$\text { I.F. }=e^{-\int d x}=e^{-x}$
$\therefore \text { solution of D.E }$
$y \cdot e^{-x}=\int\left(2 e^{-x}-e^{-2 x}\right) d x$
$\Rightarrow y=-2+\frac{e^{-x}}{2}+C \cdot e^{x}$
$\because \lim _{x \rightarrow \infty} y$ is finite
$\therefore \lim _{x \rightarrow \infty}\left(-2+\frac{ e ^{- x }}{2}+ C ^{ x }\right) \rightarrow \text { finite }$
This is possible only when $C =0$
$\therefore y = y ( x )=-2+\frac{ e ^{- x }}{2}$
$\frac{ dy }{ dx }=-\frac{1}{2} e ^{- x }$
$\left.\frac{ dy }{ dx }\right|_{ x =0}=-\frac{1}{2}= m , y (0)=-2+\frac{1}{2}=\frac{-3}{2}$
$\therefore$ equation of tangent
$y +\frac{3}{2}=-\frac{1}{2}( x -0)$
$\Rightarrow x +2 y =-3$
$a =-3, b =\frac{-3}{2}$
$a -4 b =-3+6=3$
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