Warning: "homogeneous" has two different meanings in differential equations. Here we mean a specific structural property of first-order ODEs, not the zero right-hand side of second-order equations.
A first-order ODE is homogeneous (in the sense of this page) if it can be written in the form:
Homogeneous first-order ODE — standard form
dy/dx = F(y/x)
That is, when you simplify the right-hand side completely, the only way x and y appear is as the ratio y/x. This is the defining property. Another equivalent way to state it: the ODE can be written as dy/dx = f(x, y)/g(x, y) where both f and g are homogeneous functions of the same degree.
Homogeneous Functions and Degree
A function h(x, y) is homogeneous of degree n if scaling both variables by any constant λ scales the function by λⁿ:
Definition of homogeneous function
h(λx, λy) = λⁿ h(x, y) for all λ ≠ 0
Examples:
h = x² + xy + y²: degree 2, since h(λx,λy) = λ²x² + λ²xy + λ²y² = λ²h ✓
h = x³ − y³, degree 3, since h(λx,λy) = λ³(x³−y³) = λ³h ✓
h = √(x² + y²), degree 1, since h(λx,λy) = λ√(x²+y²) = λh ✓
h = x² + y + 1, not homogeneous (terms have different degrees)
The Degree Test — Is the ODE Homogeneous?
Given dy/dx = f(x,y)/g(x,y), replace x → λx and y → λy in both numerator and denominator. If the λs cancel completely, the ODE is homogeneous.
ODE
f(x,y) degree
g(x,y) degree
Verdict
dy/dx = (x² + y²)/(xy)
2
2
✓ Homogeneous
dy/dx = (x + y)/(x − y)
1
1
✓ Homogeneous
dy/dx = y³/(x²y + x³)
3
3
✓ Homogeneous
dy/dx = (x² + 1)/y
2 (mixed)
1
✗ Not homogeneous
dy/dx = (xy + y²)/x
2
1
✗ Not homogeneous
dy/dx = (x³ + y³)/(x²y)
3
3
✓ Homogeneous
Quick shortcut: write everything in terms of y/x
Divide numerator and denominator by an appropriate power of x. If the result depends only on v = y/x (no x or y left separately), the equation is homogeneous. For example: (x + y)/(x − y) → divide top and bottom by x → (1 + y/x)/(1 − y/x) = (1 + v)/(1 − v). Purely a function of v — homogeneous confirmed.
⚠ The naming collision
In second-order ODEs, "homogeneous" means the right-hand side is zero: ay″ + by′ + cy = 0. That is a completely different use of the word. When someone says "homogeneous ODE" at first-order level, they mean the dy/dx = F(y/x) structure described on this page. Always check context.
§ 02The Substitution v = y/x — Full Derivation
The substitution v = y/x is not magic, it falls out directly from the structure of the equation. Here is exactly why it works.
If the ODE is dy/dx = F(y/x), introduce the new variable v = y/x, so y = vx. Differentiate with respect to x using the product rule:
Differentiating y = vx
dy/dx = v + x · dv/dx
Now substitute into dy/dx = F(y/x) = F(v):
DerivationReducing to a separable equation
v + x · dv/dx = F(v)
substituting y = vx into the ODE
x · dv/dx = F(v) − v
isolate the dv/dx term
dv / [F(v) − v] = dx / x
separate variables — v terms on left, x on right
∫ dv / [F(v) − v] = ∫ dx/x = ln|x| + C
integrate both sides
Back-substitute: v = y/x
replace v with y/x to express solution in original variables
The key insight is that after substituting y = vx and dy/dx = v + x(dv/dx), the variable x always factors out completely from F(v) − v, leaving a separable equation in v and x. The integral on the left, ∫dv/[F(v)−v], may be simple or may require partial fractions, but it only involves v, not x.
Why x only appears as ln|x|
The right-hand side always integrates to ln|x| + C. This is because the only x in the equation after separation is dx/x. The left-hand side integral varies from problem to problem, but the right side is always the same.
§ 03The 5-Step Method
Every homogeneous first-order ODE is solved in five steps. The method never changes: only the integrals differ.
