Board 2025 Class 12 Maths Q4 Solution: Which function of x and y is not homogeneous?

The question

Which of the following is not a homogeneous function of $x$ and $y$ ?

(A) $y^2 - xy$

(B) $x - 3y$

(C) $\sin^2\frac{y}{x} + \frac{y}{x}$

(D) $\tan x - \sec y$

Exam answer

$F(x, y)$ is homogeneous of degree n if $F(\lambda x, \lambda y) = \lambda^n F(x, y)$ ⁠.

Why?

The test: replace every x by $\lambda x$ and every y by $\lambda y$ ⁠, then simplify. If a power $\lambda^n$ comes out as a factor and leaves the original function behind, the function is homogeneous of degree n.

(A) $F(x, y) = y^2 - xy$ ⁠:

\begin{aligned} F(\lambda x, \lambda y) &= \lambda^2y^2 - \lambda^2xy \\ &= \lambda^2(y^2 - xy) \end{aligned}

Why?

Put $\lambda x$ for x and $\lambda y$ for y, then multiply out each term:

\begin{aligned} F(\lambda x, \lambda y) &= (\lambda y)^2 - (\lambda x)(\lambda y) \\ (\lambda y)^2 &= \lambda^2y^2 \\ (\lambda x)(\lambda y) &= \lambda^2xy \end{aligned}

Both terms have $\lambda^2$ ⁠, so take it out as a common factor. The same move in (B): $\lambda x - 3(\lambda y)$ ⁠, and both terms have $\lambda$ ⁠.

(B) $F(x, y) = x - 3y$ ⁠:

\begin{aligned} F(\lambda x, \lambda y) &= \lambda x - 3\lambda y \\ &= \lambda(x - 3y) \end{aligned}

(C) $F(x, y) = \sin^2\frac{y}{x} + \frac{y}{x}$ ⁠:

\begin{aligned} F(\lambda x, \lambda y) &= \sin^2\frac{\lambda y}{\lambda x} + \frac{\lambda y}{\lambda x} \\ &= \sin^2\frac{y}{x} + \frac{y}{x} \\ &= \lambda^0 F(x, y) \end{aligned}

Why?

In $\frac{\lambda y}{\lambda x}$ the $\lambda$ cancels. So any function of $\frac{y}{x}$ alone does not change at all: it is homogeneous of degree 0, because $\lambda^0 = 1$ ⁠.

(A), (B) and (C) are homogeneous of degree 2, 1 and 0.

(D) $F(x, y) = \tan x - \sec y$ ⁠:

\begin{aligned} &\tan \lambda x - \sec \lambda y \\ &\ne \lambda^n(\tan x - \sec y) \end{aligned}

Why?

Here $\lambda$ is trapped inside tan and sec, and no power of $\lambda$ can be taken out. A concrete check, with $y = 0$ ⁠:

\begin{aligned} F(\tfrac{\pi}{4}, 0) &= \tan\tfrac{\pi}{4} - \sec 0 \\ &= 1 - 1 = 0 \\[4pt] F(\tfrac{\pi}{8}, 0) &= \tan\tfrac{\pi}{8} - \sec 0 \\ &= (\sqrt{2} - 1) - 1 \\ &= \sqrt{2} - 2 \ne 0 \end{aligned}

(We used $\tan\frac{\pi}{4} = 1$ ⁠, $\sec 0 = \frac{1}{\cos 0} = 1$ and $\tan\frac{\pi}{8} = \sqrt{2} - 1$ ⁠.) With $\lambda = 2$ ⁠, the point $(\frac{\pi}{8}, 0)$ goes to $(\frac{\pi}{4}, 0)$ ⁠, so homogeneity would need $0 = 2^n(\sqrt{2} - 2)$ ⁠, which is impossible.

So $\tan x - \sec y$ is not homogeneous.

Answer: (D) $\tan x - \sec y$

Where this comes from

Chapter 9, Differential Equations: Homogeneous functions.

Question type: Homogeneous equations

How to spot it: click to see

Type 6. Homogeneous equations: put $y = vx$

Spot it: every term has the same degree, like $x^2$ ⁠, $xy$ and $y^2$ ⁠, or you see $\frac{y}{x}$ inside sin, cos or log. You cannot separate $x$ and $y$ ⁠.

Here: Replace $x, y$ by $\lambda x, \lambda y$ in each option. (A), (B) and (C) come out as $\lambda^2$ ⁠, $\lambda$ and $\lambda^0$ times themselves, but $\tan\lambda x - \sec\lambda y$ is not a power of $\lambda$ times $\tan x - \sec y$ ⁠.

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Q4: Which function of xxx and yyy is not homogeneous?

Q4: Which function of $x$ and $y$ is not homogeneous?