Question

1. If y(x) satisfies the equation x sin y + x = y, then the value of y"(
$\pi$
)=
​

Answer 1

Given:

y(x) satisfies the equation  x sin y + x = y

To find:

The value of y"( π)

Solution:

$y=x sin y + x$

differentiating on both sides, we get,

$\dfrac{dy}{dx}=\dfrac{d}{dx}(xsiny+x)$

$\dfrac{dy}{dx}=\dfrac{d}{dx}\left(x\sin \left(y\right)\right)+\dfrac{d}{dx}\left(x\right)$

$\dfrac{dy}{dx}=\sin \left(y\right)+1$

again differentiating on both sides, we get,

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(\sin \left(y\right)+1)$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\sin \left(x\right)\right)+\dfrac{d}{dx}\left(1\right)$

$\dfrac{d^2y}{dx^2}=\cos \left(x\right)+0$

$\dfrac{d^2y}{dx^2}=\cos \left(x\right)$

⇒ y" = cos (x)

y" (π) = cos (π)

∴ y" (π) = -1