Solving Homogeneous ODEs — 5 Steps
Verify homogeneity.
Confirm the equation has the form dy/dx = F(y/x) by dividing through by x^n or by checking that f and g have the same degree.
Substitute v = y/x, so y = vx.dy/dx = v + x · dv/dx
Replace every occurrence of y with vx and every dy/dx with v + x(dv/dx).
Separate variables in v and x.dv / [F(v) − v] = dx / x
Rearrange so all v terms (including dv) are on the left and all x terms (including dx) are on the right.
Integrate both sides.∫ dv / [F(v) − v] = ln|x| + C
The right side is always ln|x| + C. Evaluate the left integral: this is where most of the work happens.
Back-substitute v = y/x.
Replace v with y/x in the result to express the general solution in terms of x and y. If an initial condition is given, find C now.
§ 04Worked Examples
Ten fully solved problems: starting with clean rational cases and building to equations requiring partial fractions, IVPs, and implicit solutions.
Example 1The simplest case — linear ratioPure
Solve: dy/dx = (x + y) / x
Step 1 — Verify homogeneity
(x + y)/x = 1 + y/x = 1 + v → F(v) = 1 + v ✓
Step 2 — Substitute y = vx
v + x·dv/dx = 1 + v
Step 3 — Separate variables
x·dv/dx = 1 → dv = dx/x
Step 4 — Integrate
v = ln|x| + C
Step 5 — Back-substitute v = y/x
y/x = ln|x| + C → y = x(ln|x| + C)
y = x(ln|x| + C)
Example 2Degree 1 — fraction of linear termsPure
Solve: dy/dx = (x + y) / (x − y)
Express as F(v): divide top and bottom by x
F(v) = (1 + v)/(1 − v)
Substitute and separate
v + x·dv/dx = (1+v)/(1−v)x·dv/dx = (1+v)/(1−v) − v = (1+v − v(1−v))/(1−v) = (1+v²)/(1−v)
Separate variables
(1−v)/(1+v²) dv = dx/x
Integrate left side: split into two pieces
∫1/(1+v²) dv − ∫v/(1+v²) dv = arctan(v) − ½ln(1+v²)
Integrate right side and combine
arctan(v) − ½ln(1+v²) = ln|x| + C
Back-substitute v = y/x
arctan(y/x) − ½ln(1+y²/x²) = ln|x| + C
arctan(y/x) − ½ ln(x² + y²) = C
Note: ½ln(1+y²/x²) + ln|x| = ½ln((x²+y²)/x²) + ln|x| = ½ln(x²+y²) − ln|x| + ln|x| = ½ln(x²+y²). Absorbed into C.
Example 3Degree 2 — quadratic termsPure
Solve: dy/dx = (x² + y²) / (2xy)
Divide top and bottom by x²
F(v) = (1 + v²)/(2v)
Substitute
v + x·dv/dx = (1+v²)/(2v)x·dv/dx = (1+v²)/(2v) − v = (1+v²−2v²)/(2v) = (1−v²)/(2v)
Separate
2v/(1−v²) dv = dx/x
Integrate: left side — note d/dv[1−v²] = −2v, so ∫2v/(1−v²)dv = −ln|1−v²|
This illustrates that not every homogeneous ODE yields a neat closed-form solution. The method is always valid; the resulting integral may require numerical methods.
Example 10Reverse — constructing a homogeneous ODE from its solutionPure
The general solution of a homogeneous ODE is y² = x²(ln|x| + C). Recover the ODE.
Differentiate implicitly with respect to x
2y · dy/dx = 2x(ln|x| + C) + x²·(1/x) = 2x(ln|x| + C) + x
Replace (ln|x| + C) using the solution: ln|x| + C = y²/x²
2y · dy/dx = 2x · y²/x² + x = 2y²/x + x
Divide by 2y and simplify
dy/dx = y/x + x/(2y)
Verify: divide top and bottom by x to write as F(y/x)
= y/x + (1/2)/(y/x) = v + 1/(2v) ✓ homogeneous
dy/dx = y/x + x/(2y)
§ 05Common Mistakes
⚠ Confusing the two meanings of "homogeneous"
A first-order ODE of the form dy/dx = F(y/x) is called homogeneous. A second-order ODE with zero right-hand side (ay″ + by′ + cy = 0) is also called homogeneous — but these are entirely different concepts. Make sure you know which type you are dealing with before choosing a method.
