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(C) The given differential equation is $\frac{dy}{dx}-y=2-e^{-x}$.This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$,where $P=-1$

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(C) The given differential equation is $\frac{dy}{dx}-y=2-e^{-x}$.This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$,where $P=-1$ and $Q=2-e^{-x}$.The integrating factor is $I.F. = e^{\int -1 dx} = e^{-x}$.The solution is $y \cdot e^{-x} = \int (2-e^{-x})e^{-x} dx = \int (2e^{-x}-e^{-2x}) dx$.$y \cdot e^{-x} = -2e^{-x} + \frac{1}{2}e^{-2x} + C$.$y(x) = -2 + \frac{1}{2}e^{-x} + Ce^x$.Since $\lim_{x \rightarrow \infty} y(x)$ is finite,the coefficient of $e^x$ must be zero,so $C=0$.Thus,$y(x) = -2 + \frac{1}{2}e^{-x}$.At $x=0$,$y(0) = -2 + \frac{1}{2} = -\frac{3}{2}$.The derivative is $\frac{dy}{dx} = -\frac{1}{2}e^{-x}$.At $x=0$,the slope $m = \frac{dy}{dx}|_{x=0} = -\frac{1}{2}$.The equation of the tangent at $(0, -3/2)$ is $y - (-3/2) = -\frac{1}{2}(x - 0)$,which simplifies to $y + \frac{3}{2} = -\frac{1}{2}x$,or $x + 2y = -3$.The $x$-intercept $a$ is found by setting $y=0$,so $a = -3$.The $y$-intercept $b$ is found by setting $x=0$,so $2b = -3$,which gives $b = -\frac{3}{2}$.Finally,$a - 4b = -3 - 4(-\frac{3}{2}) = -3 + 6 = 3$.

Suppose $y=y(x)$ is the solution curve to the differential equation $\frac{dy}{dx}-y=2-e^{-x}$ 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 $a-4b$ is equal to:

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