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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]
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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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Section 2: Exercises 5
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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Section 2: Exercises 6
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
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 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
xy = C + x(1 ln x) ,Toc Back
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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 11
5. Standard integrals
f(x)
f(x)dx f(x)
f(x)dxxn x
n+1
n+1 (n = 1) [g (x)]n g (x) [g(x)]n+1
n+1 (n = 1)1x
ln |x| g(x)g(x) ln |g (x)|
ex ex ax ax
ln a (a > 0)
sin x cos x sinh x cosh xcos x sin x cosh x sinh xtan x ln |cos x| tanh x ln cosh xcosec x ln
tan x2
cosech x lntanh x2
sec x ln |
sec x + tan x
|sech x 2tan1 ex
sec2 x tan x sech2 x tanh xcot x ln |sin x| coth x ln |sinh x|sin2 x x2 sin 2x4 sinh2 x sinh 2x4 x2cos2 x x2 +
sin 2x4 cosh
2 x sinh 2x4 +x2
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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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Solutions to exercises 15
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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Solutions to exercises 16
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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S l i i 17
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Solutions to exercises 17
Exercise 3.
This is of the formdy
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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Solutions to exercises 18
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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Solutions to exercises 19
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
SET z = y1n = y3: i.e.dz
dx= 3y4 dy
dx= 3
y4dy
dx
13
dz
dx+
1
3z = ex
i.e.dz
dx z = 3ex
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Solutions to exercises 20
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Solutions to exercises 20
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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Solutions to exercises 21
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Solutions to exercises 21
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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Solutions to exercises 22
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Solutions to exercises 22
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 .
Return to Exercise 5
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Solutions to exercises 23
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Solutions to exercises 23
Exercise 6.
This is of the formdy
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
SET z = y1n = y1: i.e.dz
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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Solutions to exercises 24
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Solutions to exercises 24
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) .
Return to Exercise 6
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Solutions to exercises 25
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Solutions to exercises 25
Exercise 7.
Divide by 2 to get standard form:
dy
dx+
1
2tan x y = (4x + 5)
2
2cos xy3
This is of the formdy
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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Solutions to exercises 26
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DIVIDE by yn: i.e.1
y3dy
dx+
1
2tan x y2 = (4x + 5)
2
2cos x
SET z = y1n = y2: i.e.dz
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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Solutions to exercises 27
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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
.
Return to Exercise 7
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Solutions to exercises 28
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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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Solutions to exercises 29
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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 .
Return to Exercise 8
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Solutions to exercises 30
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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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Solutions to exercises 31
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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 .
Return to Exercise 9
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