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Question

(A) Let the curve be $y=f(x)$. The slope of the tangent at any point $(x, y)$ is $\frac{dy}{dx}$.According to the problem,the sum of the coordinates $

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the curve be $y=f(x)$. The slope of the tangent at any point $(x, y)$ is $\frac{dy}{dx}$.According to the problem,the sum of the coordinates $(x+y)$ exceeds the slope $\frac{dy}{dx}$ by $5$,so $x+y = \frac{dy}{dx} + 5$.Rearranging,we get the linear differential equation: $\frac{dy}{dx} - y = x - 5$.Here,$P = -1$ and $Q = x - 5$.The Integrating Factor $(I.F.)$ is $e^{\int P dx} = e^{\int -1 dx} = e^{-x}$.The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.$y e^{-x} = \int (x - 5) e^{-x} dx + C$.Using integration by parts,$\int (x - 5) e^{-x} dx = (x - 5)(-e^{-x}) - \int (1)(-e^{-x}) dx = -(x - 5)e^{-x} - e^{-x} + C = (5 - x - 1)e^{-x} + C = (4 - x)e^{-x} + C$.So,$y e^{-x} = (4 - x)e^{-x} + C$,which simplifies to $y = 4 - x + C e^x$.Since the curve passes through $(0, 2)$,we substitute $x=0$ and $y=2$: $2 = 4 - 0 + C e^0 \Rightarrow 2 = 4 + C \Rightarrow C = -2$.Thus,the equation of the curve is $y = 4 - x - 2e^x$.

Find the equation of a curve passing through the point $(0,2)$ given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by $5$.

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