Differential Equations
INTEGRATING FACTOR METHOD
Graham S McDonald
A Tutorial Module for learning to solve 1st
order linear differential equations
● Table of contents
● Begin Tutorial
c 2004 g.s.mcdonald@salford.ac.uk
Table of contents
1.
2.
3.
4.
5.
6.
Theory
Exercises
Answers
Standard integrals
Tips on using solutions
Alternative notation
Full worked solutions
Section 1: Theory
3
1. Theory
Consider an ordinary differential equation (o.d.e.) that we wish to
solve to find out how the variable y 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
dy
involve y either as the derivative dx
OR through a single factor of y .
Any such linear first order o.d.e. can be re-arranged to give the following standard form:
dy
+ P (x)y = Q(x)
dx
where P (x) and Q(x) are functions of x, and in some cases may be
constants.
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Section 1: Theory
4
A linear first order o.d.e. can be solved using the integrating factor
method.
After writing the equation in standard form, P (x) can be identified.
One then multiplies the equation by the following “integrating factor”:
R
IF= e
P (x)dx
This factor is defined so that the equation becomes equivalent to:
d
dx (IF y)
= IF Q(x),
whereby integrating both sides with respect to x, gives:
IF y =
R
IF Q(x) dx
Finally, division by the integrating factor (IF) gives y explicitly in
terms of x, i.e. gives the solution to the equation.
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Section 2: Exercises
5
2. Exercises
In each case, derive the general solution. When a boundary
condition is also given, derive the particular solution.
Click on Exercise links for full worked solutions (there are 10
exercises in total)
Exercise 1.
dy
+ y = x ; y(0) = 2
dx
Exercise 2.
dy
+ y = e−x ; y(0) = 1
dx
Exercise 3.
dy
x
+ 2y = 10x2 ; y(1) = 3
dx
● Theory ● Answers ● Integrals ● Tips ● Notation
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Section 2: Exercises
Exercise 4.
dy
x
− y = x2 ; y(1) = 3
dx
Exercise 5.
dy
x
− 2y = x4 sin x
dx
Exercise 6.
dy
x
− 2y = x2
dx
Exercise 7.
dy
+ y cot x = cosec x
dx
● Theory ● Answers ● Integrals ● Tips ● Notation
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Section 2: Exercises
7
Exercise 8.
dy
+ y · cot x = cos x
dx
Exercise 9.
dy
(x2 − 1)
+ 2xy = x
dx
Exercise 10.
dy
= y tan x − sec x ; y(0) = 1
dx
● Theory ● Answers ● Integrals ● Tips ● Notation
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Section 3: Answers
8
3. Answers
1. General solution is y = (x − 1) + Ce−x , and particular
solution is y = (x − 1) + 3e−x ,
2. General solution is y = e−x (x + C) , and particular solution is
y = e−x (x + 1) ,
3. General solution is y = 25 x2 +
y = 21 (5x2 + x12 ) ,
C
x2
, and particular solution is
4. General solution is y = x2 + Cx , and particular solution is
y = x2 + 2x ,
5. General solution is y = −x3 cos x + x2 sin x + Cx2 ,
6. General solution is y = x2 ln x + C x2 ,
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Section 3: Answers
9
7. General solution is y sin x = x + C ,
8. General solution is 4y sin x + cos 2x = C ,
9. General solution is (x2 − 1)y =
x2
2
+C ,
10. General solution is y cos x = C − x , and particular solution is
y cos x = 1 − x .
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Section 4: Standard integrals
10
4. 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 > 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 4: Standard integrals
f (x)
1
a2 +x2
√ 1
a2 −x2
√
a2 − x2
11
R
f (x) dx
f (x)
1
a
tan−1
1
a2 −x2
R
(a > 0)
1
x2 −a2
f (x) dx
a+x 1
2a ln a−x (0 < |x| < a)
x−a 1
ln
2a
x+a (|x| > a > 0)
sin−1
x
a
√ 1
a2 +x2
√
2
2
ln x+ aa +x (a > 0)
(−a < x < a)
√ 1
x2 −a2
√
2
2
ln x+ xa −a (x > a > 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
x2 −a2
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a2
2
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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 5: Tips on using solutions
12
5. Tips on using solutions
● When looking at the THEORY, ANSWERS, INTEGRALS, TIPS
or NOTATION 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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Section 6: Alternative notation
13
6. Alternative notation
The linear first order differential equation:
dy
+ P (x) y = Q(x)
dx
R
has the integrating factor IF=e
P (x) dx
.
