Download Homogeneous Equations

Jim Lambers
MAT 285
Spring Semester 2012-13
Lecture 8 Notes
These notes correspond to Section 2.2 in the text.
Homogeneous Equations
Now we will see that our approach to solving separable equations can be applied to certain problems
that, in their original form, are not necessarily separable. Suppose a first-order ODE can be written
in the form
dy
= f (y/x).
dx
Generally, such an equation is not separable with respect to the variables x and y. An equation of
this form is called homogeneous.
To solve this kind of equation, we use a change of dependent variable, v = y/x. Then, we have
dy
= f (v).
dx
However, before we can proceed, we must eliminate y from the equation entirely. To that end, we
differentiate the equation y = vx with respect to x, using the Product Rule, and obtain
dy
dv
=x
+ v.
dx
dx
Then, our transformed equation
dv
+ v = f (v)
dx
is separable, as we can rearrange it to obtain
x
dv
f (v) − v
=
,
dx
x
from which we can clearly see that dv/dx is the product of a function of x and a function of v.
Rewriting this equation in the usual form for separable equations,
1
1
dv
+
,
x v − f (v) dx
we can apply our solution formula to obtain
Z
Z
1
1
dx +
dv = 0.
x
v − f (v)
We can then solve this equation for v, and use the relation v = y/x to obtain the solution y of the
original equation.
Example Consider the equation
(x2 + 3xy + y 2 ) dx − x2 dy = 0,
1
which can be rewritten in the more conventional form
dy
x2 + 3xy + y 2
=
.
dx
x2
By decomposing the numerator into 3 separate terms, we obtain
dy
y2
y y 2
x2 3xy
= f (y/x),
= 2 + 2 + 2 =1+3 +
dx
x
x
x
x
x
where f (v) = 1 + 3v + v 2 . We can see that this equation is homogeneous.
Applying our solution formula, we obtain an equation for v = y/x,
Z
Z
1
1
dx +
dv = 0.
x
v − (1 + 3v + v 2 )
Evaluating the first integral and simplifying the second yields
Z
1
ln |x| −
dv = 0.
2
v + 2v + 1
As v 2 + 2v + 1 = (v + 1)2 , we can use the Power Rule for the second integral and obtain
ln |x| +
1
+ C = 0,
v+1
where C is an arbitrary constant.
Solving for v yields
v = −1 −
1
.
ln |x| + C
Finally, we use the relation v = y/x to obtain the solution of the original equation,
y = −x −
x
.
ln |x| + C
2
It is worth noting that this approach is useful for any equation of the form
ay + bx
dy
=
,
dx
cy + dx
where a, b, c and d are constants.
2