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(B) Let the point on the curve be $(x, y)$. The slope of the tangent is $\frac{dy}{dx}$.According to the problem,the sum of the ordinate $(y)$ and abs

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$y=4-x-2 e^x$

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(B) Let the point on the curve be $(x, y)$. The slope of the tangent is $\frac{dy}{dx}$.According to the problem,the sum of the ordinate $(y)$ and abscissa $(x)$ exceeds the slope by $5$,so:$y + x = \frac{dy}{dx} + 5$Rearranging the terms,we get the linear differential equation:$\frac{dy}{dx} - y = x - 5$This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -1$ and $Q = x - 5$.The integrating factor $(IF)$ is $e^{\int P dx} = e^{\int -1 dx} = e^{-x}$.The general solution is $y \cdot IF = \int Q \cdot IF dx + C$:$y e^{-x} = \int (x - 5) e^{-x} dx + C$Using integration by parts for $\int (x - 5) e^{-x} dx$:$= (x - 5)(-e^{-x}) - \int (1)(-e^{-x}) dx$$= -(x - 5)e^{-x} - e^{-x} + C$$= (-x + 5 - 1)e^{-x} + C = (4 - x)e^{-x} + C$So,$y e^{-x} = (4 - x)e^{-x} + C$,which implies $y = 4 - x + C e^x$.Given the curve passes through $(0, 2)$:$2 = 4 - 0 + C e^0 \implies 2 = 4 + C \implies C = -2$.Substituting $C = -2$ into the equation:$y = 4 - x - 2 e^x$.

The equation of the curve passing through the point $(0,2)$ given that the sum of the ordinate and abscissa of any point exceeds the slope of the tangent to the curve at that point by $5$ is

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