What is the difference between a homogeneous ODE (first-order) and a homogeneous linear ODE?

A first-order homogeneous ODE means the equation can be expressed as $dy/dx = f(y/x)$, where $M$ and $N$ are homogeneous functions of the same degree. A homogeneous linear ODE means the equation has the form $a_n y^{(n)} + \cdots + a_0 y = 0$ — the right-hand side is zero, with no forcing function. These are two distinct uses of the word 'homogeneous' in differential equations.

Homogeneous Differential Equations — Definition, Formula & Examples

A homogeneous differential equation is a first-order ODE that can be written in the form

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

, meaning the right-hand side depends only on the ratio

y/x

. The term also applies to linear differential equations where every term involves the unknown function or its derivatives (no standalone constant or forcing term).

A first-order ODE $M(x,y)\,dx + N(x,y)\,dy = 0$ is called homogeneous if $M$ and $N$ are homogeneous functions of the same degree, i.e., $M(tx, ty) = t^n M(x,y)$ and $N(tx, ty) = t^n N(x,y)$ for some integer $n$ . Separately, a linear ODE $a_n(x)y^{(n)} + \cdots + a_1(x)y' + a_0(x)y = g(x)$ is homogeneous when $g(x) = 0$ .

Key Formula

\text{Substitution: } v = \frac{y}{x}, \quad y = vx, \quad \frac{dy}{dx} = v + x\,\frac{dv}{dx}

Where:

$v$ = New dependent variable equal to y/x
$x$ = Independent variable
$y$ = Original dependent variable

How It Works

For a first-order homogeneous ODE, you solve it by substituting

v = y/x

, so

y = vx

and

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

. This substitution transforms the equation into a separable ODE in

v

and

x

, which you can integrate directly. After finding

v(x)

, you substitute back

v = y/x

to get the solution in terms of

y

and

x

. For a homogeneous linear ODE (where

g(x) = 0

), you look for solutions using characteristic equations or other standard techniques depending on the order.

Worked Example

Problem: Solve the differential equation

\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}

Step 1: Verify homogeneity: Replace

x

with

tx

and

y

with

ty

in the numerator and denominator. The numerator becomes

t^2x^2 + t^2y^2 = t^2(x^2+y^2)

and the denominator becomes

2(tx)(ty) = 2t^2xy

. Both are degree 2, so the equation is homogeneous.

Step 2: Substitute v = y/x: Let

y = vx

, so

dy/dx = v + x\,dv/dx

. Rewrite the right side using

y = vx

v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{2x(vx)} = \frac{1 + v^2}{2v}

Step 3: Separate variables: Isolate

x\,dv/dx

on one side.

x\frac{dv}{dx} = \frac{1+v^2}{2v} - v = \frac{1+v^2 - 2v^2}{2v} = \frac{1-v^2}{2v}

Step 4: Integrate both sides: Separate and integrate. The left side uses partial fractions or a direct substitution.

\int \frac{2v}{1-v^2}\,dv = \int \frac{dx}{x} \implies -\ln|1-v^2| = \ln|x| + C_1

Step 5: Back-substitute v = y/x: Replace

v

with

y/x

and simplify. Exponentiate both sides to remove logarithms.

\frac{1}{|1 - y^2/x^2|} = A|x| \implies \frac{x^2}{|x^2 - y^2|} = A|x| \implies x^2 - y^2 = Cx

Answer: The general solution is

x^2 - y^2 = Cx

, where

C

is an arbitrary constant.

Another Example

Problem: Solve

\frac{dy}{dx} = \frac{y}{x} + 1

Step 1: Recognize the form: Rewrite as

dy/dx = f(y/x)

where

f(v) = v + 1

. This depends only on

y/x

plus a constant adjustment, confirming homogeneity after rearranging:

\frac{dy}{dx} = \frac{y + x}{x}

. Both numerator and denominator are degree 1.

Step 2: Substitute v = y/x: With

y = vx

and

dy/dx = v + x\,dv/dx

v + x\frac{dv}{dx} = v + 1 \implies x\frac{dv}{dx} = 1

Step 3: Separate and integrate: This is now straightforward to integrate.

\int dv = \int \frac{dx}{x} \implies v = \ln|x| + C

Step 4: Back-substitute: Replace

v

with

y/x

\frac{y}{x} = \ln|x| + C \implies y = x\ln|x| + Cx

Answer: The general solution is

y = x\ln|x| + Cx

Why It Matters

Homogeneous differential equations appear frequently in a first course in ordinary differential equations (ODE), which is required for most engineering, physics, and applied mathematics programs. The substitution technique taught here builds the foundation for more advanced methods like variation of parameters. Recognizing homogeneity is also essential in modeling problems involving scaling, such as fluid dynamics and heat transfer.

Common Mistakes

Mistake: Confusing the two meanings of 'homogeneous' — applying the

v = y/x

substitution to a homogeneous linear ODE like

y'' + 3y' + 2y = 0

Correction: The

v = y/x

substitution only applies to first-order equations where

M

and

N

are homogeneous functions of the same degree. For linear homogeneous ODEs, use characteristic equations or other appropriate methods.

Mistake: Forgetting to substitute back from

v

y/x

after integrating, leaving the answer in terms of

v

and

x

Correction: Always replace

v

with

y/x

in your final answer so the solution is expressed in the original variables

y

and

x