Q.The value of ''n , such that the differential equation 𝒙𝒏 𝒅𝒚 𝒅𝒙 = 𝒚(𝒍𝒐𝒈𝒚 − 𝒍𝒐𝒈𝒙 + 𝟏); (𝐰𝐡𝐞𝐫𝐞 𝒙, 𝒚 ∈ 𝑹+) is homogeneous, is
(A) 0
(B) 1
(C) 2
(D) 3

Concept: Homogeneous Differential Equation — an equation of the form $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$.

Step 1: Rewrite the given equation:

Step 2: For homogeneity, the right-hand side must be expressible as a function of $\frac{y}{x}$ alone. Divide both sides by $x^n$:

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

We need to find $n$ so that

is homogeneous for $x, y \in \mathbb{R}^+$.

Why homogeneity matters: A first-order differential equation is homogeneous if it can be expressed as $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$. This means the right-hand side depends only on the ratio $y/x$, not on $x$ and $y$ separately. The test is: replace $x$ with $tx$ and $y$ with $ty$; if the equation remains unchanged in form (the $t$ cancels out), it's homogeneous.

Let's apply this step by step.

1. Rewrite the equation in standard form Divide both sides by $x^n$ (valid since $x>0$):

2. Simplify the logarithmic term Using $\log y - \log x = \log\left(\frac{y}{x}\right)$, we get:

3. Check homogeneity condition Replace $x$ by $tx$ and $y$ by $ty$ (with $t>0$). Then $\frac{y}{x}$ becomes $\frac{ty}{tx} = \frac{y}{x}$, so the logarithmic part $\log\frac{y}{x} + 1$ is unchanged. The numerator becomes $(ty)\left(\log\frac{y}{x} + 1\right) = t \cdot y\left(\log\frac{y}{x} + 1\right)$. The denominator becomes $(tx)^n = t^n x^n$. So the transformed right-hand side is:

4. For homogeneity, the $t$ factor must vanish The original equation had no $t$ factor. For the transformed equation to be identical in form to the original, we need $t^{1-n} = 1$ for all $t>0$. This forces the exponent to be zero: $1-n = 0$, so $n=1$. …

Method: Finding the parameter that makes a DE homogeneous

Use this when a differential equation carries an unknown power (or constant) and you must choose it so the equation becomes homogeneous — i.e. so $\dfrac{dy}{dx}$ can be written as a function of $\dfrac{y}{x}$ alone.

Steps

Step 1: Solve for the derivative

Isolate $\dfrac{dy}{dx}$ so the equation reads

Step 2: Force every group into the ratio $y/x$ …

Common Mistakes

Mistake 1: Thinking any equation with $\log y-\log x$ is automatically homogeneous

Why it's wrong: homogeneity requires the entire right side to reduce to a function of $\frac{y}{x}$; a leftover free power of $x$ (from $x^n$) breaks it. Correct approach: demand the exponent of the bare $x$ be zero. …