Recognising homogeneous equations
A recipe that is doubled in every ingredient tastes the same, because the ratio of the ingredients stays the same. A homogeneous equation depends only on the ratio y/x.
A function f(x, y) is homogeneous of degree n if f(λx, λy) = λⁿ f(x, y). The equation dy/dx = f(x, y) / g(x, y) is homogeneous if f and g are homogeneous of the same degree, so dy/dx can be written as a function of y/x. Then put y = vx.
Formula: y = vx, dy/dx = v + x dv/dx
Rule 1: Test with λ — Replace x by λx and y by λy.
- Easy: (x + y)/x is homogeneous, because both numerator and denominator have degree 1.
- Medium: (x² + y²)/(xy) is homogeneous, with degree 2 in both parts.
- Exam-type: (x² + y)/x is not homogeneous, because x² and y have different degrees.
Rule 2: Degree and constants — Count the total degree of each term.
- Easy: For dy/dx = (x + y)/x, v + x dv/dx = 1 + v, so dv = dx/x.
- Medium: x² + 3xy is homogeneous of degree 2.
- Exam-type: (y + 3)/(x + 1) is not homogeneous, because of the constants 3 and 1.
Every term must have the same total degree. A constant added to x or y destroys homogeneity.
Writing dy/dx = v when y = vx. The derivative needs the product rule: v + x dv/dx.
Same degree on top and bottom, then y equals v x.
Practice questions
- 1Is dy/dx = (x + y)/x homogeneous?
- 2Is dy/dx = (x² + y)/x homogeneous?
- 3Is dy/dx = (x² + y²)/(xy) homogeneous?
- 4Is dy/dx = (x³ + y³)/(x y²) homogeneous?
- 5What is the degree of homogeneity of x² + 3xy?
- 6If y = vx, what is dy/dx?
- 7For dy/dx = (x + y)/x, which separable equation is obtained by y = vx?
- 8Is dy/dx = (y + 3)/(x + 1) homogeneous?
Show answers (try them yourself first)
- 1Yes, both parts have degree 1
- 2No, x² has degree 2 and y has degree 1
- 3Yes, degree 2 in both parts
- 4Yes, degree 3 in both parts
- 52
- 6dy/dx = v + x dv/dx
- 7v + x dv/dx = 1 + v gives x dv/dx = 1, so dv = dx/x
- 8No, because of the constants 3 and 1
Solving homogeneous equations
To solve a puzzle in terms of a ratio, we count in units of x. Writing y as v times x gives an equation in v and x that can be separated.
Put y = vx and dy/dx = v + x dv/dx. Simplify so that v-terms and x-terms separate, integrate, and finally replace v by y/x.
Formula: v + x dv/dx = F(v), so dv / (F(v) − v) = dx / x
Rule 1: Substitute and separate — Collect all v terms on one side.
- Easy: dy/dx = (x + y)/x gives v = ln|x| + C, so y = x (ln|x| + C).
- Medium: dy/dx = (x² + y²)/(xy) gives v dv = dx/x, so y² = 2x² (ln|x| + C).
- Exam-type: dy/dx = (x − y)/(x + y) gives x² − 2xy − y² = K.
Rule 2: Special forms — Look for a trigonometric ratio of y/x.
- Easy: x dy/dx = y + x tan(y/x) gives cot v dv = dx/x, so sin(y/x) = Cx.
- Medium: dy/dx = y/x + 2 gives v = 2 ln|x| + C, so y = x (2 ln|x| + C).
- Exam-type: dy/dx = (x² + y²)/(2xy) gives x² − y² = Kx.
Check your answer by differentiating it. It must return the original dy/dx.
Forgetting to replace v by y/x in the final answer.
Let y be v x, separate, integrate, return to y/x.
Practice questions
- 1Solve dy/dx = (x + y)/x.
- 2Solve dy/dx = (x² + y²)/(xy).
- 3Solve dy/dx = (x − y)/(x + y).
- 4Solve x dy/dx = y + x tan(y/x).
- 5Solve dy/dx = y/x + 2.
- 6Solve dy/dx = (x + y)/x with y(1) = 0.
- 7Solve dy/dx = y/x + (y/x)².
- 8Solve dy/dx = (x² + y²)/(2xy).
Show answers (try them yourself first)
- 1y = x (ln|x| + C)
- 2y² = 2x² (ln|x| + C)
- 3x² − 2xy − y² = K
- 4sin(y/x) = Cx
- 5y = x (2 ln|x| + C)
- 6y = x (ln|x| + C) and y(1) = 0 give C = 0, so y = x ln x
- 7v + x dv/dx = v + v², so dv/v² = dx/x and y = −x/(ln|x| + C)
- 8x² − y² = Kx