Animated Solution for Mathematics - Differential Equations: Let $y=y(x)$ be the solution of the differential equation, $\frac{dy}{dx}+y\tan x=2x+x^2\tan x$, $x\in (-\pi /2,\pi /2)$, such that $y(0)=1$. Then :

When you first look at the differential equation

it is natural to feel a bit overwhelmed. However, in the world of JEE Advanced, complexity is often just a mask for elegance. Our goal is to peel back that mask.

Every great journey begins with recognition. We compare our equation to the standard form:

By aligning the terms, we identify $P(x)=\tan x$ and $Q(x)=2x+x^2\tan x$. This identification is our compass.

We define the Integrating Factor (I.F.) as $e^{\int P(x)dx}$. Substituting our $P(x)$, we calculate:

We know that the integral of $\tan x$ is $\ln |\sec x|$. Thus, our I.F. becomes $e^{\ln |\sec x|}$, which simplifies beautifully to just $\sec x$.

Now, we multiply our entire original equation by this I.F. to obtain:

Distributing the $\sec x$ inside the integral gives us:

Here is where most students get stuck, trying to perform integration by parts on each term. But look closer at the structure. If we let $u=x^2$ and $v=\sec x$, then the derivative $\frac{d}{dx}(u\cdot v)$ is exactly $2x\sec x+x^2\sec x\tan x$.

The entire integral collapses into the derivative of $x^2\sec x$. This is the 'Aha!' moment that separates the masters from the novices. The integral of a derivative is simply the original function, so our equation simplifies to:

We are given the initial condition $y(0)=1$. Plugging in $x=0$ and $y=1$, we find:

Since $\sec (0)=1$, we find that $C=1$. Our specific solution is $y\sec x=x^2\sec x+1$.

Dividing by $\sec x$ (or multiplying by $\cos x$), we arrive at the beautiful, clean result:

To solve the final part of the problem, we differentiate to find $y^{\prime }(x)=2x-\sin x$. Evaluating this at $\frac{\pi }{4}$ and $-\frac{\pi }{4}$ and taking the difference leads us directly to $\pi -\sqrt{2}$.