Q.Find a particular solution of the differential equation $\frac{dy}{dx} + y \cot x = 4x \operatorname{cosec} x\ (x \neq 0)$, given that $y = 0$ when $x = \frac{\pi}{2}$.

The key idea is the Integrating Factor Method for a first-order linear ODE.

Step 1 – Identify the standard form

The equation is already in the form $\frac{dy}{dx} + P(x) y = Q(x)$, with $P(x) = \cot x$ and $Q(x) = 4x \csc x$.

Step 2 – Find the integrating factor

Step 3 – Multiply and integrate

Multiplying through: $\sin x \frac{dy}{dx} + y \cos x = 4x$.

The left side is $\frac{d}{dx}(y \sin x)$. Integrate both sides:

$y \sin x = \int 4x \, dx = 2x^2 + C$ …

The equation $\frac{dy}{dx} + y \cot x = 4x \operatorname{cosec} x$ is in the standard linear form $\frac{dy}{dx} + P(x) y = Q(x)$. The key idea: we multiply through by an integrating factor (IF) that turns the left side into the derivative of a product, making it directly integrable.

Why does this work? The IF is $e^{\int P\,dx}$. When we multiply the ODE by this factor, the left side becomes $\frac{d}{dx}(y \cdot \text{IF})$, because the derivative of the IF itself gives the $P$ term we need. This is the core trick — it converts a sum into a single derivative.

Let’s apply it step by step.

1. Identify $P(x)$ and $Q(x)$

   Here $P(x) = \cot x$ and $Q(x) = 4x \operatorname{cosec} x$.

2. Compute the integrating factor

(Since $x \neq 0$ and we’re near $x = \pi/2$, $\sin x > 0$, so we drop the absolute value.)

3. Multiply the ODE by the IF

Simplify: $\sin x \cot x = \cos x$, and $\operatorname{cosec} x \cdot \sin x = 1$. So we get:

4. Recognize the left side as a derivative Notice that $\frac{d}{dx}(y \sin x) = \sin x \frac{dy}{dx} + y \cos x$. Exactly our left side! So:

5. Integrate both sides

6. Apply the initial condition Given $y = 0$ when $x = \frac{\pi}{2}$: $0 \cdot \sin\left(\frac{\pi}{2}\right) = 2\left(\frac{\pi}{2}\right)^2 + C \quad \Rightarrow \quad 0 = \frac{\pi^2}{2} + C$ …

Method: Linear IVP with a $\cot x$ coefficient (I.F. $=\sin x$)

Apply the integrating factor for $P=\cot x$, then use the given point.

Steps

Step 1: Confirm standard form.

The equation is already $\dfrac{dy}{dx}+\cot x\,y=4x\csc x$.

Step 2: Integrating factor.

Step 3: Multiply and integrate. …

Common Mistakes

Mistake 1: Not simplifying after multiplying by $\sin x$.

Why it's wrong: $\sin x\cot x=\cos x$ and $\csc x\cdot\sin x=1$, so the right side becomes $4x$; skipping this leaves a messy integral. Correct approach: simplify to $\dfrac{d}{dx}(y\sin x)=4x$.

Mistake 2: Re-multiplying the constant by the I.F.

Why it's wrong: from $y\sin x=2x^2+C$, the $C$ is not multiplied by $\sin x$ again; only $y$ is divided out. Correct approach: solve $y=\dfrac{2x^2+C}{\sin x}$ directly. …