“JUST THE MATHS”
UNIT NUMBER
15.2
ORDINARY
DIFFERENTIAL EQUATIONS 2
(First order equations (B))
by
A.J.Hobson
15.2.1
15.2.2
15.2.3
15.2.4
Homogeneous equations
The standard method
Exercises
Answers to exercises
UNIT 15.2 - ORDINARY DIFFERENTIAL EQUATIONS 2
FIRST ORDER EQUATIONS (B)
15.2.1 HOMOGENEOUS EQUATIONS
A differential equation of the first order is said to be “homogeneous” if, on replacing x
dy
by λx and y by λy in all the parts of the equation except dx
, λ may be removed from the
n
equation by cancelling a common factor of λ , for some integer n.
Note:
Some examples of homogeneous equations would be
(x + y)
dy
dy
+ (4x − y) = 0, and 2xy
+ (x2 + y 2 ) = 0,
dx
dx
where, from the first of these, a factor of λ could be cancelled and, from the second, a factor
of λ2 could be cancelled.
15.2.2 THE STANDARD METHOD
It turns out that the substitution
!
y = vx
dy
dv
giving
=v+x
,
dx
dx
always converts a homogeneous differential equation into one in which the variables can be
separated. The method will be illustrated by examples.
EXAMPLES
1. Solve the differential equation
x
dy
= x + 2y,
dx
subject to the condition that y = 6 when x = 6.
Solution
dy
dv
If y = vx, then dx
= v + x dx
, so that the differential equation becomes
dy
x v+x
dx
1
!
= x + 2vx.
That is,
v+x
dv
= 1 + 2v
dx
or
x
dv
= 1 + v.
dx
On separating the variables,
Z
Z
1
1
dv =
dx,
1+v
x
giving
ln(1 + v) = ln x + ln A,
where A is an arbitrary constant.
An alternative form of this solution, without logarithms, is
Ax = 1 + v
and, substituting back v = xy , the solution becomes
Ax = 1 +
y
x
or
y = Ax2 − x.
Finally, if y = 6 when x = 1, we have 6 = A − 1 and, hence, A = 7.
The required particular solution is thus
y = 7x2 − x.
2. Determine the general solution of the differential equation
(x + y)
dy
+ (4x − y) = 0.
dx
2
Solution
If y = vx, then
dy
dx
dv
= v + x dx
, so that the differential equation becomes
!
dv
+ (4x − vx) = 0.
(x + vx) v + x
dx
That is,
!
dv
(1 + v) v + x
+ (4 − v) = 0
dx
or
v+x
dv
v−4
=
.
dx
v+1
On further rearrangement, we obtain
x
dv
v−4
−4 − v 2
=
−v =
;
dx
v+1
v+1
and, on separating the variables,
Z
Z
v+1
1
dv
=
−
dx
2
4+v
x
or
1Z
2
2v
2
+
2
4+v
4 + v2
dv = −
Z
1
dx.
x
Hence,
1
v
ln(4 + v 2 ) + tan−1 = − ln x + C,
2
2
where C is an arbitrary constant.
Substituting back v = xy , gives the general solution
1
y2
ln 4 + 2
2
x
"
!
+ tan
3
−1
y
2x
#
= − ln x + C.
3. Determine the general solution of the differential equation
2xy
Solution
If y = vx, then
dy
dx
dy
+ (x2 + y 2 ) = 0.
dx
dv
= v + x dx
, so that the differential equation becomes
!
2vx
2
dv
v+x
+ (x2 + v 2 x2 ) = 0.
dx
That is,
!
dv
2v v + x
+ (1 + v 2 ) = 0
dx
or
2vx
dv
= −(1 + 3v 2 ).
dx
On separating the variables, we obtain
Z
Z
2v
1
dx
=
−
dx,
1 + 3v 2
x
which gives
1
ln(1 + 3v 2 ) = − ln x + ln A,
3
where A is an arbitrary constant.
Hence,
1
(1 + 3v 2 ) 3 =
A
x
or, on substituting back v = xy ,
x2 + 3y 2
x2
!1
3
= Ax,
which can be written
x2 + 3y 2 = Bx5 ,
where B = A3 .
4
15.2.3 EXERCISES
Use the substitution y = vx to solve the following differential equations subject to the given
boundary condition:
1.
(2y − x)
dy
= 2x + y,
dx
where y = 3 when x = −2.
2.
(x2 − y 2 )
dy
= xy,
dx
where y = 5 when x = 0.
3.
x3 + y 3 = 3xy 2
dy
,
dx
where y = 1 when x = 2.
4.
x(x2 + y 2 )
dy
= 2y 3 ,
dx
where y = 2 when x = 1.
5.
x
q
dy
− (y + x2 − y 2 ) = 0,
dx
where y = 0 when x = 1.
5
15.2.4 ANSWERS TO EXERCISES
1.
y 2 − xy − x2 = 11.
2.
−
y = 5e
x2
2y 2
.
3.
x3 − 2y 3 = 3x.
4.
3x2 y = 2(y 2 − x2 ).
5.
esin
−1 y
x
6
= x.