Let y = y(x) be the solution of the differential equation dy/dx = 1 + xe^(y-x), -√2 < x < √2, y(0) = 0, then the minimum value of y(x), x ∈ (-√2,√2) is equal to:

 

$y(ydx+xdy)=x^5\left(\frac{xdy-ydx}{x^2}\right)$

$\Rightarrow d\left(xy\right)=x^5{y}^{-1}\left(\frac{xdy-ydx}{x^2}\right)\Rightarrow d\left(xy\right)=(xy{)}^{\alpha }{\left(\frac{y}{x}\right)}^{\beta }d\left(\frac{y}{x}\right)$

$\alpha -\beta =5\mathrm{}\alpha +\beta =-1$

$2\alpha =4\mathrm{}\alpha =2\mathrm{}\beta =-3$

$\Rightarrow \mathrm{}\left(xy{)}^{-2}d\right(xy)={\left(\frac{y}{x}\right)}^{-3}d\left(\frac{y}{x}\right)\Rightarrow -(xy{)}^{-1}=-\frac{1}{2}{\left(\frac{y}{x}\right)}^{-2}+c$

$\Rightarrow \mathrm{}-(xy{)}^{-1}=-\frac{1}{2}{\left(\frac{y}{x}\right)}^{-2}+c\mathrm{}$ Passing $\left(\mathrm{1,1}\right)$

$\Rightarrow \mathrm{}-1=-\frac{1}{2}+c\Rightarrow c=-\frac{1}{2};-(xy{)}^{-1}=-\frac{1}{2}{\left(\frac{y}{x}\right)}^{-2}-\frac{1}{2}$

$\underset{x\to 0}{lim}{\left(2-cosx.\sqrt[]{cos2x}\right)}^{\frac{x+2}{x^2}}$ $\left(1^{\infty }\right)$

= $\underset{x\to 0}{lim}{e}^{2\left(\frac{sinx\sqrt[]{cos2x}-cosx.\frac{1\left(-2sin2x\right)}{2\sqrt[]{cos2x}}}{2x}\right)}$

$=\underset{x\to 0}{lim}{e}^{2\left(\frac{1}{2}+1\right)}=e^3=e^a$

a = 3