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

Vedclass · Accepted Answer

$x+y+1=e^{x}$

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(A) Let the curve be $y = f(x)$. The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx}$.According to the problem,$\frac{dy}{dx} = x + y$.This is a linear differential equation of the form $\frac{dy}{dx} - y = x$.Here,$P = -1$ and $Q = x$.The integrating factor $(I.F.)$ is $e^{\int P dx} = e^{\int -1 dx} = e^{-x}$.The general solution is $y(I.F.) = \int Q(I.F.) dx + C$.$y e^{-x} = \int x e^{-x} dx + C$.Using integration by parts for $\int x e^{-x} dx$:$\int x e^{-x} dx = x(-e^{-x}) - \int 1(-e^{-x}) dx = -x e^{-x} - e^{-x} = -e^{-x}(x+1)$.Thus,$y e^{-x} = -e^{-x}(x+1) + C$.Multiplying by $e^{x}$,we get $y = -(x+1) + C e^{x}$,which simplifies to $x + y + 1 = C e^{x}$.Since the curve passes through the origin $(0, 0)$,we substitute $x=0$ and $y=0$:$0 + 0 + 1 = C e^{0} \Rightarrow 1 = C(1) \Rightarrow C = 1$.Substituting $C=1$ into the equation,we get $x + y + 1 = e^{x}$.

Find the equation of a curve passing through the origin,given that the slope of the tangent to the curve at any point $(x, y)$ is equal to the sum of the coordinates of the point.

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