# Homogeneous differential equation

A differential equation can be homogeneous in either of two respects.

A first order differential equation is said to be homogeneous if it may be written

${\displaystyle f(x,y)dy=g(x,y)dx,}$

where f and g are homogeneous functions of the same degree of x and y. In this case, the change of variable y = ux leads to an equation of the form

${\displaystyle {\frac {dx}{x}}=h(u)du,}$

which is easy to solve by integration of the two members.

Otherwise, a differential equation is homogeneous if it is a homogeneous function of the unknown function and its derivatives. In the case of linear differential equations, this means that there are no constant terms; the solutions of any linear ordinary differential equation of any order may be deduced by integration from the solution of the homogeneous equation obtained by removing the constant term.

## Homogeneous first-order differential equations

A first-order ordinary differential equation in the form:

${\displaystyle M(x,y)\,dx+N(x,y)\,dy=0}$

is a homogeneous type if both functions M(x, y) and N(x, y) are homogeneous functions of the same degree n.[1] That is, multiplying each variable by a parameter  ${\displaystyle \lambda }$, we find

${\displaystyle M(\lambda x,\lambda y)=\lambda ^{n}M(x,y)\,}$     and     ${\displaystyle N(\lambda x,\lambda y)=\lambda ^{n}N(x,y)\,.}$

Thus,

${\displaystyle {\frac {M(\lambda x,\lambda y)}{N(\lambda x,\lambda y)}}={\frac {M(x,y)}{N(x,y)}}\,.}$

### Solution method

In the quotient   ${\displaystyle {\frac {M(tx,ty)}{N(tx,ty)}}={\frac {M(x,y)}{N(x,y)}}}$, we can let   ${\displaystyle t=1/x}$   to simplify this quotient to a function ${\displaystyle f}$ of the single variable ${\displaystyle y/x}$:

${\displaystyle {\frac {M(x,y)}{N(x,y)}}={\frac {M(tx,ty)}{N(tx,ty)}}={\frac {M(1,y/x)}{N(1,y/x)}}=f(y/x)\,.}$

That is

${\displaystyle {\frac {dy}{dx}}=-f(y/x).}$

Introduce the change of variables ${\displaystyle y=ux}$; differentiate using the product rule:

${\displaystyle {\frac {dy}{dx}}={\frac {d(ux)}{dx}}=x{\frac {du}{dx}}+u{\frac {dx}{dx}}=x{\frac {du}{dx}}+u.}$

This transforms the original differential equation into the separable form

${\displaystyle x{\frac {du}{dx}}=-f(u)-u,}$

or

${\displaystyle {\frac {1}{x}}{\frac {dx}{du}}={\frac {-1}{f(u)+u}},}$

which can now be integrated directly: log x equals the antiderivative of the right-hand side (see ordinary differential equation).

### Special case

A first order differential equation of the form (a, b, c, e, f, g are all constants)

${\displaystyle (ax+by+c)dx+(ex+fy+g)dy=0\,}$

where afbe can be transformed into a homogeneous type by a linear transformation of both variables (${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are constants):

${\displaystyle t=x+\alpha ;\,\,\,\,z=y+\beta \,.}$

## Homogeneous linear differential equations

A linear differential equation is homogeneous if it is a homogeneous linear equation in the unknown function and its derivatives. It follows that, if ${\displaystyle \phi (x)}$ is a solution, so is ${\displaystyle c\phi (x)}$, for any (non-zero) constant c. In order for this condition to hold, each nonzero term of the linear differential equation must depend on the unknown function or any derivative of it. A linear differential equation that fails this condition is called inhomogeneous.

A linear differential equation can be represented as a linear operator acting on y(x) where x is usually the independent variable and y is the dependent variable. Therefore, the general form of a linear homogeneous differential equation is

${\displaystyle L(y)=0\,}$

where L is a differential operator, a sum of derivatives (defining the "0th derivative" as the original, non-differentiated function), each multiplied by a function  ${\displaystyle f_{i}}$  of x:

${\displaystyle L=\sum _{i=0}^{n}f_{i}(x){\frac {d^{i}}{dx^{i}}}\,,}$

where  ${\displaystyle f_{i}}$  may be constants, but not all  ${\displaystyle f_{i}}$  may be zero.

For example, the following differential equation is homogeneous:

${\displaystyle \sin(x){\frac {d^{2}y}{dx^{2}}}+4{\frac {dy}{dx}}+y=0\,,}$

whereas the following two are inhomogeneous:

${\displaystyle 2x^{2}{\frac {d^{2}y}{dx^{2}}}+4x{\frac {dy}{dx}}+y=\cos(x)\,;}$
${\displaystyle 2x^{2}{\frac {d^{2}y}{dx^{2}}}-3x{\frac {dy}{dx}}+y=2\,.}$

The existence of a constant term is a sufficient condition for an equation to be inhomogeneous, as in the above example.