Bernoulli Differential Equations

7/31/2019 Bernoulli Differential Equations

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Differential Equations

BERNOULLI EQUATIONS

Graham S McDonald

A Tutorial Module for learning how to solveBernoulli differential equations

q Table of contents

q Begin Tutorial

c 2004 [email protected]
mailto:[email protected]://www.cse.salford.ac.uk/


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Table of contents

1. Theory

2. Exercises

3. Answers

4. Integrating factor method

5. Standard integrals

6. Tips on using solutions

Full worked solutions


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Section 1: Theory 3

1. Theory

A Bernoulli differential equation can be written in the following

standard form: dy

dx+ P(x)y = Q(x)yn ,

where n = 1 (the equation is thus nonlinear).To find the solution, change the dependent variable from y to z, wherez = y1n. This gives a differential equation in x and z that islinear, and can be solved using the integrating factor method.

Note: Dividing the above standard form by yn gives:

1yn

dydx

+ P(x)y1n = Q(x)

i.e.1

(1 n)dz

dx+ P(x)z = Q(x)

where we have used

dzdx = (1 n)y

n dydx

.

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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

dx +P

(x

)y

=Q

(x

)yn ,

where P and Q are functions of x, and n is a constant. Show thatthe transformation to a new dependent variable z = y1n reducesthe equation to one that is linear in z (and hence solvable using theintegrating factor method).

Solve the following Bernoulli differential equations:

Exercise 2.

dy

dx

1

x

y = xy2

q Theory q Answers q IF method q Integrals q Tips

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Exercise 3.

dy

dx

+y

x

= y2

Exercise 4.

dy

dx+

1

3y = exy4

Exercise 5.

xdy

dx+ y = xy3

Exercise 6.

dy

dx+

2

xy = x2 cos x y2


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Exercise 7.

2dy

dx+ tan x

y =

(4x + 5)2

cos xy3

Exercise 8.

xdy

dx+ y = y2x2 ln x

Exercise 9.

dy

dx= y cot x + y3cosecx


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Section 3: Answers 7

3. Answers

1. HINT: Firstly, divide each term by yn. Then, differentiate zwith respect to x to show that 1(1n)

dzdx

= 1yn

dydx

,

2. 1y

= x23 + Cx ,

3.1

y = x(C ln x) ,4. 1

y3= ex(C 3x) ,

5. y2 = 12x+Cx2 ,

6. 1y

= x2(sin x + C) ,

7. 1y2

= 112 cosx (4x + 5)3 + Ccosx ,

8.1

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Section 3: Answers 8

9. y2 = sin2 x

2 cosx+C .

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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 isa first derivative.

If it is also a linear equation then this means that each term caninvolve z either as the derivative dz

dxOR through a single factor of z .

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

dz

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

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 integratingfactor method.

After writing the equation in standard form, P1(x) can be identified.One then multiplies the equation by the following integratingfactor:

IF= eP1(x)dx

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

ddx

(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 12

f(x)

f(x) dx f(x)

f(x) dx

1

a2+x21

a tan1 x

a

1

a2x21

2a lna+xax (0 < |x|< a)

(a > 0) 1x2a2

12a ln

xax+a (|x| > a > 0)

1a2x2 sin

1 xa

1a2+x2 ln

x+a2+x2a

(a > 0)

(a < x < a) 1x2a2 ln

x+x2a2a (x > a > 0)

a2 x2 a22

sin1xa

a2 +x2 a22 sinh1 xa + xa2+x2a2

+xa2x2a2

x2a2 a22

cosh1 x

a

+ x

x2a2a2

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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 ofthe 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 yourintermediate results are correct.

q Try to make less use of the full solutions as you work your waythrough the Tutorial.

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Solutions to exercises 14

Full worked solutions

Exercise 1.dy

dx + P(x)y = Q(x)yn

DIVIDE by yn: 1yn

dydx

+ P(x)y1n = Q(x)

SET z = y1n

: i.e.dz

dx = (1 n)y(1n

1) dy

dx

i.e. 1(1n)dzdx

= 1yn

dydx

SUBSTITUTE 1(1n)

dzdx

+ P(x)z = Q(x)

i.e. dzdx

+ 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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Exercise 2.

This is of the formdy

dx

+ P(x)y = Q(x)yn where

where P(x) = 1x

Q(x) = x

and n = 2

DIVIDE by yn: i.e.1

y2dy

dx 1

xy1 = x

SET z = y1n = y1: i.e.dz

dx= y2 dy

dx= 1

y2dy

dx

dzdx

1x

z = x

i.e.dz

dx+

1

xz = x

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Integrating factor, IF = e

1xdx = eln x = x

x

dz

dx + z = x2

i.e.d

dx[x z] = x2

i.e. xz =

x2dx

i.e. xz = x3

3+ C

Use z = 1y

: xy

= x33 + C

i.e. 1y

= x2

3+ C

x.

Return to Exercise 2

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Exercise 3.


dx+ P(x)y = Q(x)yn

where P(x) = 1x ,

Q(x) = 1,

and n = 2

DIVIDE by yn: i.e. 1y2

dydx

+ 1x

y1 = 1

SET z = y1n = y1: i.e. dzdx

=

1

y2 dy

dx=

1y2

dydx

dzdx

+ 1x

z = 1

i.e. dzdx 1

xz = 1

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Integrating factor, IF = e

dxx = e ln x = eln x

1

=1

x

1xdzdx 1

x2z = 1

x

i.e. ddx

1x z = 1

x

i.e. 1x z = dx

x

i.e. zx

= ln x + C

Use z = 1y

: 1yx

= C ln x

i.e. 1y

= x(C ln x) .Return to Exercise 3

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Exercise 4.

