Suppose y = y(x) be the solution curve to the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}=2-{\mathrm{e}}^{-\mathrm{x}}$ such that $\underset{\mathrm{x}\to \mathrm{\infty }}{lim} \mathrm{y}\left(\mathrm{x}\right)$ is finite. If a and b are respectively the x - and y - intercepts of the tangent to the curve at x = 0, then the value of a - 4b is equal to _________.

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}=2-{\mathrm{e}}^{-\mathrm{x}} \mathrm{is} \mathrm{a} \mathrm{linear} \mathrm{differential} \mathrm{equation} \mathrm{in} \mathrm{y}\\ \text{ I.F. }={\mathrm{e}}^{-\int \mathrm{dx}}={\mathrm{e}}^{-\mathrm{x}}\\ \therefore \text{ solution of D.E is }\\ \mathrm{y}\cdot {\mathrm{e}}^{-\mathrm{x}}=\int \left(2{\mathrm{e}}^{-\mathrm{x}}-{\mathrm{e}}^{-2\mathrm{x}}\right)\mathrm{dx}\\ \Rightarrow \mathrm{y}=-2+\frac{{\mathrm{e}}^{-\mathrm{x}}}{2}+\mathrm{C}\cdot {\mathrm{e}}^{\mathrm{x}}\\ \because \underset{\mathrm{x}\to \mathrm{\infty }}{lim} \mathrm{y}\text{ is finite }\\ \therefore \underset{\mathrm{x}\to \mathrm{\infty }}{lim} \left(-2+\frac{{\mathrm{e}}^{-\mathrm{x}}}{2}+\mathrm{C}.{\mathrm{e}}^{\mathrm{x}}\right)\to \text{ finite }\end{array}$

This is possible only when C = 0

$\begin{array}{l}\therefore \mathrm{y}=\mathrm{y}\left(\mathrm{x}\right)=-2+\frac{{\mathrm{e}}^{-\mathrm{x}}}{2}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{2}{\mathrm{e}}^{-\mathrm{x}}\\ {\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|}_{\mathrm{x}=0}=-\frac{1}{2}=\mathrm{m},\mathrm{y}\left(0\right)=-2+\frac{1}{2}=\frac{-3}{2}\end{array}$

$\therefore$ equation of tangent at (0, -3/2)

$\begin{array}{l}\mathrm{y}+\frac{3}{2}=-\frac{1}{2}(\mathrm{x}-0)\\ \Rightarrow \mathrm{x}+2\mathrm{y}=-3\\ \mathrm{a}=-3,\mathrm{b}=\frac{-3}{2}\\ \mathrm{a}-4\mathrm{b}=-3+6=3\end{array}$