differential equation introduce

Solving Differential Equationswith Integrating Factors
mccp-dobson-0111
Introduction
Suppose we have the first order differential equation
dy
dx
+ Py = Q
where P and Q are functions involving x only. For example
dy
dx
+
3y
x
=
ex
x3
or
dy
dx
−
3y
x + 1
= (x + 1)4.
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
I = eR P dx.
General Solution
Multiplying our original differential equation by I we get that
dy
dx
+ Py = Q , I
dy
dx
+ IPy = IQ
, Z (I
dy
dx
+ IPy) dx = Z IQdx
, Iy = Z IQdx since
d
dx
(Iy) = I
dy
dx
+ IPy by the product rule.
As both I and Q are functions involving only x in most of the problems you are likely to meet, R IQdx
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.
Example
To find the general solution of the differential equation
dy
dx
+
3y
x
=
ex
x3
www.mathcentre.ac.uk

c Katy Dobson
University of Leeds
Alan Slomson
University of Leeds
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we first find the integrating factor
I = eR P dx = eR 3
x
dx
now Z 3
x
dx = 3 ln x = ln x3
hence I = eln x3
= x3.
Then we multiply the differential equation by I to get
x3 dy
dx
+ 3x2y = ex
so integrating both sides we have x3y = 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
dx
−
3y
x + 1
= (x + 1)4
we first find the integrating factor
I = eR P dx = eR −3
x+1
dx
now Z −3
x + 1
dx = −3 ln(x + 1) = ln(x + 1)−3
hence I = eln(x+1)−3
= (x + 1)−3 =
1
(x + 1)3
.
Then multiplying the differential equation by I we get
1
(x + 1)3
dy
dx
−
3y
(x + 1)4 = (x + 1)
so integrating both sides we have
y
(x + 1)3 =
1
2
x2 + x + c where c is a constant.
Thus the general solution is
y = (x + 1)3(
1
2
x2 + x + c).
Exercises
Find the general solution of
1.
dy
dx
+
2y
x
=
sin x
x2 2.
dy
dx
−
y
x
= −xe−x 3.
dy
dx
+ 2xy = x 4.
dy
dx
−
2y
x
= 3x3
Answers
1. y =
c − cos x
x2 2. y = x(e−x + c) 3. y =
1
2
+ ce−x2
4. y =
3
2
x4 + cx2
www.mathcentre.ac.uk

c Katy Dobson
University of Leeds
Alan Slomson
University of Leeds