Suppose y = y(x) be the solution curve to the differential equation...

Given,

$\dfrac{dy}{dx} - y = 2 - {e}^{- x}$

Linear differential equation,

So, $I . F . = {e}^{- ∫ dx} = {e}^{- x}$

Now solution of differential equation is given by

$y \times IF = ∫ \left(2 - {e}^{- x}\right) \times IFdx$

$⇒ y{e}^{- x} = ∫ \left(2 - {e}^{- x}\right){e}^{- x}dx$

$⇒ y{e}^{- x} = ∫ \left(2{e}^{- x} - {e}^{- 2x}\right)dx$

$⇒ y{e}^{- x} = - 2{e}^{- x} + \dfrac{{e}^{- 2x}}{2} + C$

$⇒ y = - 2 + \dfrac{{e}^{- x}}{2} + C{e}^{x}$

Given $\lim_{x \to \infty}y$ is finite

So, $\lim_{x \to \infty}\left(- 2 + \dfrac{{e}^{- x}}{2} + C \cdot {e}^{x}\right) \to$finite

This is possible only when $C = 0$

Hence$y = y\left(x\right) = - 2 + \dfrac{{e}^{- x}}{2}$

Now finding the slope $\dfrac{dy}{dx} = - \dfrac{1}{2}{e}^{- x}$

${\dfrac{dy}{dx}}_{x = 0} = - \dfrac{1}{2} = m, y\left(0\right) = - 2 + \dfrac{1}{2} = \dfrac{- 3}{2}$

Now equation of tangent will be,

$y + \dfrac{3}{2} = - \dfrac{1}{2}\left(x - 0\right)$

$⇒ x + 2y = - 3$

x-intercept $a = - 3,$ y-intercept $b = \dfrac{- 3}{2}$

So, $a - 4b = - 3 + 6 = 3$