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

The key idea is that this is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x) y = Q(x)$, solved using an integrating factor.

Step 1: Identify $P(x)$ and find the integrating factor (I.F.)

Here $P(x) = \cot x$.

I.F. $= e^{\int \cot x \, dx} = e^{\log|\sin x|} = \sin x$ (since $x \neq 0$, $\sin x > 0$ near $x = \pi/2$).

Step 2: Multiply the equation by the I.F.

$\sin x \frac{dy}{dx} + y \cos x = (2x + x^2 \cot x) \sin x$

The left side is $\frac{d}{dx}(y \sin x)$. The right side simplifies:

$(2x + x^2 \cot x) \sin x = 2x \sin x + x^2 \cos x$.

Step 3: Integrate both sides

$\frac{d}{dx}(y \sin x) = 2x \sin x + x^2 \cos x$

Integrate: $y \sin x = \int (2x \sin x + x^2 \cos x) \, dx$

Notice that $\frac{d}{dx}(x^2 \sin x) = 2x \sin x + x^2 \cos x$. …

The problem gives us a differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, which is a classic first-order linear ODE. The key idea: we can multiply both sides by an integrating factor that turns the left-hand side into the derivative of a product, making it directly integrable.

Let’s see why this works. If we have $\frac{dy}{dx} + Py = Q$, we want to find a function $\mu(x)$ such that $\mu \frac{dy}{dx} + \mu P y = \frac{d}{dx}(\mu y)$. Expanding the right side gives $\mu \frac{dy}{dx} + \frac{d\mu}{dx} y$, so we need $\frac{d\mu}{dx} = \mu P$. This is a separable equation: $\frac{d\mu}{\mu} = P\,dx$, so $\mu = e^{\int P\,dx}$.

Once we have $\mu$, the equation becomes $\frac{d}{dx}(\mu y) = \mu Q$, and we integrate both sides.

Step 1: Identify $P(x)$ and $Q(x)$

The given equation is:

So $P(x) = \cot x$ and $Q(x) = 2x + x^2 \cot x$.

Step 2: Compute the integrating factor $\mu(x)$

Since $x \neq 0$ and we are given $x = \pi/2$ (where $\sin x > 0$), we can take $\mu = \sin x$ for the interval containing this point.

Step 3: Multiply the differential equation by $\mu$

Notice $\sin x \cot x = \sin x \cdot \frac{\cos x}{\sin x} = \cos x$. So the left side becomes:

And the right side simplifies:

So we have:

Step 4: Recognize the left side as a derivative

The left side is exactly $\frac{d}{dx}(y \sin x)$, because:

So the equation becomes:

Step 5: Integrate both sides

$y \sin x = \int (2x \sin x + x^2 \cos x) \, dx$ …

Method: Linear equation with an initial condition (integrating factor)

Use this when a linear equation $\dfrac{dy}{dx} + P(x)y = Q(x)$ comes with a data point, so a particular solution is required.

Steps

Step 1: Read off $P(x)$ and form $\mu(x) = e^{\int P\,dx}$.

For trigonometric coefficients recall $\displaystyle\int \cot x\,dx = \log|\sin x|$, so the integrating factor can simplify to something like $\sin x$.

Step 2: Multiply through; the left side is $\dfrac{d}{dx}\big(\mu y\big)$.

Step 3: Integrate the right side, looking for a product-rule pattern. …

Common Mistakes

Mistake 1: Wrong integrating factor from $\int \cot x\,dx$.

Why it's wrong: $\displaystyle\int \cot x\,dx = \log|\sin x|$, so $\mu = \sin x$; students sometimes write $\cos x$ or $\log(\sin x)$ as the factor. Correct approach: exponentiate to get $\mu = \sin x$.

Mistake 2: Missing that the right side is a neat derivative.

Why it's wrong: after multiplying, $2x\sin x + x^2\cos x = \dfrac{d}{dx}(x^2\sin x)$, so the integral is immediate; not spotting this leads to long, error-prone by-parts work. Correct approach: recognise the product-rule pattern. …