Let y = y ( x ) be the solution of the differential equation secxdy...

Given: $secxdy + \left\{2\left(1 - x\right)\tan x + x\left(2 - x\right)\right\}dx = 0$

$⇒ secxdy = \left\{2\left(x - 1\right)\tan x + \left({x}^{2} - 2x\right)\right\}dx$

$⇒ \dfrac{dy}{\cos x} = \left\{2\left(x - 1\right)\dfrac{\sin x}{\cos x} + \left({x}^{2} - 2x\right)\right\}dx$

$⇒ dy = \left[2\left(x - 1\right)\sin x + \left({x}^{2} - 2x\right)\cos x\right]dx$

$⇒ ∫ dy = ∫ \left[2\left(x - 1\right)\sin x + \left({x}^{2} - 2x\right)\cos x\right]dx$

$⇒ ∫ dy = ∫ 2\left(x - 1\right)\sin x + ∫ \left({x}^{2} - 2x\right)cosxdx$

$⇒ y ( x ) = ∫ 2 ( x - 1 ) sinxdx + \left[\left({x}^{2} - 2x\right) ( \sin x ) - ∫ ( 2x - 2 ) sinxdx\right]$

$⇒ y ( x ) = \left({x}^{2} - 2x\right)\sin x + \lambda$

It is given that, $y\left(0\right) = 2$

$⇒ y ( 0 ) = \left(0 - 0\right)\sin 0 + \lambda$

$⇒ 2 = 0 + \lambda$

$⇒ \lambda = 2$

$⇒ y ( x ) = \left({x}^{2} - 2x\right)\sin x + 2$

$⇒ y ( 2 ) = \left({2}^{2} - 2 \times 2\right)\sin \left(2\right) + 2$

$⇒ y\left(2\right) = 2$