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

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$ye^x-x=0$

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(C) The slope of the tangent is given by $\frac{dy}{dx} = -y + e^{-x}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 1$ and $Q = e^{-x}$.The Integrating Factor ($I$.$F$.) is $e^{\int P dx} = e^{\int 1 dx} = e^x$.The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.$y e^x = \int e^{-x} \cdot e^x dx + C$.$y e^x = \int 1 dx + C$.$y e^x = x + C$.Since the curve passes through the origin $(0,0)$,we substitute $x=0$ and $y=0$:$0 \cdot e^0 = 0 + C \Rightarrow C = 0$.Thus,the equation of the curve is $y e^x = x$,which can be written as $y e^x - x = 0$.

If the slope of the tangent of the curve at any point is equal to $-y+e^{-x}$,then the equation of the curve passing through the origin is

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