Suppose $y=y(x)$ be the solution curve to the differential equation $\frac{d y}{d x}-y=2-e^{-x}$ suc

Question

Suppose $y=y(x)$ be the solution curve to the differential equation $\frac{d y}{d x}-y=2-e^{-x}$ such that $\lim _{x \rightarrow \infty} y(x)$ 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 $a-4 b$ is equal to

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(3) : The given differential equation is\ $\frac{d y}{d x}-y=2-e^{-x} \quad 	herefore$ I.F. $=e^{-\int d x}=e^{-x}$\ So, the solution is given by,\ $y \cdot e^{-x}=\int 2 e^{-x} d x-\int e^{-2 x} d x$\ $\Rightarrow y \cdot e^{-x}=-2 e^{-x}+\frac{1}{2} e^{-2 x}+c$\ $\Rightarrow \quad y=-2+\frac{1}{2} e^{-x}+c e^{x}$\ But $\lim _{x ightarrow \infty} y(x)$ is finite. So, $c=0$\ $	herefore \quad y=-2+\frac{1}{2} e^{-x} \Rightarrow \frac{d y}{d x}=\frac{-1}{2} e^{-x}$\ So, equation of tangent at $x=0$,\ $y=-2+\frac{1}{2}=\frac{-3}{2}$ is,\ $y+\frac{3}{2}=\frac{-1}{2}(x-0) \Rightarrow \frac{x}{2}+y+\frac{3}{2}=0$\ $\Rightarrow \frac{x}{(-3)}+\frac{y}{(-3 / 2)}=1$\ So, $a=-3, b=\frac{-3}{2} 	herefore a-4 b=-3+4 \cdot \frac{3}{2}=3$\