Animated Solution for Mathematics - Differential Equations: Let $y=y(x)$ be the solution of the differential equation $\sec xdy+\{2(1-x)\tan x+x(2-x)\}dx=0$ such that $y(0)=2$. Then $y(2)$ is equal to :

Welcome, fellow traveler on the JEE Advanced journey. Today, we confront a differential equation that, at first glance, looks like a tangled mess of trigonometric functions and polynomials.

You might be tempted to reach for complex integration techniques or panic at the sight of $\sec x$ and $\tan x$ mixed with $x(2-x)$. But pause. Take a breath.

In the world of competitive physics and mathematics, the most complex-looking problems often hide the most elegant, simple solutions. Our goal is to find $y(2)$ given $y(0)=2$.

The Cleanup

We begin with the given equation:

Our first instinct is to isolate the derivative, $\frac{dy}{dx}$. By moving the $dx$ term to the right and dividing, we get:

Here is the 'JEE trick': multiply the entire equation by $\cos x$. Since $\sec x\cdot \cos x=1$ and $\tan x\cdot \cos x=\sin x$, the left side becomes clean, and the right side transforms into:

The Epiphany

Now, let's distribute that $\cos x$ inside the curly braces:

Stop here and look at this expression. Let $u=2x-x^2$ and $v=\sin x$.

If we differentiate $u$, we get $\frac{du}{dx}=2-2x=2(1-x)$. If we differentiate $v$, we get $\frac{dv}{dx}=\cos x$.

The expression inside the braces is exactly $u^{\prime }v+u{v}^{\prime }$. This is the Product Rule in its full glory, allowing us to condense the equation to:

The Integration and The Result

With the equation in the form $\frac{dy}{dx}=-\frac{d}{dx}[(2x-x^2)\sin x]$, the integration becomes trivial. Integrating both sides with respect to $x$, we get:

To make it cleaner, we rewrite this as:

Now, we use our initial condition $y(0)=2$. Substituting $x=0$ and $y=2$:

Our particular solution is:

Finally, to find $y(2)$, we substitute $x=2$:

Since $4-4=0$, the entire term involving $\sin (2)$ vanishes. The final result is: $y(2)=2$