Q.The general solution of the differential equation $\frac{y\, dx - x\, dy}{y} = 0$ is (A) $xy = C$ (B) $x = Cy^2$ (C) $y = Cx$ (D) $y = Cx^2$

The key idea is to rewrite the given equation in a standard linear form and then apply the Integrating Factor Method.

First, simplify the expression:

This gives:

This is a first-order linear ODE in $x(y)$. The integrating factor is:

Multiply through: …

The key here is to see what the equation is really saying. You have $\frac{y\, dx - x\, dy}{y} = 0$. Before diving into any method, simplify: dividing term-by-term gives $dx - \frac{x}{y}\, dy = 0$. That’s a differential equation where $x$ is a function of $y$ (or vice versa). It’s not in the standard $dy/dx$ form, but that’s fine — we can treat $y$ as the independent variable.

1. Rewrite in standard linear form From $dx - \frac{x}{y}\, dy = 0$, bring the $dy$ term to the other side:

This is a first-order linear ODE in $x(y)$: $\frac{dx}{dy} + P(y)x = Q(y)$ with $P(y) = -\frac{1}{y}$ and $Q(y) = 0$.

2. Why the Integrating Factor works The idea: if we multiply the whole equation by some function $\mu(y)$, the left side becomes the derivative of $\mu(y) \cdot x$ with respect to $y$. That turns the problem into a simple integration. The formula for the integrating factor is $\mu(y) = e^{\int P(y)\, dy}$. Here $P(y) = -1/y$, so

Hence

Since we usually work with a positive integrating factor, we take $\mu(y) = \frac{1}{y}$ (assuming $y \neq 0$; the constant sign can be absorbed later).

3. Multiply and simplify Multiply the ODE $\frac{dx}{dy} - \frac{x}{y} = 0$ by $\frac{1}{y}$:

Notice that the left side is exactly $\frac{d}{dy}\left(\frac{x}{y}\right)$ — check by differentiating:

So the equation becomes

4. Integrate Integrating both sides with respect to $y$:

where $C$ is an arbitrary constant. Multiply through by $y$:

$x = C y$ …

Method: Recognising $y\,dx-x\,dy$ as a $d\!\left(\tfrac{x}{y}\right)$ pattern

For a compact form like $\dfrac{y\,dx-x\,dy}{y}=0$, simplify first — it is exact in $\dfrac{x}{y}$.

Steps

Step 1: Simplify the given expression.

Divide term by term: $\dfrac{y\,dx-x\,dy}{y}=dx-\dfrac{x}{y}\,dy$.

Step 2: Write the linear-in-$x$ form (or spot the differential). …

Common Mistakes

Mistake 1: Separating before clearing the $y$ in the denominator.

Why it's wrong: $\dfrac{y\,dx-x\,dy}{y}=0$ must first be multiplied through (or simplified to $dx-\tfrac{x}{y}dy=0$); skipping this confuses the separation. Correct approach: simplify to $dx-\tfrac{x}{y}dy=0$ first.

Mistake 2: Not recognising the $d\!\left(\tfrac{x}{y}\right)$ structure. …