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

Vedclass · Accepted Answer

$y = -1 + 2e^{\frac{x^2}{2}}$

Vedclass · Answer

(A) The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx}$.According to the problem,$\frac{dy}{dx} = x + xy$.This can be written as $\frac{dy}{dx} - xy = x$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -x$ and $Q = x$.The Integrating Factor ($I$.$F$.) is $e^{\int P dx} = e^{\int -x dx} = e^{-\frac{x^2}{2}}$.The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.$y e^{-\frac{x^2}{2}} = \int x e^{-\frac{x^2}{2}} dx + C$.Let $t = -\frac{x^2}{2}$,then $dt = -x dx$,so $x dx = -dt$.Thus,$\int x e^{-\frac{x^2}{2}} dx = -\int e^t dt = -e^t = -e^{-\frac{x^2}{2}}$.Substituting this back,$y e^{-\frac{x^2}{2}} = -e^{-\frac{x^2}{2}} + C$.Dividing by $e^{-\frac{x^2}{2}}$,we get $y = -1 + C e^{\frac{x^2}{2}}$.Since the curve passes through $(0, 1)$,we substitute $x = 0$ and $y = 1$:$1 = -1 + C e^0 \implies 1 = -1 + C \implies C = 2$.Therefore,the equation of the curve is $y = -1 + 2e^{\frac{x^2}{2}}$.

Find the equation of a curve passing through the point $(0,1)$. If the slope of the tangent to the curve at any point $(x, y)$ is equal to the sum of the $x$ coordinate (abscissa) and the product of the $x$ coordinate and $y$ coordinate (ordinate) of that point.

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