You learn very early in your study of mathematics that, many times, the best way to solve a new problem is to reduce it, in some way, into the form of a problem that you already know how to solve. This is what you do with homogeneous differential equations. If you recognize the fact that an equation is homogeneous you can, in some cases, perform a substitution which will allow you to use separation of variables to solve the equation.
At this point you may be asking yourself, what is a homogeneous differential equation? It is simply an equation where both coefficients of the differentials dx and dy are homogeneous. To test this you first place the equation into the differential form:
If after testing for homogeneity and finding that the equation is homogeneous, you call it a first order differential equation with homogeneous coefficients. Remember, coefficients, in the context of differential equations, can be functions. In the equation above, the functions M (x,y) and N (x,y) are the coefficients. The next thing you may want to know is, what makes a function homogeneous? Below is a review of what you may have studied in Algebra.
Very simply put, homogeneous functions are functions where the sum of the powers of every term are the same. So the first function below is homogeneous of degree 3, the second and third are not homogeneous.
This method of recognizing a homogeneous function does not work all of the time, but is useful for many examples. A more formal definition states that if a function is homogeneous it can be written in the following form:
In the following example you will look at a simple function that you can determine by inspection. Then you will look at a more complicated one to show how LiveMath can help when inspection is more difficult.
Part 1:
The wildcard function notation is necessary for the method to work. For the target Prop, wildcard variables are not necessary.
Notice how the last conclusion is achieved by substituting the initial function (with the WC variables) into the right side of the 5th Prop. The result confirms what is obvious by inspection, that the function is homogeneous of degree 2 (i.e. t is to the second power).
The homogeneity of the function in the next example is not as easy to determine by inspection.
Part 2:
Solution (make sure Auto-Simplify is turned OFF):
The function is homogeneous of degree 1.
NOTE: You could have also done Part 2 by just changing the function in Part 1, rather than performing the exercise from the beginning. Before you do, make sure you have Auto ReManipulation turned on and only change the RHS of the first Prop (this will maintain the link established in the first manipulation).
Another method of determining homogeneity uses definition #1. By using the substitution u=y/x or v=x/y the x's or y's can be isolated from a function of u. The degree x or y determines the degree of the function.
Expl #6 Another Test for Homogeniety
Since the variable x is isolated from a new function of u, it shows that the function is homogeneous (of degree 4). In like fashion, you could have used the substitution v = x/y and achieved the same answer in terms of y.
For the most part, homogeneous differential equations are solved by means of an algebraic substitution. This substitution is the same as was used in Example 6. As you can see in this example, by making the substitution, it looks like you can separate the variables, and if so, you should be able to solve the equation using the method of Separation of Variables.
After an intermediate solution is determined using the substitution, the equation will be solved and the substitution equation will be placed back in to achieve the answer.
Expl #7 Homogeneous Solution #1
You can place this equation into a form to test for homogeneity by using the method discussed at the start of the section on Separation of Variables.
By inspection you can see that both 
and
are homogeneous of degree 2. This is necessary
for the equation to be considered homogeneous. That is to say,
both M(x,y) and N(x,y) must be homogeneous of the same degree.
For the substitution itself, a rule of thumb is that you should use x = vy if M(x,y) is simpler than N(x,y) and y = u x if N(x,y) is simpler than M(x,y) . It looks like you should use y = ux for this problem.
NOTE: Make sure you sub in dy first. Otherwise you may have problems with the y Prop going into the y of the dy.
There is another way of solving this problem. It is a method that is derived from the fact that you can write a homogeneous equation in the form:
The following example demonstrates this method.
Expl #8 Homogeneous Solution #2
Below is the resulting Prop after the substitutions:
First Order Ordinary
