Understanding Homogeneous Differential Equations

A homogeneous differential equation is a central type in first-order ordinary differential equations, chiefly identified by the presence of homogeneous functions of the same degree in both variables. The structure and solution method for such equations relies critically on the property of homogeneity under simultaneous scaling of variables.

Homogeneity in Functions of Two Variables

A function $f(x, y)$ is said to be homogeneous of degree $n$ if, for any nonzero scalar $\lambda$, the identity $f(\lambda x, \lambda y) = \lambda^{n} f(x, y)$ holds. The integer $n$ is called the degree of homogeneity. For example, the function $f(x, y) = x^2 + xy$ is homogeneous of degree $2$ since $f(\lambda x, \lambda y) = (\lambda x)^2 + (\lambda x)(\lambda y) = \lambda^2 x^2 + \lambda^2 x y = \lambda^2 (x^2 + x y) = \lambda^2 f(x, y)$.

This property ensures that every term in $f(x, y)$ has the same combined degree (sum of powers of $x$ and $y$ in each term). Homogeneous functions play a central role in the classification and solution of differential equations of the type discussed herein.

Characterisation of Homogeneous Differential Equations of First Order

A first-order ordinary differential equation of the form \[ \frac{dy}{dx} = f(x, y) \] is called a homogeneous differential equation if $f(x, y)$ is a homogeneous function of degree zero. This means that $f(x, y)$ can be rewritten solely as a function of the ratio $\dfrac{y}{x}$ (or equivalently $\dfrac{x}{y}$) alone: \[ f(x, y) = F\left(\frac{y}{x}\right) \] for some function $F$.

Alternatively, any differential equation that can be written in the differential form \[ M(x, y)\,dx + N(x, y)\,dy = 0 \] where $M(x, y)$ and $N(x, y)$ are both homogeneous functions of the same degree, qualifies as a homogeneous equation. In such cases, division by $dx$ (assuming $dx \neq 0$) reduces the equation to a form compatible with the scaling property above.

It is crucial to recognise that the definition restricts $f(x, y)$ to exclude terms involving only $x$ or only $y$ without corresponding terms in the opposite variable, as such terms would alter the degree or violate the scaling property. Additionally, the absence of constant (degree zero) terms is mandatory.

Structural and Solution Properties of Homogeneous Differential Equations

The invariance of $f(x, y)$ under simultaneous scaling allows the reduction of every homogeneous equation to a separable form via substitution. Specifically, for equations where $f(x, y)$ is homogeneous of degree zero, introduce the substitution $y = v x$ with $v$ being a function of $x$. Through this substitution, the original differential equation is transformed into an equation involving $v$ and $x$ only, which can be further separated and integrated.

The distinctive feature of homogeneous equations is that the derivatives of $y$ with respect to $x$ only depend on their ratio, instilling a certain symmetry under scaling. This characteristic is what enables universal reduction by substitution, irrespective of the explicit form of $f(x, y)$, provided the homogeneity condition is satisfied.

Derivation of the Standard Method for Solution via Variable Substitution

Consider the homogeneous differential equation: \[ \frac{dy}{dx} = f(x, y) \] with $f(x, y)$ homogeneous of degree zero. By the property of homogeneity, $f(x, y) = F\left(\frac{y}{x}\right)$ for some function $F$.

Let $y = v x$, where $v$ is a differentiable function of $x$. According to the product rule: \[ \frac{dy}{dx} = \frac{d}{dx}(v x) = v \frac{d}{dx} (x) + x \frac{d}{dx} (v) = v + x \frac{dv}{dx} \] Substituting $y$ and $\frac{dy}{dx}$ into the original equation yields: \[ v + x \frac{dv}{dx} = F(v) \] Subtract $v$ from both sides to obtain: \[ x \frac{dv}{dx} = F(v) - v \] Rewriting: \[ \frac{dv}{dx} = \frac{F(v) - v}{x} \] This expression may now be rearranged to group the variables: \[ \frac{dv}{F(v) - v} = \frac{dx}{x} \]

Both sides are now expressed in terms of a single variable. Integrating each side gives: \[ \int \frac{1}{F(v) - v} \, dv = \int \frac{1}{x} \, dx \] \[ \int \frac{1}{F(v) - v} \, dv = \ln |x| + C \] where $C$ is the constant of integration.

To obtain the final solution in terms of the original variables, it is necessary to reverse the substitution and express $v$ in terms of $y$ and $x$. Recall $v = \frac{y}{x}$, so the solution is completed by back-substituting: \[ \int \frac{1}{F\left(\frac{y}{x}\right) - \frac{y}{x}} \, d\left(\frac{y}{x}\right) = \ln|x| + C \]

Conditions, Limitations, and Validity of the Substitution Approach

The outlined substitution method applies only when the function $f(x, y)$ can be written solely in terms of the ratio $\dfrac{y}{x}$ or is homogeneous of degree zero. If $f(x, y)$ is not homogeneous, or presents additional nonhomogeneous terms, or involves $x$ and $y$ inside nonlinearities (such as logarithmic or trigonometric arguments not reducible to functions of $\dfrac{y}{x}$), this substitution does not lead to separability, and alternative solution methods must be used.

