12 min readPathway sheetUsually a 3-mark answer, often with an initial condition; sometimes a 1-mark MCQ on the substitution

Homogeneous differential equations

dy/dx = F(x, y) is homogeneous when F is a homogeneous function of degree zero. Put y = vx to turn it into a separable equation.

Homogeneous function of degree nF(λx, λy) = λⁿF(x, y) for every non-zero λ

Homogeneous DEdy/dx = F(x, y), where F is homogeneous of degree 0, so F depends only on y/x

Substitutiony = vx, so dy/dx = v + x(dv/dx)

Other formdx/dy = G(x, y) with G of degree 0: put x = vy, dx/dy = v + y(dv/dy)

Homogeneous functions

A function F(x, y) is homogeneous of degree n if replacing x by λx and y by λy multiplies it by λⁿ. For example, F = x² + 3xy gives (λx)² + 3(λx)(λy) = λ²(x² + 3xy), so its degree is 2. A DE dy/dx = F(x, y) is a homogeneous DE when F has degree zero. Then F(x, y) can be written as a function of y/x alone, and the substitution y = vx works.

Test these functions

Function	F(λx, λy)	Homogeneous?
x² + xy	λ²(x² + xy)	Yes, degree 2
(x² + y²)/(xy)	λ⁰ × (x² + y²)/(xy)	Yes, degree 0
x + y + 1	λx + λy + 1	No, the constant 1 does not scale
sin(y/x)	sin(y/x)	Yes, degree 0
log y − log x	log(λy) − log(λx) = log(y/x)	Yes, degree 0
x² + y	λ²x² + λy	No, the terms scale differently

The method (y = vx)

1
Write the DE as dy/dx = F(x, y). Show F(λx, λy) = λ⁰F(x, y), so it is homogeneous.
2
Put y = vx. By the product rule, dy/dx = v + x(dv/dx).
3
Substitute: v + x(dv/dx) = F(x, vx). The x cancels from the right side, leaving a function of v.
4
Separate: move v to the left, so dv/[F(1, v) − v] = dx/x.
5
Integrate both sides and add one constant.
6
Put v = y/x back to get the general solution in x and y.

Example 1: direct

Problem

Solve x(dy/dx) = y + xe^(y/x), x ≠ 0.

Working

Divide by x: dy/dx = y/x + e^(y/x). The right side is a function of y/x only, so the DE is homogeneous.
Put y = vx, dy/dx = v + x(dv/dx).
v + x(dv/dx) = v + eᵛ, so x(dv/dx) = eᵛ.
Separate: e⁻ᵛ dv = dx/x.
Integrate: −e⁻ᵛ = log|x| + C.
Put v = y/x and write K = −C: log|x| + e^(−y/x) = K.

Answer

log|x| + e^(−y/x) = K, where K is an arbitrary constant.

Example 2: board-style, with a condition

Problem

Find the particular solution of 2xy(dy/dx) = x² + y², given that y = 2 when x = 1.

Working

dy/dx = (x² + y²)/(2xy). Top and bottom both have degree 2, so F has degree 0: the DE is homogeneous.
Put y = vx: v + x(dv/dx) = (x² + v²x²)/(2vx²) = (1 + v²)/(2v).
x(dv/dx) = (1 + v²)/(2v) − v = (1 − v²)/(2v).
Separate: 2v dv/(1 − v²) = dx/x.
Integrate (put u = 1 − v², du = −2v dv): −log|1 − v²| = log|x| + log|C|.
So x(1 − v²) = K, where K is a new constant.
Put v = y/x: x(1 − y²/x²) = K, that is x² − y² = Kx.
Use x = 1, y = 2: 1 − 4 = K, so K = −3. Then x² − y² = −3x.

Answer

y² = x² + 3x

Quick tips

Terms like y/x inside sin, log or e^( ) are a strong hint that the DE is homogeneous.
If the DE looks neater as dx/dy with x/y inside, put x = vy and dx/dy = v + y(dv/dy).
Always put v = y/x back at the end. An answer left in v loses a mark.
A constant term, as in x + y + 1, makes the DE non-homogeneous.

In real life

A boatman crossing the Ganga who always steers towards a fixed ghat is pushed sideways by the current. The curved path his boat follows comes from a homogeneous DE.

Exam angle

1 marks: “Find the degree of the homogeneous function F(x, y) = (x³ + y³)/(x + y).” F(λx, λy) = λ³(x³ + y³)/(λ(x + y)) = λ²F(x, y). Degree = 2.

3 marks: “Solve x²(dy/dx) = x² + xy + y².” dy/dx = 1 + y/x + (y/x)², homogeneous. y = vx: v + x(dv/dx) = 1 + v + v², so dv/(1 + v²) = dx/x. tan⁻¹ v = log|x| + C, so tan⁻¹(y/x) = log|x| + C.

Remember it

Same degree on top and bottom? Put y = vx. Then dy/dx = v + x(dv/dx).