Animated Solution for Mathematics - Differential Equations: If $y=\left(x-\frac{\pi }{2}-1\right)\mathrm{cosec}x$ is the solution of the differential equation, $\frac{dy}{dx}+p(x)y=\frac{\pi }{2}\mathrm{cosec}x$ $0<x<\frac{\pi }{2}$, then the function $p(x)$ is equal to :

We are tasked with finding the differential equation that corresponds to the solution $y=(x-\frac{\pi }{2}-1)\csc x$. In JEE Advanced problems, working backward requires identifying the most efficient algebraic structure before differentiating.

When you see $y=(x-\frac{\pi }{2}-1)\csc x$, your instinct might be to use the quotient rule. However, the quotient rule is often a trap that leads to messy calculations and potential sign errors.

Instead, we utilize the identity $\csc x=\frac{1}{\sin x}$. By rewriting the equation as:

We can clear the denominator by multiplying both sides by $\sin x$. This yields the simplified form:

Now that we have a clean expression, we differentiate both sides with respect to $x$. Applying the product rule to the left side, where the derivative of $uv$ is $u^{\prime }v+u{v}^{\prime }$, we obtain:

On the right side, the derivative of $x$ is $1$, while the derivatives of the constants $-\frac{\pi }{2}$ and $-1$ are $0$. This equation represents the differential equation in its raw form.

To match the standard linear form $\frac{dy}{dx}+p(x)y=Q(x)$, the coefficient of $\frac{dy}{dx}$ must be $1$. We divide the entire equation by $\sin x$:

Using fundamental trigonometric identities, we substitute $\frac{\cos x}{\sin x}=\cot x$ and $\frac{1}{\sin x}=\csc x$. This results in:

Comparing this to the standard form, we identify that $p(x)=\cot x$. By choosing the path of algebraic elegance, we have successfully reconstructed the differential equation.