A function *f*( *x,y*) is said to be **homogeneous of degree n **if the equation

*x,y*, and *z* (for which both sides are defined).

**Example 1**: The function *f*( *x,y*) = *x* ^{2} + *y* ^{2} is homogeneous of degree 2, since

**Example 2**: The function is homogeneous of degree 4, since

**Example 3**: The function *f*( *x,y*) = 2 *x* + *y* is homogeneous of degree 1, since

**Example 4**: The function *f*( *x,y*) = *x* ^{3} – *y* ^{2} is not homogeneous, since

*z* ^{n }*f*( *x,y*) for any *n*.

**Example 5**: The function *f*( *x,y*) = *x* ^{3} sin ( *y/x*) is homogeneous of degree 3, since

A first‐order differential equation **homogeneous** if *M*( *x,y*) and *N*( *x,y*) are both homogeneous functions of the same degree.

**Example 6**: The differential equation

*M*( *x,y*) = *x* ^{2} – *y* ^{2} and *N*( *x,y*) = *xy* are homogeneous functions of the same degree (namely, 2).

The method for solving homogeneous equations follows from this fact:

The substitution *y* = *xu* (and therefore *dy* = *xdu* + *udx*) transforms a homogeneous equation into a separable one.

**Example 7**: Solve the equation ( *x* ^{2} – *y* ^{2}) *dx* + *xy dy* = 0.

This equation is homogeneous, as observed in Example 6. Thus to solve it, make the substitutions *y* = *xu* and *dy* = *x dy* + *u dx*:

This final equation is now separable (which was the intention). Proceeding with the solution,

Therefore, the solution of the separable equation involving *x* and *v* can be written

To give the solution of the original differential equation (which involved the variables *x* and *y*), simply note that

Replacing *v* by *y*/ *x* in the preceding solution gives the final result:

This is the general solution of the original differential equation.

**Example 8:** Solve the IVP

*y* = *xv* and *dy* = *x dv* + *v dx* transform the equation into

The equation is now separable. Separating the variables and integrating gives

The integral of the left‐hand side is evaluated after performing a partial fraction decomposition:

Therefore,

The right‐hand side of (†) immediately integrates to

Therefore, the solution to the separable differential equation (†) is

Now, replacing *v* by *y*/ *x* gives

*y*(1) = 0 determines the value of the constant *c*:

Thus, the particular solution of the IVP is

Technical note: In the separation step (†), both sides were divided by ( *v* + 1)( *v* + 2), and *v* = –1 and *v* = –2 were lost as solutions. These need not be considered, however, because even though the equivalent functions *y* = – *x* and *y* = –2 *x* do indeed satisfy the given differential equation, they are inconsistent with the initial condition.