Additionally, if after substitution $F(v) - v = 0$ for some values of $v$, the denominator in the integral vanishes and special care may be required to account for singular solutions. In standard cases, however, this obstacle does not inhibit the main integration method.

A detailed study of other types of first-order equations can be found in the main article on Differential Equations.

Worked Example 1: Verification of Homogeneity

Given: $(x - y)\dfrac{dy}{dx} = x + 2y$

Substitution: Express the right-hand side as a function of $(x, y)$, then check homogeneity: \[ f(x, y) = \frac{x + 2y}{x - y} \] Substitute $x \rightarrow \lambda x$, $y \rightarrow \lambda y$ for arbitrary nonzero $\lambda$: \[ f(\lambda x, \lambda y) = \frac{\lambda x + 2\lambda y}{\lambda x - \lambda y} = \frac{\lambda (x + 2y)}{\lambda (x - y)} = \frac{x + 2y}{x - y} = f(x, y) \]

Final result: $f(x, y)$ is homogeneous of degree zero; hence, the equation is homogeneous.

Worked Example 2: Solving a Homogeneous Equation

Given: $x \cos\left(\dfrac{y}{x}\right)\dfrac{dy}{dx} = y \cos\left(\dfrac{y}{x}\right) + x$

Substitution: Divide both sides by $x \cos\left(\frac{y}{x}\right)$: \[ \frac{dy}{dx} = \frac{y \cos\left(\frac{y}{x}\right) + x}{x \cos\left(\frac{y}{x}\right)} = \frac{y}{x} + \frac{1}{\cos\left(\frac{y}{x}\right)} \] Let $v = \frac{y}{x}$, so $y = v x$.

Reduction to Separable Form: Compute $\dfrac{dy}{dx}$ in terms of $v$ and $x$: \[ \frac{dy}{dx} = v + x \frac{dv}{dx} \] Substitute: \[ v + x \frac{dv}{dx} = v + \frac{1}{\cos v} \] \[ x \frac{dv}{dx} = \frac{1}{\cos v} \] \[ \cos v \, dv = \frac{dx}{x} \]

Integration: \[ \int \cos v \, dv = \int \frac{dx}{x} \] \[ \sin v = \ln |x| + C \] Recall $v = \frac{y}{x}$, \[ \sin\left(\frac{y}{x}\right) = \ln |x| + C \]

Final result: The general solution is $\sin\left(\dfrac{y}{x}\right) = \ln |x| + C$.

Worked Example 3: General Solution in Differential Form

Given: $(x^2 + y^2)dy - x y dx = 0$

Substitution: First, rearrange the equation: \[ (x^2 + y^2)\, dy - x y\, dx = 0 \] \[ (x^2 + y^2)\, dy = x y\, dx \] \[ \frac{dy}{dx} = \frac{x y}{x^2 + y^2} \]

Let $v = \frac{y}{x}$, so $y = v x$, and thus \[ \frac{dy}{dx} = v + x \frac{dv}{dx} \] Substitute back: \[ v + x \frac{dv}{dx} = \frac{x v x}{x^2 + v^2 x^2} \] \[ v + x \frac{dv}{dx} = \frac{v}{1 + v^2} \] \[ x \frac{dv}{dx} = \frac{v}{1 + v^2} - v = -\frac{v^3}{1 + v^2} \] \[ \frac{dv}{dx} = -\frac{v^3}{x (1 + v^2)} \] \[ \frac{1 + v^2}{v^3} dv = -\frac{dx}{x} \] \[ \left( \frac{1}{v^3} + \frac{v^2}{v^3} \right) dv = -\frac{dx}{x} \] \[ \left( v^{-3} + v^{-1} \right) dv = -\frac{dx}{x} \] \[ \int \left( v^{-3} + v^{-1} \right) dv = - \int \frac{dx}{x} \] \[ \int v^{-3} dv + \int v^{-1} dv = - \ln|x| + C \] \[ -\frac{1}{2} v^{-2} + \ln |v| = - \ln|x| + C \] \[ -\frac{1}{2} \frac{x^2}{y^2} + \ln \left| \frac{y}{x} \right| = - \ln|x| + C \]

Final result: This is the implicit general solution: \[ - \frac{1}{2} \frac{x^2}{y^2} + \ln \left| \frac{y}{x} \right| + \ln|x| = C \]

Characteristic Difference: Homogeneous vs. Nonhomogeneous Differential Equations

A first-order differential equation is homogeneous only if all function terms meet the homogeneity property. In contrast, an equation such as \[ \frac{dy}{dx} + P(x) y = Q(x) \] with $Q(x)$ not identically zero, is not homogeneous as $Q(x)$ does not scale with the dependent variable in a way required by homogeneity. Such equations are known as linear but nonhomogeneous.

Further Study and Practice

Homogeneous differential equations form a significant sub-class of first-order ODEs, with solution approaches generalizable to other related forms by suitable transformation. For broader coverage, refer to the Differential Equations Important Questions page.

Additional worked problems and solved examples can be found on the Differential Equations Practice Paper and in the article Homogeneous Differential Equation for further mastery.