Let y = y ( x ) be the solution of the differential equation sec x d y + 2 1 - x tan x + x 2 - x d x = 0…

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

$\Rightarrow \sec xdy=\left\{2\left(x-1\right)\tan x+\left(x^2-2x\right)\right\}dx$

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

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

$\Rightarrow \int dy=\int \left[2\left(x-1\right)\sin x+\left(x^2-2x\right)\cos x\right]dx$

$\Rightarrow \int dy=\int 2\left(x-1\right)\sin x+\int \left(x^2-2x\right)\cos xdx$

$\Rightarrow y\left(x\right)=\int 2(x-1)\sin xdx+\left[\left(x^2-2x\right)\left(\sin x\right)-\int (2x-2)\sin xdx\right]$

$\Rightarrow y\left(x\right)=\left(x^2-2x\right)\sin x+\lambda$

It is given that, $y\left(0\right)=2$

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

$\Rightarrow 2=0+\lambda$

$\Rightarrow \lambda =2$

$\Rightarrow y\left(x\right)=\left(x^2-2x\right)\sin x+2$

$\Rightarrow y\left(2\right)=\left(2^2-2\times 2\right)\sin \left(2\right)+2$

$\Rightarrow y\left(2\right)=2$