# Bernoulli-differential-equations

```Differential Equations
BERNOULLI EQUATIONS
Graham S McDonald
A Tutorial Module for learning how to solve
Bernoulli differential equations
● Begin Tutorial
c 2004 [email protected]
1.
2.
3.
4.
5.
6.
Theory
Exercises
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 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 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
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= 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
● Theory ● Answers ● IF method ● Integrals ● Tips
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Section 2: Exercises
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 &middot; 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
+ tan x &middot; 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
● Theory ● Answers ● IF method ● Integrals ● Tips
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7
1. HINT: Firstly, divide each term by y n . Then, differentiate z
dz
1 dy
1
with respect to x to show that (1−n)
dx = y n dx ,
2
2.
1
y
= − x3 +
C
x
3.
1
y
= x(C − ln x) ,
4.
1
y3
,
= ex (C − 3x) ,
5. y 2 =
1
2x+Cx2
,
6.
1
y
7.
1
y2
=
8.
1
xy
= C + x(1 − ln x) ,
= x2 (sin x + C) ,
1
12 cos x (4x
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+ 5)3 +
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C
cos x
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9. y 2 =
2
8
sin x
2 cos x+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 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”:
R
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 =
R
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)
n
x
1
x
x
e
sin x
cos x
tan x
cosec x
sec x
sec2 x
cot x
sin2 x
cos2 x
R
f (x)dx
xn+1
n+1
(n 6= −1)
ln |x|
ex
− cos x
sin x
− ln |cos x|
ln tan x2
ln |sec x + tan x|
tan x
ln |sin x|
x
sin 2x
2 −
4
x
sin 2x
2 +
4
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f (x)
n
0
[g (x)] g (x)
g 0 (x)
g(x)
x
a
sinh x
cosh x
tanh x
cosech x
sech x
sech2 x
coth x
sinh2 x
cosh2 x
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f (x)dx
[g(x)]n+1
n+1
(n 6= −1)
ln |g (x)|
ax
(a &gt; 0)
ln a
cosh x
sinh x
ln cosh x
ln tanh x2
2 tan−1 ex
tanh x
ln |sinh x|
sinh 2x
− x2
4
sinh 2x
+ x2
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Section 5: Standard integrals
f (x)
1
a2 +x2
√ 1
a2 −x2
√
a2 − x2
12
R
f (x) dx
f (x)
R
1
a
tan−1
1
a2 −x2
1
2a
ln
a+x
a−x
(0 &lt; |x| &lt; a)
(a &gt; 0)
1
x2 −a2
1
2a
ln
x−a
x+a
(|x| &gt; a &gt; 0)
sin−1
x
a
√ 1
a2 +x2
ln
√
x+ a2 +x2
a
(a &gt; 0)
(−a &lt; x &lt; a)
√ 1
x2 −a2
ln
√
x+ x2 −a2
a
(x &gt; a &gt; 0)
a2
2
−1
sin
√
+x
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x
a
x
a
a2 −x2
a2
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√
i √
a2 +x2
a2
2
a2
2
x2 −a2
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f (x) dx
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h
h
sinh−1
− cosh−1
I
x
a
i
√
x a2 +x2
2
a
i
√
2
2
+ x xa2−a
+
x
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
● 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
y n dx
DIVIDE by y n :
SET z = y 1−n :
dy
= (1 − n)y (1−n−1) dx
i.e.
dz
dx
i.e.
1
dz
(1−n) dx
=
1
dz
(1−n) dx
+ P (x)z = Q(x)
SUBSTITUTE
i.e.
where
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+ P (x)y 1−n = Q(x)
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dz
dx
1 dy
y n dx
+ P1 (x)z = Q1 (x)
linear in z
P1 (x) = (1 − n)P (x)
Q1 (x) = (1 − n)Q(x) .
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Solutions to exercises
15
Exercise 2.
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
This is of the form
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Solutions to exercises
16
R
Integrating factor,
IF = e
1
x dx
= eln x = x
i.e.
dz
+ z = −x2
dx
d
[x &middot; z] = −x2
dx
Z
xz = − x2 dx
i.e.
xz = −
∴
i.e.
Use z = y1 :
x
x
y
i.e.
x3
+C
3
3
= − x3 + C
1
x2
C
=− +
.
y
3
x
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Solutions to exercises
17
Exercise 3.
This is of the form
dy
+ P (x)y = Q(x)y n
dx
where P (x) = x1 ,
Q(x) = 1,
and
n=2
DIVIDE by y n :
i.e.
1 dy
y 2 dx
SET z = y 1−n = y −1 :
i.e.
dz
dx
∴
dz
− dx
+ x1 z = 1
i.e.
dz
dx
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+ x1 y −1 = 1
dy
dy
= −1 &middot; y −2 dx
= − y12 dx
− x1 z = −1
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Solutions to exercises
18
Integrating factor, IF = e−
∴
1 dz
x dx
−
d
dx
i.e.
1
x
&middot;z =−
i.e.
z
x
= − ln x + C
i.e.
1
yx
1
y
x
dx
x
−1
= e− ln x = eln x
=
1
x
= − x1
i.e.
Use z = y1 :
1
1
x2 z
R
&middot; z = − x1
R
dx
x
= C − ln x
= x(C − ln x) .
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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 :
i.e.
SET z = y 1−n = y −3 :
i.e.
∴
i.e.
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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
IF = e−
Integrating factor,
∴
dx
= e−x
dz
− e−x z = −3e−x &middot; ex
dx
i.e.
d −x
[e &middot; z] = −3
dx
Z
e−x &middot; z = −3 dx
i.e.
e−x &middot; z = −3x + C
i.e.
