“JUST THE MATHS”
UNIT NUMBER
15.3
ORDINARY
DIFFERENTIAL EQUATIONS 3
(First order equations (C))
by
A.J.Hobson
15.3.1
15.3.2
15.3.3
15.3.4
Linear equations
Bernouilli’s equation
Exercises
Answers to exercises
UNIT 15.3 - ORDINARY DIFFERENTIAL EQUATIONS 3
FIRST ORDER EQUATIONS (C)
15.3.1 LINEAR EQUATIONS
For certain kinds of first order differential equation, it is possible to multiply the equation
throughout by a suitable factor which converts it into an exact differential equation.
For instance, the equation
dy 1
+ y = x2
dx x
may be multiplied throughout by x to give
x
dy
+ y = x3 .
dx
It may now be written
d
(xy) = x3
dx
and, hence, it has general solution
xy =
x4
+ C,
4
where C is an arbitrary constant.
Notes:
(i) The factor, x which has multiplied both sides of the differential equation serves as an
“integrating factor”, but such factors cannot always be found by inspection.
(ii) In the discussion which follows, we shall develop a formula for determining integrating
factors, in general, for what are known as “linear differential equations”.
1
DEFINITION
A differential equation of the form
dy
+ P (x)y = Q(x)
dx
is said to be “linear”.
RESULT
Given the linear differential equation
dy
+ P (x)y = Q(x),
dx
the function
R
e
P (x) dx
is always an integrating factor; and, on multiplying the differential equation throughout by
this factor, its left hand side becomes
R
d h
y×e
dx
P (x) dx
i
.
Proof
Suppose that the function, R(x), is an integrating factor; then, in the equation
R(x)
dy
+ R(x)P (x)y = R(x)Q(x),
dx
the left hand side must be the exact derivative of some function of x.
Using the formula for differentiating the product of two functions of x, we can make it the
derivative of R(x)y provided we can arrange that
R(x)P (x) =
2
d
[R(x)].
dx
But this requirement can be interpreted as a differential equation in which the variables R(x)
and x may be separated as follows:
Z
Z
1
dR(x) =
P (x) dx.
R(x)
Hence,
ln R(x) =
Z
P (x) dx.
That is,
R
R(x) = e
P (x) dx
,
as required.
The solution is obtained by integrating the formula
d
[y × R(x)] = R(x)P (x).
dx
Note:
There is no need to include an arbitrary constant, C, when P (x) is integrated, since it
would only serve to introduce a constant factor of eC in the above result, which would then
immediately cancel out on multiplying the differential equation by R(x).
EXAMPLES
1. Determine the general solution of the differential equation
dy 1
+ y = x2 .
dx x
Solution
An integrating factor is
R
e
1
x
dx
3
= eln x = x.
On multiplying throughout by the integrating factor, we obtain
d
[y × x] = x3 ;
dx
and so,
yx =
x4
+ C,
4
where C is an arbitrary constant.
2. Determine the general solution of the differential equation
dy
2
+ 2xy = 2e−x .
dx
Solution
An integrating factor is
R
e
2x dx
2
= ex .
Hence,
i
d h
2
y × ex = 2,
dx
giving
2
yex = 2x + C,
where C is an arbitrary constant.
15.3.2 BERNOUILLI’S EQUATION
A similar type of differential equation to that in the previous section has the form
dy
+ P (x)y = Q(x)y n .
dx
4
It is called “Bernouilli’s Equation” and may be converted to a linear differential equation
by making the substitution
z = y 1−n .
Proof
The differential equation may be rewritten as
y −n
dy
+ P (x)y 1−n = Q(x).
dx
Also,
dz
dy
= (1 − n)y −n .
dx
dx
Hence the differential equation becomes
1 dz
+ P (x)z = Q(x).
1 − n dx
That is,
dz
+ (1 − n)P (x)z = (1 − n)Q(x),
dx
which is a linear differential equation.
Note:
It is better not to regard this as a standard formula, but to apply the method of obtaining
it in the case of particular examples.
EXAMPLES
1. Determine the general solution of the differential equation
xy −
dy
2
= y 3 e−x .
dx
5
Solution
The differential equation may be rewritten
−y −3
Substituting z = y −2 , we obtain
dz
dx
dy
2
+ x.y −2 = e−x .
dx
dy
= −2y −3 dx
and, hence,
1 dz
2
+ xz = e−x
2 dx
or
dz
2
+ 2xz = 2e−x .
dx
An integrating factor for this equation is
R
e
2x dx
2
= ex .
Thus,
d x2 ze
= 2,
dx
giving
2
zex = 2x + C,
where C is an arbitrary constant.
Finally, replacing z by y −2 ,
2
ex
y =
.
2x + C
2
2. Determine the general solution of the differential equation
dy y
+ = xy 2 .
dx x
6
Solution
The differential equation may be rewritten
y −2
On substituting z = y −1 we obtain
dy 1 −1
+ .y = x.
dx x
dz
dx
−
dy
= −y −2 dx
so that
dz
1
+ .z = x
dx x
or
dz
1
− .z = −x.
dx x
An integrating factor for this equation is
R
1
e (− x )
dx
= e− ln x =
1
.
x
Hence,
d
1
z×
= −1,
dx
x
giving
z
= −x + C,
x
where C is an arbitrary constant.
The general solution of the given differential equation is therefore
1
1
= −x + C or y =
.
xy
Cx − x2
15.3.3 EXERCISES
Use an integrating factor to solve the following differential equations subject to the given
boundary condition:
1.
3
dy
+ 2y = 0,
dx
where y = 10 when x = 0.
7
2.
3
dy
− 5y = 10,
dx
where y = 4 when x = 0.
3.
dy y
+ = 3x,
dx x
where y = 2 when x = −1.
4.
dy
y
+
= 1 − x2 ,
dx 1 − x
where y = 0 when x = −1.
5.
dy
+ y cot x = cos x,
dx
where y =
5
2
when x = π2 .
6.
(x2 + 1)
dy
− xy = x,
dx
where y = 0 when x = 1.
7.
3y − 2
dy
= y 3 e4x ,
dx
where y = 1 when x = 0.
8.
2y − x
dy
= x(x − 1)y 4 ,
dx
where y 3 = 14 when x = 1.
8
15.3.4 ANSWERS TO EXERCISES
1.
2
y = 10e− 3 x .
2.
5
y = 6e 3 x − 2.
3.
yx = x3 − 1.
4.
1
y = (1 − x)(1 + x)2 .
2
5.
y=
sin x
2
+
.
2
sin x
6.
y = 1 + x2 −
q
2(1 + x2 ).
7.
y2 =
7e3x
.
e7x + 6
8.
y3 =
56x6
.
21x6 − 24x7 + 7
9