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

Given,

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

Linear differential equation,

So, $I.F. ={\mathrm{e}}^{-\int \mathrm{dx}}={\mathrm{e}}^{-\mathrm{x}}$

Now solution of differential equation is given by 

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

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

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

$\Rightarrow y{e}^{-x}=-2e^{-x}+\frac{e^{-2x}}{2}+C$

$\Rightarrow y=-2+\frac{e^{-x}}{2}+C{e}^x$

Given $\underset{x\to \infty }{\lim }y$ is finite

So, $\underset{x\to \infty }{\lim }\left(-2+\frac{e^{-x}}{2}+C·e^x\right)\to$finite

This is possible only when $C=0$

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

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

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

Now equation of tangent will be,

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

$\Rightarrow x+2y=-3$

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

So, $a-4b=-3+6=3$