 12 (4x

+ 5)3 + C

i.e.
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1 C 1 = (4x + 5)3 + . 2 y 12 cos x cos x Return to Exercise 7
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Solutions to exercises

28

Exercise 8. Standard form: i.e. DIVIDE by y 2 : SET z = y −1 : ∴ i.e.
dy dx

+

1 x

y = (x ln x)y 2

1 P (x) = x , Q(x) = x ln x , n = 2 1 dy y 2 dx 1 x

+

y −1 = x ln x

dz dx

dy dy = −y −2 dx = − y12 dx 1 x

dz − dx + dz dx

z = x ln x

−

1 x

· z = −x ln x

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Solutions to exercises

29

Integrating factor:

IF = e− ∴ i.e. i.e.

dx x

= e− ln x = eln x −
1 xz 1 x2 z

−1

=

1 x

1 dz x dx d dx 1 xz

= − ln x

= − ln x

= − ln x dx + C v du dx, dx =1]
1 x · x dx + C

[ Use integration by parts:

dv u dx dx = uv −

with u = ln x , i.e.
1 Use z = y : 1 xz 1 xy

dv dx

= − x ln x −

= x(1 − ln x) + C . Return to Exercise 8

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Solutions to exercises

30

Exercise 9. Standard form: DIVIDE by y 3 : SET z = y −2 : ∴ i.e.
dy dx

− (cot x) · y = (cosec x) y 3 − (cot x) · y −2 = cosec x
1 dy y 3 dx

1 dy y 3 dx

dz dx

dy = −2y −3 dx = −2 ·

dz − 1 dx − cot x · z = cosec x 2 dz dx

+ 2 cot x · z = −2 cosec x

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Solutions to exercises

31
cos x sin x dx

Integrating factor: IF = e2 ∴ i.e. i.e. Use z =
1 y2 :

≡ e2

f (x) dx f (x)

= e2 ln(sin x) = sin2 x.

sin2 x ·
d dx

dz dx

+ 2 sin x · cos x · z = −2 sin x

sin2 x · z = −2 sin x

z sin2 x = (−2) · (− cos x) + C
sin2 x y2

= 2 cos x + C
sin2 x 2 cos x+C

i.e.

y2 =

. Return to Exercise 9

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