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Solving Differential Equations with Integrating Factors
mccp-dobson-0111
Introduction
Suppose we have the first order differential equation
dy
+ Py = Q
dx
where P and Q are functions involving x only. For example
dy 3y
ex
+
= 3
dx
x
x
dy
3y
−
= (x + 1)4 .
dx x + 1
or
We can solve these differential equations using the technique of an integrating factor.
Integrating Factor
We multiply both sides of the differential equation by the integrating factor I which is defined as
R
I=e
P dx
.
General Solution
Multiplying our original differential equation by I we get that
dy
dy
+ Py = Q ⇔ I
+ IP y = IQ
dx
Z
Z dx
dy
+ IP y) dx = IQ dx
⇔ (I
dx
Z
⇔ Iy = IQ dx
d
dy
(Iy) = I
+ IP y by the product rule.
dx
dx
R
As both I and Q are functions involving only x in most of the problems you are likely to meet, IQ dx
can usually be found. So the general solution to the differential equation is found by integrating IQ
and then re-arranging the formula to make y the subject.
since
Example
To find the general solution of the differential equation
ex
dy 3y
+
= 3
dx
x
x
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c
Katy
Dobson
University of Leeds
Alan Slomson
University of Leeds
we first find the integrating factor
R
R
3
I = e P dx = e x dx
Z
3
dx = 3 ln x = ln x3
x
3
I = eln x = x3 .
mathcentre community project
now
hence
encouraging academics
to share maths support resources
All mccp resources are released under an Attribution Non-commerical Share Alike licence
Then we multiply the differential equation by I to get
x3
dy
+ 3x2 y = ex
dx
so integrating both sides we have x3 y = ex + c
where c is a constant. Thus the general solution is
y=
ex + c
.
x3
Example
To find the general solution of the differential equation
dy
3y
−
= (x + 1)4
dx x + 1
we first find the integrating factor
R
R
now
hence
I = e P dx = e x+1 dx
Z
−3
dx = −3 ln(x + 1) = ln(x + 1)−3
x+1
1
−3
.
I = eln(x+1) = (x + 1)−3 =
(x + 1)3
−3
Then multiplying the differential equation by I we get
1
3y
dy
−
= (x + 1)
3
(x + 1) dx (x + 1)4
so integrating both sides we have
1
y
= x2 + x + c
3
(x + 1)
2
Thus the general solution is
where c is a constant.
1
y = (x + 1)3 ( x2 + x + c).
2
Exercises
Find the general solution of
1.
dy 2y
sin x
+
= 2
dx
x
x
2.
dy
y
− = −xe−x
dx x
2.
y = x(e−x + c)
3.
dy
+ 2xy = x
dx
4.
dy 2y
−
= 3x3
dx
x
Answers
1.
y=
c − cos x
x2
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3.
y=
c
Katy
Dobson
University of Leeds
1
2
+ ce−x
2
4.
3
y = x4 + cx2
2
Alan Slomson
University of Leeds