Sample 2024-25 Class 12 Maths Q5 Solution: n for which xⁿ dy/dx = y(log y − log x + 1) is homogeneous

The question

The value of $'n'$ ⁠, such that the differential equation $x^n\frac{dy}{dx} = y(\log y - \log x + 1)$ ⁠; (where $x, y \in R^+$ ⁠) is homogeneous, is

(A) $0$

(B) $1$

(C) $2$

(D) $3$

Exam answer

\begin{aligned} x^n\frac{dy}{dx} &= y\left(\log\frac{y}{x} + 1\right) \\ \Rightarrow \frac{dy}{dx} &= \frac{y}{x^n}\left(\log\frac{y}{x} + 1\right) \end{aligned}

Why?

The quotient law of logarithms joins the two logs:

\log y - \log x = \log\frac{y}{x}

Then divide both sides by $x^n$ ⁠, which is not 0 because $x > 0$ ⁠.

Let the right side be $f(x, y)$ ⁠. The equation is homogeneous if $f(x, y)$ is a homogeneous function of degree 0.

\begin{aligned} &f(\lambda x, \lambda y) \\ &= \frac{\lambda y}{\lambda^n x^n}\left(\log\frac{\lambda y}{\lambda x} + 1\right) \\ &= \lambda^{1 - n} \cdot \frac{y}{x^n}\left(\log\frac{y}{x} + 1\right) \end{aligned}

Why?

A function is homogeneous of degree $k$ if replacing $x, y$ by $\lambda x, \lambda y$ brings out exactly $\lambda^k$ ⁠:

f(\lambda x, \lambda y) = \lambda^k f(x, y)

Inside the log, $\frac{\lambda y}{\lambda x} = \frac{y}{x}$ ⁠, so the bracket does not change. In front, $\frac{\lambda}{\lambda^n} = \lambda^{1 - n}$ ⁠.

\begin{aligned} \frac{\lambda y}{\lambda x} &= \frac{y}{x} \quad (\text{cancel } \lambda) \\[4pt] \frac{\lambda y}{\lambda^n x^n} &= \frac{\lambda}{\lambda^n} \cdot \frac{y}{x^n} \\ &= \lambda^{1 - n} \cdot \frac{y}{x^n} \end{aligned}

(Law of indices: $\frac{\lambda^1}{\lambda^n} = \lambda^{1 - n}$ ⁠.)

That is,

f(\lambda x, \lambda y) = \lambda^{1 - n}f(x, y)

So the degree is 0 only if $1 - n = 0$ ⁠, that is $n = 1$ ⁠.

Why?

Check: with $n = 1$ the equation is

\frac{dy}{dx} = \frac{y}{x}\left(\log\frac{y}{x} + 1\right)

The right side depends on $\frac{y}{x}$ alone, as in every homogeneous equation.

The other options give degree $1 - n = 1$ ⁠, $-1$ and $-2$ ⁠, not 0.

\begin{aligned} n = 0&: 1 - 0 = 1 \\ n = 2&: 1 - 2 = -1 \\ n = 3&: 1 - 3 = -2 \end{aligned}

Answer: (B) $1$

Where this comes from

Chapter 9, Differential Equations: Homogeneous differential equations.

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: Write the right side as $\frac{y}{x^n}\left(\log\frac{y}{x} + 1\right)$ ⁠. Putting in $\lambda x, \lambda y$ gives $\lambda^{1 - n}$ times it, and degree 0 needs $n = 1$ ⁠.

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Q5: nnn for which xndydx=y(log⁡y−log⁡x+1)x^n\frac{dy}{dx} = y(\log y - \log x + 1)xndxdy​=y(logy−logx+1) is homogeneous

Q5: $n$ for which $x^n\frac{dy}{dx} = y(\log y - \log x + 1)$ is homogeneous