Homogeneous Differential Equations Made Simple

A homogeneous differential equation is a first-order equation dy/dx = F(x, y) where F is unchanged when x and y are scaled by the same factor, which means F really depends only on the ratio y/x. The substitution y = vx then collapses it into a separable equation.

This matters because many geometric and physical problems involve only ratios: slopes determined by direction rather than position, or rates depending on relative rather than absolute amounts. Recognising homogeneity unlocks an otherwise stubborn equation with one substitution.

Key Concepts

• Homogeneous function of degree n: F(tx, ty) = tⁿF(x, y). For differential equations we need the right-hand side to be homogeneous of degree 0, meaning scaling changes nothing.
• Ratio test: replace every x by 1 and every y by y/x (equivalently set x = 1, or divide numerator and denominator by the highest power of x); if only y/x remains, the equation is homogeneous.
• The substitution: set y = vx, where v is a new function of x. By the product rule, dy/dx = v + x·dv/dx.
• Reduction to separable: after substitution the equation becomes v + x·dv/dx = F(v), which rearranges to the separable form dv/(F(v) − v) = dx/x.
• Back-substitution: the final answer must be written in x and y by replacing v with y/x.

Formulas

• Standard setup: for dy/dx = F(y/x), substitute y = vx to get x·dv/dx = F(v) − v.
• Separated form: dv/(F(v) − v) = dx/x, then integrate both sides.

Worked Example

Solve dy/dx = (x² + y²)/(2xy).

1. Check homogeneity: scaling x → tx, y → ty gives (t²x² + t²y²)/(2tx·ty) = t²(x² + y²)/(t²·2xy), and the t² cancels, leaving the original. Degree 0 confirmed.
2. Substitute y = vx: the left side becomes v + x·dv/dx. The right side becomes (x² + v²x²)/(2x·vx) = x²(1 + v²)/(2vx²) = (1 + v²)/(2v).
3. Simplify: v + x·dv/dx = (1 + v²)/(2v), so x·dv/dx = (1 + v²)/(2v) − v = (1 + v² − 2v²)/(2v) = (1 − v²)/(2v).
4. Separate and integrate: 2v·dv/(1 − v²) = dx/x. Integrating gives −ln|1 − v²| = ln|x| + C, which combines to x(1 − v²) = C₁.
5. Back-substitute v = y/x: x(1 − y²/x²) = C₁, i.e. (x² − y²)/x = C₁, or x² − y² = Cx.
6. Verify: differentiating x² − y² = Cx gives 2x − 2y·y′ = C = (x² − y²)/x; multiplying by x and rearranging yields 2xy·y′ = x² + y², so y′ = (x² + y²)/(2xy). Correct.

Common Misconceptions

• Writing dy/dx = dv/dx after substituting. Since y = vx is a product, dy/dx = v + x·dv/dx; dropping the v term breaks everything downstream.
• Skipping the homogeneity check. The substitution only helps when the equation is genuinely homogeneous; otherwise it creates a worse mess.
• Leaving the answer in v. The problem asks for y as a function of x, so always replace v with y/x at the end and simplify.

Real-World Applications

• Geometry: families of curves defined by directional properties and orthogonal trajectories often lead to homogeneous equations.
• Fluid flow: certain velocity fields depending only on direction from a source are modelled with homogeneous equations.
• Economics: production functions with constant returns to scale are homogeneous, and related rate equations inherit the property.

Homogeneous equations depend only on the ratio y/x, and the substitution y = vx exposes that structure, reducing the problem to a separable equation. Check homogeneity first, apply the product rule carefully, and never forget to substitute back to y.