Homogeneous differential equation

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. That is, multiplying each variable by a parameter λ, we find

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

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 $\frac{M(tx,ty)}{N(tx,ty)}=\frac{M(x,y)}{N(x,y)}$, we can let t = ⁠1/x⁠ to simplify this quotient to a function f of the single variable ⁠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 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: ln 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) $${\displaystyle \left(ax+by+c\right)dx+\left(ex+fy+g\right)dy=0}$$ where af ≠ be can be transformed into a homogeneous type by a linear transformation of both variables (α and β are constants): $${\displaystyle t=x+\alpha ;z=y+\beta ,}$$ where $${\displaystyle \alpha =\frac{cf-bg}{af-be};\beta =\frac{ag-ce}{af-be}.}$$ For cases where af = be, introduce the change of variables u = ax + by or u = ex + fy; differentiation yields: $${\displaystyle \frac{du}{dx}=a-b(\frac{ac+au}{ag+eu}),}$$ or $${\displaystyle \frac{du}{dx}=e-f(\frac{ec+au}{eg+eu}),}$$ for each respective substitution. Both may be solved via separation of variables.

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 φ(x) is a solution, so is cφ(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 f_i of x:

$${\displaystyle L=\sum _{i=0}^n{f}_i(x)\frac{d^i}{d{x}^i},}$$ where f_i may be constants, but not all f_i may be zero.

For example, the following linear differential equation is homogeneous:

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

whereas the following two are inhomogeneous:

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

$${\displaystyle 2x^2\frac{d^{2}y}{d{x}^2}-3x\frac{dy}{dx}+y=2.}$$ The existence of a constant term is a sufficient condition for an equation to be inhomogeneous, as in the above example.