The integrating factor method is sometimes explained in terms of
simpler forms of differential equation. For example, when constant
coefficients a and b are involved, the equation may be written as:
dy
+ b y = Q(x)
a
dx
In our standard form this is:
dy
b
Q(x)
+ y=
dx a
a
with an integrating factor of:
R
IF = e
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b
a
dx
bx
=ea
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Solutions to exercises
14
Full worked solutions
Exercise 1.
Compare with form:
dy
+ P (x)y = Q(x) (P, Q are functions of x)
dx
Integrating factor:
P (x) = 1.
Integrating factor,
R
IF = e P (x)dx
R
= e dx
= ex
Multiply equation by IF:
dy
+ ex y = ex x
dx
d x
i.e.
[e y] = ex x
dx
ex
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Solutions to exercises
15
Integrate both sides with respect to x:
ex y
Z
{ Note:
dv
u dx
dx
u
i.e. y
=
ex (x − 1) + C
Z
=
uv −
v
du
dx i.e. integration by parts with
dx
dv
≡ ex
dxZ
→ xex − ex dx
≡
x,
→ xex − ex = ex (x − 1) }
= (x − 1) + Ce−x .
Particular solution with y(0) = 2:
2
= (0 − 1) + Ce0
= −1 + C i.e. C = 3 and y = (x − 1) + 3e−x .
Return to Exercise 1
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Solutions to exercises
16
Exercise 2.
Integrating Factor:
P (x) = 1 ,
R
IF = e
P dx
R
=e
dx
= ex
Multiply equation:
ex
i.e.
dy
+ ex y = ex e−x
dx
d x
[e y] = 1
dx
Integrate:
ex y
y
i.e.
= x+C
= e−x (x + C) .
Particular solution:
y=1
x=0
gives
1
and y
= e0 (0 + C)
= 1.C
i.e. C = 1
−x
= e (x + 1) .
Return to Exercise 2
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Solutions to exercises
17
Exercise 3.
Equation is linear, 1st order
i.e.
dy
dx
+ x2 y = 10x,
Integrating factor : IF =
dy
+ P (x)y = Q(x)
dx
P (x) = x2 , Q(x) = 10x
i.e.
where
R
e
P (x)dx
= e2
2
= e2 ln x = eln x = x2 .
dy
+ 2xy =
dx
d 2 x ·y =
dx
i.e.
Integrate:
i.e.
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dx
x
x2
Multiply equation:
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x2 y
=
y
=
I
10x3
10x3
5 4
x +C
2
C
5 2
x + 2
2
x
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Solutions to exercises
18
Particular solution
i.e.
3 =
5
2
·1+
i.e.
6
2
=
5
2
+C
i.e.
C=
1
2
∴
y = 52 x2 +
y(1) = 3
i.e. y(x) = 3
when x = 1.
C
1
1
2x2
=
1
2
5x2 +
1
x2
.
Return to Exercise 3
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Solutions to exercises
19
Exercise 4.
dy
−
dx
Standard form:
Compare with
1
y=x
x
dy
+ P (x)y = Q(x) , giving
dx
R
Integrating Factor: IF = e
P (x)dx
= e−
R
dx
x
P (x) = −
1
x
−1
= e− ln x = eln(x
Multiply equation:
1
1 dy
− 2 y =
x dx x d 1
i.e.
y
=
dx x
1
1
Integrate:
i.e.
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y
x
y
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= x+C
= x2 + Cx .
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)
=
1
.
x
Solutions to exercises
20
Particular solution with y(1) = 3:
i.e.
Particular solution is
3 = 1+C
C = 2
y
= x2 + 2x .
Return to Exercise 4
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Solutions to exercises
Exercise 5.
