General Contents
Detailed Contents
Index
Homogeneous Differential Equations: Example 2
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Show that the differential equation,
, is homogeneous, and obtain its general solution.
General Contents
Detailed Contents
Index
How do we show that it is homogeneous?
One way is to substitute
tx
for
x
and
ty
for
y
, and then compare the resulting equation with the original equation.
If the equation is homogeneous, what do we get for the comparison?
The new equation will be
t
n
times the original.
We use the phrase “homogeneous of degree ___.”
What is the degree here?
The exponent of t, or n.
Substitute
tx
for
x
and
ty
for
y
in the original equation.
We get
Can we factor out a power of t?
Yes.
Do it.
What is our conclusion about homogeneity?
The equation is homogeneous.
What is the degree?
One.
How can we use the homogeneity to get the general solution?
We can introduce a new variable,
v
.
How is
v
defined?
We’ll need to substitute for
y
.
Express
y
in terms of
v
and
x
.
Would we use the same substitution for other degrees of homogeneity?
Yes.
We’ll also need to determine
dy
from
y
=
vx
. What rule of differentiation do we need to use?
The Product Rule.
Find
dy
.
dy
=
vdx
+
xdv
Substitute our result for
y
and
dy
in the original equation.
, or
Simplify by collecting the
vxdx
terms.
We get
, or
moving
dv
to the other side, we get
Can we finally separate variables?
Yes.
How?
By dividing by
on both sides.
Do it.
, or
Rewrite this with another trig function which will make it easier to integrate.
Are the variables separated now?
Yes.
Set up the integration.
Do the integration.
Putting the constant on the left side, we get
Combine the logs on the left.
We get
Take the inverse on both sides of the equation.
Simplify.
Are we done?
No.
What do we do next?
Substitute
to get a result in terms of
y
and
x
.
Do it.
Equation 1
How can we check this result?
We could determine
from Equation 1 and compare it to the original problem,
Do that, using the Chain Rule.
From Equation 1, we get
Equation 2
:
How does this compare?
It doesn’t yet look like the original problem.
Equation 2 contains the constant C and the exponential.
It also looks too complicated.
Well, we have differentiated Equation 1 to get this far.
Can we solve Equation 1 to find C?
Yes.
Do that.
We had
, so we get
Substitute this into Equation 2.
We get
Are we making progress?
It is still hard to decide.
Can we cancel the exponential now?
Yes.
Do it.
We get
What do we do now?
Multiply both sides by
We get
What is the result for the product of the
csc
and the sin?
We get 1, so we have finally,
,
which checks!
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, is homogeneous, and obtain its general solution.