Use z =
e−x
R
e−x &middot;
1
y3 :
i.e.
1
y3
= −3x + C
1
= ex (C − 3x) .
y3
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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
y 3 dx
DIVIDE by y n i.e. y 3 :
SET z = y 1−n i.e.
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z = y −2 :
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dz
dx
+ x1 y −2 = 1
dy
= −2y −3 dx
1 dy
y 3 dx
i.e.
dz
− 12 dx
=
∴
dz
− 12 dx
+ x1 z = 1
i.e.
dz
dx
J
− x2 z = −2
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Solutions to exercises
22
IF = e−2
Integrating factor,
∴
Use z =
1 dz
x2 dx
2
x3 z
−
1
x2 z
d
dx
i.e.
1
x2 z
i.e.
z = 2x + Cx2
1
y2 :
y2 =
dx
x
−2
= e−2 ln x = eln x
=
1
x2
= − x22
i.e.
R
= − x22
= (−2) &middot; (−1) x1 + C
1
2x+Cx2
.
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Solutions to exercises
23
Exercise 6.
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
y dx x
dz
1 dy
dy
SET z = y 1−n = y −1 : i.e.
= −1 &middot; 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
This is of the form
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Solutions to exercises
24
R
Integrating factor,
IF = e
∴
i.e.
i.e.
i.e.
Use z = y1 :
2
−x
dx
dx
x
−2
= e−2 ln x = eln x
=
1
x2
1 dz
2
x2
− 3 z = 2 cos x
2
x dx x
x
d 1
&middot; z = cos x
dx x2
Z
1
&middot; z = cos x dx
x2
1
&middot; z = sin x + C
x2
1
x2 y
i.e.
= e−2
R
= sin x + C
1
= x2 (sin x + C) .
y
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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 &middot; 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
Q(x)
=
(4x + 5)2
2 cos x
n
=
3
and
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Solutions to exercises
26
DIVIDE by y n :
i.e.
1 dy
1
(4x + 5)2
+ tan x &middot; y −2 =
3
y dx 2
2 cos x
SET z = y 1−n = y −2 :
i.e.
dz
dy
2 dy
= −2y −3
=− 3
dx
dx
y dx
∴
i.e.
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−
1 dz
1
(4x + 5)2
+ tan x &middot; z =
2 dx 2
2 cos x
(4x + 5)2
dz
− tan x &middot; z =
dx
cos x
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Solutions to exercises
Integrating factor,
27
R
IF = e
− tan x&middot;dx
R
=e
sin x
− cos
x dx
R
≡e
f 0 (x)
f (x)
dx
= eln cos x = cos x
∴
i.e.
i.e.
i.e.
i.e.
Use z =
1
y2 :
cos x
y2
i.e.
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dz
(4x+5)2
−cos x tan x &middot; z = cos x
dx
cos x
dz
cos x
− sin x &middot; z = (4x + 5)2
dx
d
[cos x &middot; z] = (4x + 5)2
dx
Z
cos x &middot; z = (4x + 5)2 dx
1
1
cos x &middot; z =
&middot; (4x + 5)3 + C
4
3
cos x
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=
1
12 (4x
+ 5)3 + C
1
C
1
=
(4x + 5)3 +
.
2
y
12 cos x
cos x
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Solutions to exercises
28
Exercise 8.
dy
dx
Standard form:
i.e.
1 dy
y 2 dx
SET z = y −1 :
dz
dx
+
y = (x ln x)y 2
1
x
y −1 = x ln x
dy
dy
= −y −2 dx
= − y12 dx
∴
dz
− dx
+
i.e.
dz
dx
JJ
P (x) = x1 , Q(x) = x ln x , n = 2
DIVIDE by y 2 :
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1
x
+
−
1
x
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1
x
z = x ln x
&middot; z = −x ln x
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Solutions to exercises
Integrating factor:
29
IF = e−
∴
R
dx
x
1 dz
x dx
−1
= e− ln x = eln x
−
1
x2 z
d
dx
1 x z = − ln x
i.e.
1
xz
R
= − ln x dx + C 0
R
dv
u dx
dx = uv −
with u = ln x ,
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1
x
= − ln x
i.e.
[ Use integration by parts:
=
R
dv
dx
v du
dx dx,
=1]
i.e.
1
xz
R
= − x ln x − x &middot; x1 dx + C
Use z = y1 :
1
xy
= x(1 − ln x) + C .
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Solutions to exercises
30
Exercise 9.
Standard form:
dy
dx
DIVIDE by y 3 :
1 dy
y 3 dx
SET z = y −2 :
dz
dx
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− (cot x) &middot; y = (cosec x) y 3
− (cot x) &middot; y −2 = cosec x
dy
= −2y −3 dx
= −2 &middot;
1 dy
y 3 dx
∴
dz
− 12 dx
− cot x &middot; z = cosec x
i.e.
dz
dx
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+ 2 cot x &middot; z = −2 cosec x
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Solutions to exercises
31
Integrating factor: IF = e2
∴
Use z =
sin2 x &middot;
dz
dx
R
cos x
sin x dx
≡ e2
R
f 0 (x)
f (x)
dx
= e2 ln(sin x) = sin2 x.
+ 2 sin x &middot; cos x &middot; z = −2 sin x
2
sin x &middot; z = −2 sin x
i.e.
d
dx
i.e.
z sin2 x = (−2) &middot; (− cos x) + C
1
y2 :
sin2 x
y2
= 2 cos x + C
i.e.
y2 =
sin2 x
2 cos x+C
.