Linear in y :
21
dy
dx
− x2 y = x3 sin x
Integrating factor: IF = e−2
R
dx
x
−2
= e−2 ln x = eln x
=
1
x2
1 dy
2
− y = x sin x
x2 dx x3 d 1
y
= x sin x
dx x2
Z
y
= −x cos x− 1 · (− cos x)dx+C 0
x2
Multiply equation:
i.e.
Integrate:
[Note: integration by parts,
R dv
R
u dx dx = uv − v du
dx dx, u = x,
i.e.
i.e.
y
x2
y
dv
dx
= sin x]
= −x cos x + sin x + C
= −x3 cos x + x2 sin x + Cx2 .
Return to Exercise 5
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Solutions to exercises
22
Exercise 6.
2
y=x
x
Standard form:
dy
−
dx
Integrating Factor:
P (x) = −
R
IF = e
P dx
Multiply equation:
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= e−2
R
dx
x
2
x
−2
= e−2 ln x = eln x
=
1
x2
1 dy
2
1
−
y=
x2 dx x3
x
d 1
1
i.e.
y =
dx x2
x
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Solutions to exercises
23
1
y=
x2
Integrate:
Z
dx
x
i.e.
1
y
x2
= ln x + C
i.e.
y
= x2 ln x + Cx2 .
Return to Exercise 6
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Solutions to exercises
24
Exercise 7.
Of the form:
where
dy
+ P (x)y = Q(x) (i.e. linear, 1st order o.d.e.)
dx
P (x) = cot x.
R
Integrating factor :
IF = e
P (x)dx
cos x
sin x dx
R
=e
R
≡e
f 0 (x)
dx
f (x)
= eln(sin x) = sin x
sin x ·
dy
dx
+ sin x
i.e.
sin x ·
dy
dx
+ cos x · y = 1
i.e.
d
dx
Multiply equation :
Integrate:
cos x
sin x
y=
[sin x · y] = 1
(sin x)y = x + C .
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sin x
sin x
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Solutions to exercises
25
Exercise 8.
Integrating factor:
P (x) = cot x =
R
IF = e
Note:
P dx
cos x
f 0 (x)
≡
sin x
f (x)
R
=e
cos x
sin x dx
cos x
sin x
= eln(sin x) = sin x
Multiply equation:
sin x ·
Integrate:
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cos x
dy
+ sin x · y ·
= sin x · cos x
dx
sin x
d
i.e.
[sin x · y] = sin x · cos x
dx
Z
y sin x = sin x · cos x dx
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Solutions to exercises
26
Z
{ Note:
Z
sin x cos x dx ≡
f (x)f (x)dx ≡
Z
≡
i.e.
i.e.
Z
0
f df =
y sin x =
1
2
sin2 x + C
=
1
2
· 12 (1 − cos 2x) + C
f
df
· dx
dx
1 2
f +C }
2
4y sin x + cos 2x = C 0
where C 0 = 4C + 1
= constant
.
Return to Exercise 8
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Solutions to exercises
27
Exercise 9.
dy
+
dx
Standard form:
2x
x2 − 1
P (x) =
Integrating factor:
R
IF = e
Multiply equation:
P dx
y=
x
x2 − 1
2x
x2 − 1
R
=e
(x2 − 1)
2x
dx
x2 −1
2
= eln(x −1)
= x2 − 1
dy
+ 2x y = x
dx
(the original form of the equation was half-way there!)
d 2
i.e.
(x − 1)y = x
dx
Integrate:
(x2 − 1)y =
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1 2
x +C .
2
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Solutions to exercises
28
Exercise 10.
P (x) = − tan x
Q(x) = − sec x
IF = e−
R
tan x dx
= e−
R
sin x
cos x dx
= e+
R
− sin x
cos x dx
= eln(cos x) = cos x
Multiply by IF:
i.e.
y(0) = 1
d
dx
dy
cos x dx
− cos x ·
[cos x · y] = −1
sin x
cos x y
= − cos x · sec x
i.e. y cos x = −x + C .
i.e. y = 1 when x = 0 gives
cos(0) = 0 + C ∴ C = 1
i.e. y cos x = −x + 1 .
Return to Exercise 10
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