Q.Solve the differential equation $dy=\cos x\,(2-y\csc x)\,dx$ given that $y=2$ when $x=\frac{\pi}{2}$.

We are solving an Initial Value Problem. The equation is:

Step 1 – Rewrite in standard form.

Divide through by $dx$:

So:

This is a linear first-order ODE.

Step 2 – Find the integrating factor.

Step 3 – Multiply and integrate.

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

Integrate:

Step 4 – Apply the initial condition.

At $x = \frac{\pi}{2}$, $y = 2$: …

The problem gives us $dy = \cos x \, (2 - y \csc x) \, dx$, with the condition $y = 2$ when $x = \frac{\pi}{2}$. At first glance, this looks like a separable equation might work, but the $y$ term is tangled inside the parentheses. Let's rewrite it to see its true nature.

Divide both sides by $dx$ (treating it as a differential form, which is valid here):

Expand the right-hand side:

Since $\csc x = \frac{1}{\sin x}$, we have $\cos x \csc x = \cot x$. So:

Now bring the term involving $y$ to the left side:

This is a first-order linear differential equation in the standard form $\frac{dy}{dx} + P(x) y = Q(x)$, where $P(x) = \cot x$ and $Q(x) = 2\cos x$.

1. Find the integrating factor.

   Compute $\int \cot x \, dx = \int \frac{\cos x}{\sin x} dx = \log|\sin x|$.

   So $\mu(x) = e^{\log|\sin x|} = |\sin x|$. Since we are near $x = \frac{\pi}{2}$ where $\sin x > 0$, we can drop the absolute value and take $\mu(x) = \sin x$.

2. Multiply the entire equation by $\sin x$.

But $\sin x \cot x = \cos x$, so the left side becomes:

Notice that this is exactly the derivative of $y \sin x$ with respect to $x$ (by the product rule: $\frac{d}{dx}(y \sin x) = \frac{dy}{dx} \sin x + y \cos x$).

The right side simplifies: $2 \cos x \sin x = \sin 2x$.

So the equation becomes:

3. Integrate both sides.

4. Solve for $y$. …

Method: Linear DE with a $\cot x$ coefficient, plus a condition

Use this when a DE with trigonometric coefficients rearranges to linear form $\dfrac{dy}{dx}+P(x)y=Q(x)$.

Steps

Step 1: Rewrite in standard linear form

Expand and divide so the coefficient of $\dfrac{dy}{dx}$ is $1$; identify $P(x)$ (here $\cot x$) and $Q(x)$.

Step 2: Integrating factor

$\mathrm{IF}=e^{\int\cot x\,dx}=e^{\log|\sin x|}=\sin x.$ …

Common Mistakes

Mistake 1: Not rewriting $\cos x(2-y\csc x)$ into linear form

Why it's wrong: expanding gives $\frac{dy}{dx}+(\cot x)y=2\cos x$; leaving it unexpanded hides the linear structure. Correct approach: distribute, using $\cos x\csc x=\cot x$.

Mistake 2: Wrong integrating factor

Why it's wrong: $\int\cot x\,dx=\log|\sin x|$, so $\mathrm{IF}=\sin x$. Correct approach: use this standard integral. …