⚠ Forgetting dy/dx = v + x·dv/dx
The most common algebraic slip: after writing y = vx, students differentiate to get dy/dx = v·(dv/dx) instead of dy/dx = v + x·(dv/dx). The product rule gives two terms, not one. Always write out y = vx → dy/dx = v + x·dv/dx before substituting.
⚠ Not simplifying F(v) − v before separating
The expression x·dv/dx = F(v) − v must be fully simplified algebraically before you separate variables. Skipping this step and trying to integrate a complex fraction that could be simplified leads to unnecessarily hard integrals.
⚠ Forgetting to back-substitute at the end
After solving the separable equation in v and x, the answer is still in terms of v = y/x. You must replace v with y/x to express the solution in the original variables. An answer left in terms of v is incomplete.
⚠ Applying this method to non-homogeneous equations
If the degrees of f(x,y) and g(x,y) differ, the method fails. For example dy/dx = (x² + y)/x has f of degree 2 and the y term of degree 1 — it is not homogeneous, and substituting v = y/x will not separate the variables.
§ 06Practice Quiz — 10 Questions
Identify, set up, and solve. Some questions test classification; others require carrying out steps of the method.
Score: 0 / 0
Question 1 — Classification
Which of the following ODEs is homogeneous (in the first-order sense)?
(a) dy/dx = x + y: substitute y = vx, so x + y = x + vx = x(1 + v). This still depends on x itself, not just on v = y/x, so it cannot be written as a pure function of v. Not homogeneous.
(b) (x² + y²)/(xy): divide by x²/x → (1 + v²)/v, purely a function of v = y/x. Homogeneous ✓
(c) dy/dx = y + sin x, sin x depends only on x, not the ratio y/x. Not homogeneous.
(d) x²y + 1, the +1 breaks homogeneity. Not homogeneous.
Question 2 — Substitution
After substituting y = vx into dy/dx = F(y/x), dy/dx becomes:
Question 3 — Degree check
The function f(x, y) = x⁴ − 3x²y² + y⁴ is homogeneous of degree n. What is n?
Replace x → λx, y → λy and factor out λ.
Question 4 — Setting up the separable equation
For dy/dx = (x + y)/x, after substituting v = y/x, x·dv/dx equals:
F(v) = (x + y)/x = 1 + y/x = 1 + v
v + x·dv/dx = 1 + v
x·dv/dx = 1
Question 5 — Right-hand side integral
After separating variables in a homogeneous ODE, the right-hand side is always ∫dx/x. What is this integral?
Question 6 — Applying the method
For the ODE dy/dx = (x² + y²)/(2xy), after substituting v = y/x, x·dv/dx = (1 − v²)/(2v). The separated equation is (2v)/(1 − v²) dv = dx/x. Integrating gives −ln|1 − v²| = ln|x| + C. So 1 − v² = A/x. Using the IVP y(2) = 2, what is A?
At x = 2, y = 2 → v = y/x = 1. Substitute v = 1, x = 2 into 1 − v² = A/x.
v = y/x = 2/2 = 1
1 − 1² = A/2 → 0 = A/2 → A = 0
Solution: x² − y² = 0 → y = x (the particular solution is the line y = x)
Question 7 — Back-substitution
After solving a homogeneous ODE you obtain v + ln|v| = ln|x| + C. Replacing v = y/x gives:
Question 8 — IVP constant
The ODE dy/dx = y/x + tan(y/x) has solution sin(y/x) = Cx. If y(π) = π/2, what is C? Give your answer to 3 decimal places.
At x = π, y = π/2 → v = y/x = 1/2. sin(1/2) = Cπ → C = sin(0.5)/π.