Mathematics · Ch 9 — Differential Equations

Homogeneous Differential Equations

9.4.2

Homogeneous Differential Equations

9.4.2 Homogeneous Differential Equations

What Makes a Function Homogeneous?

A function $F(x, y)$ is homogeneous of degree $n$ if replacing $x, y$ by $\lambda x, \lambda y$ (for any nonzero $\lambda$ ) gives

F(\lambda x, \lambda y) = \lambda^n F(x, y)

Function	$F(\lambda x, \lambda y)$	Degree $n$
$F_1(x, y) = y^2 + 2xy$	$\lambda^2(y^2 + 2xy) = \lambda^2 F_1(x, y)$	2
$F_2(x, y) = 2x - 3y$	$\lambda(2x - 3y) = \lambda F_2(x, y)$	1
$F_3(x, y) = \cos\left(\frac{y}{x}\right)$	$\cos\left(\frac{\lambda y}{\lambda x}\right) = \cos\left(\frac{y}{x}\right) = \lambda^0 F_3(x, y)$	0
$F_4(x, y) = \sin x + \cos y$	$\sin(\lambda x) + \cos(\lambda y) \neq \lambda^n F_4(x, y)$ for any $n$	Not homogeneous

Note

The degree $n$ can be any real number — it need not be an integer. In $F_3$ the degree is $0$ because the function depends only on the ratio $\frac{y}{x}$ .

Alternative Form of Homogeneous Functions

A homogeneous function of degree $n$ can always be written in one of two equivalent forms:

F(x, y) = x^n \, g\left(\frac{y}{x}\right) \quad \text{or} \quad F(x, y) = y^n \, h\left(\frac{x}{y}\right)

For example:

$F_1 = y^2 + 2xy = x^2\left[\left(\frac{y}{x}\right)^2 + 2\frac{y}{x}\right] = y^2\left[1 + 2\frac{x}{y}\right]$
$F_2 = 2x - 3y = x\left[2 - 3\frac{y}{x}\right] = y\left[2\frac{x}{y} - 3\right]$
$F_3 = \cos\left(\frac{y}{x}\right) = x^0 \cos\left(\frac{y}{x}\right)$

Watch out

$F_4 = \sin x + \cos y$ cannot be written as $x^n g\left(\frac{y}{x}\right)$ for any $n$ — the quickest way to check that a function is not homogeneous.

Defining a Homogeneous Differential Equation

The equation $\frac{dy}{dx} = F(x, y)$ is homogeneous if $F(x, y)$ is homogeneous of degree zero, in which case $F(x, y) = x^0 g\left(\frac{y}{x}\right) = g\left(\frac{y}{x}\right)$ :

\frac{dy}{dx} = g\left(\frac{y}{x}\right)

Method of Solution: Substitution $y = vx$

To solve $\frac{dy}{dx} = g\left(\frac{y}{x}\right)$ , substitute $y = vx$ , where $v$ is a function of $x$ .

›Proof

Step 1: Differentiate $y = vx$ with respect to $x$ :

$\frac{dy}{dx} = v + x\frac{dv}{dx}$

Step 2: Substitute into the equation:

$v + x\frac{dv}{dx} = g(v)$

Steps 3–4: Rearrange and separate variables:

$x\frac{dv}{dx} = g(v) - v \;\Rightarrow\; \frac{dv}{g(v) - v} = \frac{dx}{x}$

Step 5: Integrate both sides:

$\int \frac{dv}{g(v) - v} = \int \frac{dx}{x} + C$

Step 6: Replace $v$ by $\frac{y}{x}$ to get the general solution.

Tip

Check that $g(v) - v \neq 0$ before dividing. If $g(v) = v$ , then $\frac{dy}{dx} = \frac{y}{x}$ , a special case solved by direct integration.

Alternative Form: $\frac{dx}{dy} = h\left(\frac{x}{y}\right)$

If the equation is given as $\frac{dx}{dy} = F(x, y)$ with $F$ homogeneous of degree zero, substitute $x = vy$ (where $v$ is a function of $y$ ) instead:

›Proof

Differentiating $x = vy$ gives $\frac{dx}{dy} = v + y\frac{dv}{dy}$ . Substituting into $\frac{dx}{dy} = h(v)$ and separating: …

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Homogeneous Differential Equations