Homogeneous Differential Equations

A homogeneous differential equation is one that can be written in the form dy/dx = f(y/x), where the right-hand side depends only on the ratio y/x rather than on x and y separately. This special structure means the equation has a hidden symmetry: if you scale both x and y by the same constant, the equation is unchanged. Recognising this structure is the key to unlocking a powerful substitution technique. Not every equation wearing a 'homogeneous' label looks immediately like f(y/x). Many equations appear in the form dy/dx = (ax + by + c)/(px + qy + r), where the presence of constants c and r means the equation is not quite homogeneous — but it is reducible to homogeneous form. By shifting the origin with a suitable substitution, we remove those constants and reveal the underlying homogeneous equation, which we can then solve with the standard v = y/x substitution. Mastering these two layers — the substitution v = y/x for genuinely homogeneous equations, and the origin-shifting trick for the reducible case — gives you a systematic toolkit for an important family of first-order differential equations. These ideas connect to deeper themes in mathematics such as similarity solutions, dimensional analysis, and the geometry of vector fields.

Key takeaways
  • Identify the Type
  • Find the Intersection Point (Reducible Case)
  • Shift the Origin
  • Substitute v = Y/X
  • Separate and Integrate
  • Back-Substitute

Identify the Type

Write the equation as dy/dx = (ax + by + c)/(px + qy + r). If c = 0 and r = 0, the equation is already homogeneous. If not, check whether a/p ≠ b/q (the lines ax + by + c = 0 and px + qy + r = 0 intersect), which means it is reducible to homogeneous form.

Find the Intersection Point (Reducible Case)

Solve the simultaneous equations ax + by + c = 0 and px + qy + r = 0 to find the point (h, k). This is the new origin. If a/p = b/q (parallel lines), the method is different — use the substitution u = ax + by instead.

Shift the Origin

Substitute x = X + h and y = Y + k. Since dx = dX and dy = dY, the equation becomes dY/dX = (aX + bY)/(pX + qY), which is now homogeneous because the constants have cancelled.

Substitute v = Y/X

Let Y = vX, so dY/dX = v + X(dv/dX). The equation becomes v + X(dv/dX) = (a + bv)/(p + qv). Rearrange to separate variables: X(dv/dX) = (a + bv)/(p + qv) − v = (a + bv − pv − qv²)/(p + qv).

Separate and Integrate

Rearrange into the form [(p + qv)/(a + (b−p)v − qv²)] dv = dX/X. Integrate both sides. The left side often requires partial fractions. Add an arbitrary constant C after integrating.

Back-Substitute

Replace v with Y/X, then replace X with x − h and Y with y − k to express the general solution in terms of the original variables x and y. Apply initial conditions if given to find C.

The Key Equation

dydx=ax+by+cpx+qy+r→x=X+hy=Y+kdYdX=aX+bYpX+qY→Y=vXv+XdvdX=a+bvp+qv\frac{dy}{dx} = \frac{ax + by + c}{px + qy + r} \xrightarrow{\substack{x = X+h \\ y = Y+k}} \frac{dY}{dX} = \frac{aX + bY}{pX + qY} \xrightarrow{Y = vX} v + X\frac{dv}{dX} = \frac{a + bv}{p + qv}
  • a, b, p, q — Coefficients (dimensionless): Constants from the original differential equation determining the slopes of the two linear expressions.
  • c, r — Constant terms (dimensionless): The constants that prevent the original equation from being homogeneous; they vanish after the origin shift.
  • (h, k) — Intersection point (dimensionless): The solution of ax + by + c = 0 and px + qy + r = 0; used as the new origin.
  • v — Substitution variable (dimensionless): Defined as v = Y/X; transforms the homogeneous equation into a separable one.

Worked Example

Problem. Solve the differential equation dy/dx = (x + y − 1)/(x − y − 2).

arctan⁡ ⁣(y+12x−32)=12ln⁡ ⁣[(x−32)2+(y+12)2]+C\arctan\!\left(\frac{y + \tfrac{1}{2}}{x - \tfrac{3}{2}}\right) = \frac{1}{2}\ln\!\left[\left(x - \tfrac{3}{2}\right)^2 + \left(y + \tfrac{1}{2}\right)^2\right] + C

Answer. arctan((y + 1/2)/(x − 3/2)) = (1/2) ln((x − 3/2)² + (y + 1/2)²) + C

Figure
Diagram
Origin Shift: Reducing to Homogeneous FormxyL₁: ax+by+c=0L₂: px+qy+r=0(h, k)hkX = x−hY = y−kNew origin (h,k) shifts axes so constant terms vanish
Origins · 1 minAlexis Claude Clairaut and how this was discovered
AC
Alexis Claude Clairaut
1713–1765 · French

Clairaut made fundamental contributions to the theory of differential equations in the 18th century, including the study of singular solutions and special families of first-order ODEs. His work on the geometric interpretation of differential equations — viewing them as conditions on families of curves — helped establish the framework within which homogeneous equations are understood as equations whose solution curves have a self-similar, scaling structure. Clairaut was admitted to the French Academy of Sciences at just 18 years old, making him one of the youngest members ever accepted — he had already written a significant paper on curves at the age of 12.

  • Separable Differential Equations — After the substitution v = Y/X, a homogeneous equation always becomes separable, so mastering separation of variables is essential to completing the solution.
  • First-Order Linear Differential Equations — First-order linear equations form another major class of solvable ODEs alongside homogeneous equations; knowing both extends your problem-solving repertoire significantly.
  • Simultaneous Linear Equations — Finding the intersection point (h, k) requires solving a 2×2 linear system, a skill that is central to the reducible-to-homogeneous method.
  • Integration by Partial Fractions — The integral that appears after separating variables in a homogeneous equation frequently requires partial fraction decomposition to evaluate.
  • Exact Differential Equations — Some homogeneous equations can also be solved as exact equations; understanding exactness gives an alternative perspective on the same solution curves.