Q.Solve the following differential equation: $\frac{dy}{dx} - 3y \cot x = \sin 2x; \ y=2 \text{ when } x=\frac{\pi}{2}$

This is a linear first-order equation. Write it as $\frac{dy}{dx}+Py=Q$ with $P=-3\cot x$, $Q=\sin 2x$, and use an integrating factor.

Integrating factor: $\text{IF}=e^{\int -3\cot x\,dx}=e^{-3\log|\sin x|}=\csc^3 x$.

Multiply through: the left side becomes an exact derivative,

Integrate (put $u=\sin x$): …

The idea

The equation is linear: $y$ and $\frac{dy}{dx}$ appear only to the first power. Every linear equation $\frac{dy}{dx}+P(x)y=Q(x)$ can be multiplied by an integrating factor that turns the left-hand side into the derivative of a single product, after which we just integrate.

Set up

Here

so $P(x)=-3\cot x$ and $Q(x)=\sin 2x$.

Integrating factor

so

Multiply and integrate

Multiplying by $\csc^3 x$ makes the left side exact:

Simplify the right side with $\sin 2x=2\sin x\cos x$:

Integrate; on the right put $u=\sin x$, $du=\cos x\,dx$: …

Method: Linear IVP with a $\cot x$ coefficient (I.F. a power of $\csc x$)

Use the integrating factor when $P=-n\cot x$, which gives I.F. $=\csc^{n}x$, then apply the condition.

Steps

Step 1: Standard form.

Identify $P=-3\cot x$, $Q=\sin 2x$.

Step 2: Integrating factor.

Step 3: Simplify and integrate. …

Common Mistakes

Mistake 1: Wrong I.F. from $P=-3\cot x$.

Why it's wrong: $\int-3\cot x\,dx=-3\log|\sin x|$, so I.F. $=\csc^3x$, not $\sin^3x$. Correct approach: the negative exponent gives $\csc^3x$.

Mistake 2: Not simplifying $\sin 2x$ before integrating.

Why it's wrong: writing $\sin 2x\csc^3x=\dfrac{2\cos x}{\sin^2x}$ (via $\sin 2x=2\sin x\cos x$) makes the substitution $u=\sin x$ work. Correct approach: expand $\sin 2x$ first. …