Homogeneous functions & ODEs

Before we start this section, we make a note on terminology. The term homogeneous has more than one meaning with regard to differential equations. For the linear differential equation (of any order), homogeneity implies that the equation does not contain a term that is a function of the independent variable alone. For example, in the case of the first order linear ODE given by (12.19), the equation is homogeneous if $f(x)=0$. A linear ODE that is not homogeneous is referred to as inhomogeneous or nonhomogeneous.

The second meaning, which is the focus of this section, requires the coefficients of the differentials ${ }^{20}$ in first order ODEs be homogeneous functions. A homogeneous function is one whose terms are all of the same degree with respect to all the variables taken together. Put more mathematically, a homogeneous function of degree $n$ satisfies the condition

for some integer $n$. For example, consider the function

For $f(t x, t y)$, we have,

Hence, $f(x, y)$ is homogeneous function of degree 4. A differential equation of the form

${ }^{20}$ Given a function $y=f(x)$, the differential terms are $d y$ and $d x$ and the relationship between them is $d y=f^{\prime} d x$. is said to be homogeneous if the functions $P(x, y)$ and $Q(x, y)$ are homogeneous functions of the same degree. An equation of this form can be transformed into one in which the variables are separable through a transformation $y=v x$, where $v=v(x)$ is a new variable.

The reason why the substitution $y=v x$ transforms the equation into a separable ODE is that if, $P(x, y)$ and $Q(x, y)$ are homogeneous functions of the same degree, then substituting $y=v x$ in Eq. (12.40) makes the RHS of the following equation to be a function of $v$ alone,

The LHS is given by

Equations (12.41) and (12.42) give the following separable ODE expressed in differential form,

Example 12.5 Solve the following differential equation using the substitution $y=v x$, where $v=v(x)$,

Solution In differential form, Eq. (12.44) is expressed as

with $P(x, y)=x^{2}-3 y^{2}$ and $Q(x, y)=2 x y$, which are homogeneous functions of degree 2. If $y=v x$, then

Using Eq. (12.45) and $y=v x$ in the original ODE and after some algebra, yields

The above is a separable ODE in the variables $v$ and $x$ which, when integrated, gives us the following implicit, general solution for $v(x)$,

Finally, we replace $v=y / x$ to obtain an implicit solution for $y(x)$,

Exercises

1. Solve the following IVP,

2. Solve the following ODE giving your answer in implicit form,