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
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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. Toc
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Section 1: Theory
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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. Toc
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Section 2: Exercises
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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 Toc
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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 Toc JJ II J I Back
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Section 2: Exercises
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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 Toc
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Section 3: Answers
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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
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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
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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 4
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Section 4: Standard integrals
f (x) 1 a2 +x2
√ 1 a2 −x2
√
a2 − x2
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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
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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
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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 Toc
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b a
dx
bx
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Solutions to exercises
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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
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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 Toc
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Solutions to exercises
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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
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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
=
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10x3 10x3 5 4 x +C 2 C 5 2 x + 2 2 x
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Solutions to exercises
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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
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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. Toc
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1 y x y II
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)
=
1 . x
Solutions to exercises
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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 :
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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
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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
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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
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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 â&#x2030;Ąe
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 . Toc
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sin x sin x
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Solutions to exercises
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Exercise 8. Integrating factor:
P (x) = cot x = R
IF = e Note:
P dx
cos x f 0 (x) â&#x2030;Ą 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
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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
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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 II
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Solutions to exercises
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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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