$\text{ If }x\frac{dy}{dx}=x^2+y-2,y\left(1\right)=1\text{ then }y\left(2\right)\text{ is }$

$\frac{dy}{dx}=x+\frac{y}{x}-\frac{2}{x}$

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{1}{\mathrm{x}}\mathrm{y}=(\mathrm{x}-\frac{2}{x}),\text{ I.F. }={\mathrm{e}}^{-\int \frac{1}{x} \mathrm{dx}}=e^{-\ln x}=x^{-1}=\frac{1}{x}\\ \text{ GS of the D.E is }\frac{\mathrm{y}}{\mathrm{x}}=\int (\mathrm{x}-\frac{2}{x})\frac{1}{\mathrm{x}}\mathrm{dx}\\ \frac{\mathrm{y}}{\mathrm{x}}=\int 1dx-2\int \frac{1}{x^2}dx\\ \frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x}+\frac{2}{x}+\mathrm{c}\\ \text{ Put x=1, y=1}\\ 1=3+c\to c=-2\\ \frac{y}{x}=\frac{x^2+2-2x}{x}\\ \to y=x^2-2x+2\\ \to \mathrm{y}\left(2\right)=2^2-4+2=2\end{array}$