QJEEAdv05Ma6c0a: JEE Advanced 2005 Mathematics - Class 12 JEEAdv05Ma6c0a Solution: If and it follows the relation then

To find $y''(0)$ , we need to differentiate the given equation twice with respect to $x$ and then substitute $x=0$ .

Differentiating the equation $x\cos \,y + y\,cos\,x = \pi$ with respect to $x$ , we get:

$\cos\,y - x\sin\,y \cdot y' + y'\cos\,x - y\sin\,x = 0$

Now, let's differentiate this equation again with respect to $x$ :

$- \sin\,y \cdot y' - \sin\,y \cdot y' - x\cos\,y \cdot (y')^2 - x\sin\,y \cdot y'' + y''\cos\,x - y'\sin\,x - y'\sin\,x - y\cos\,x = 0$

Simplifying the equation, we get:

$-2\sin\,y \cdot y' - x\cos\,y \cdot (y')^2 - x\sin\,y \cdot y'' + y''\cos\,x - 2y'\sin\,x - y\cos\,x = 0$

Now, let's substitute $x = 0$ in the original equation and the first derivative equation. This will help us find the values of $y(0)$ and $y'(0)$ respectively:

From the original equation, $x\cos \,y + y\,cos\,x = \pi$ , substituting $x = 0$ , we get:

$y(0) \cdot 1 = \pi$

Therefore, $y(0) = \pi$ .

Now, substituting $x = 0$ in the first derivative equation, we get:

$\cos\,y(0) + y'(0) \cdot 1 = 0$

Substituting $y(0) = \pi$ , we get:

$\cos \pi + y'(0) = 0$

Therefore, $y'(0) = 1$ .

Finally, let's substitute $x = 0$ , $y(0) = \pi$ , and $y'(0) = 1$ in the second derivative equation:

$-2\sin\pi \cdot 1 - 0 \cdot \cos \pi \cdot (1)^2 - 0 \cdot \sin \pi \cdot y''(0) + y''(0)\cos 0 - 2 \cdot 1 \cdot \sin 0 - \pi \cdot \cos 0 = 0$

Simplifying the equation, we get:

$y''(0) - \pi = 0$

Therefore, $y''(0) = \pi$ .