This of the formdy

dx

+ P(x)y = Q(x)yn

where P(x) =1

3Q(x) = ex

and n = 4

DIVIDE by yn: i.e. 1y4

dy

dx+ 1

3y3 = ex


dx= 3y4 dy

dx= 3

y4dy

dx

13

dz

dx+

1

3z = ex

i.e.dz

dx z = 3ex

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Integrating factor, IF = edx = ex

ex dz

dx exz

= 3ex

ex

i.e.d

dx[ex z] = 3

i.e. ex z = 3 dxi.e. ex z = 3x + C

Use z = 1y3

: ex 1y3

= 3x + C

i.e.1

y3 = ex(C 3x) .Return to Exercise 4

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Exercise 5.

Bernoulli equation:dy

dx+

y

x= y3 with P(x) =

1

x, Q(x) = 1, n = 3

DIVIDE by yn i.e. y3: 1y3

dydx

+ 1x

y2 = 1

SET z = y1n i.e. z = y2: dzdx

= 2y3 dydx

i.e. 12 dzdx = 1y3 dydx

12 dzdx + 1xz = 1

i.e. dzdx 2

xz = 2

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Integrating factor, IF = e2

dxx = e2 ln x = eln x

2

=1

x2

1x2

dzdx 2

x3z = 2

x2

i.e. ddx

1x2

z

= 2x2

i.e. 1x2

z = (2) (1) 1x

+ C

i.e. z = 2x + Cx2

Use z = 1y2 : y2 = 12x+Cx2 .


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Exercise 6.


dx

+ P(x)y = Q(x)yn where

where P(x) =2

x

Q(x) = x2 cos xand n = 2

DIVIDE by yn: i.e.1

y2dy

dx+

2x

y1 = x2 cos x


dx= 1 y2 dy

dx= 1

y2dy

dx

dzdx

+ 2x

z = x2 cos x

i.e.dz

dx 2

xz = x2 cos x

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Integrating factor, IF = e2xdx = e2

dxx = e2 ln x = eln x

2

=1

x2

1

x2dz

dx 2

x3 z =

x2

x2 cos x

i.e.d

dx

1

x2 z

= cos x

i.e.1

x2 z = cos x dx

i.e.1

x2 z = sin x + C

Use z = 1y

: 1x2y

= sin x + C

i.e.1

y= x2(sin x + C) .


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Exercise 7.

Divide by 2 to get standard form:

dy

dx+

1

2tan x y = (4x + 5)

2

2cos xy3


dx+ P(x)y = Q(x)yn

where P(x) =1

2tan x

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

2cos x

and n = 3

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DIVIDE by yn: i.e.1

y3dy

dx+

1

2tan x y2 = (4x + 5)

2

2cos x


dx= 2y3 dy

dx= 2

y3dy

dx

12

dz

dx+

1

2tan x z = (4x + 5)

2

2cos x

i.e.dz

dx tan x z = (4x + 5)

2

cos x

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Integrating factor, IF = e tanxdx = e

sinxcosx dx

e

f(x)f(x) dx

= eln cos x = cos x

cos xdz

dxcos x tan x z = cos x (4x+5)

2

cos x

i.e. cos xdz

dx sin x z = (4x + 5)2

i.e.d

dx[cos x z] = (4x + 5)2

i.e. cos x z =

(4x + 5)2dx

i.e. cos x z = 14 1

3(4x + 5)3 + C

Use z = 1y2

: cosxy2

= 112 (4x + 5)3 + C

i.e.1

y2

=1

12cos x

(4x + 5)3 +C

cos x

.


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Exercise 8.

Standard form: dy

dx

+ 1x y = (x ln x)y2

i.e. P(x) = 1x

, Q(x) = x ln x , n = 2

DIVIDE by y2: 1

y

2dy

dx

+ 1x y1 = x ln x

SET z = y1: dzdx

= y2 dydx

= 1y2

dydx

dz

dx + 1x z

=x

lnx

i.e. dzdx 1

x z = x ln x

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Integrating factor: IF = e

dxx = e ln x = eln x

1

=1

x

1

x

dz

dx 1

x2 z =

ln x

i.e. ddx

1x

z

= ln x

i.e. 1x

z = ln x dx + C[Use integration by parts:

u dvdx

dx = uv v dudx

dx,

with u = ln x , dvdx

= 1 ]

i.e. 1

x

z = x ln x x

1

x

dx + CUse z = 1

y: 1

xy= x(1 ln x) + C .


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Exercise 9.

Standard form: dydx

(cot x)

y = (cosec x) y3

DIVIDE by y3: 1y3

dydx (cot x) y2 = cosec x

SET z = y

2: dzdx

=

2y

3 dy

dx=

2

1

y3dy

dx

12 dzdx cot x z = cosec x

i.e.

dz

dx + 2 cotx

z= 2 cosec

x

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Integrating factor: IF= e2

cos xsinx dx e2

f(x)f(x)

dx= e2 ln(sinx) = sin2 x.

sin2 x dzdx + 2 sin x cos x z = 2 sin x

i.e. ddx

sin2 x z = 2 sin x

i.e. z sin2 x = (2) ( cos x) + C

Use z = 1y2

: sin2 xy2

= 2 cos x + C

i.e. y2 = sin2 x

2cos x+C .


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Bernoulli Differential Equations

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