At any point on a curve, the slope of the tan | vedclass

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(A) According to the question, the slope of the tangent is $\frac{dy}{dx} = x + xy$. Rearranging the terms, we get the linear differential equation: $

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

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(A) According to the question, the slope of the tangent is $\frac{dy}{dx} = x + xy$. Rearranging the terms, we get the linear differential equation: $\frac{dy}{dx} - xy = x$. This is of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = -x$ and $Q(x) = x$. The integrating factor $(IF)$ is $e^{\int P(x) dx} = e^{\int -x dx} = e^{-\frac{x^2}{2}}$. The solution is given by $y \cdot IF = \int Q(x) \cdot IF dx + c$. $y \cdot e^{-\frac{x^2}{2}} = \int x \cdot e^{-\frac{x^2}{2}} dx + c$. Let $t = -\frac{x^2}{2}$, then $dt = -x dx$, so $x dx = -dt$. $y \cdot e^{-\frac{x^2}{2}} = -\int e^t dt + c = -e^t + c = -e^{-\frac{x^2}{2}} + c$. Since the curve passes through $(0, 1)$, we substitute $x=0$ and $y=1$: $1 \cdot e^0 = -e^0 + c \Rightarrow 1 = -1 + c \Rightarrow c = 2$. Thus, $y \cdot e^{-\frac{x^2}{2}} = -e^{-\frac{x^2}{2}} + 2$. Dividing by $e^{-\frac{x^2}{2}}$, we get $y = -1 + 2e^{\frac{x^2}{2}}$ or $y = 2e^{\frac{x^2}{2}} - 1$.

At any point on a curve, the slope of the tangent is equal to the sum of the abscissa and the product of the ordinate and abscissa of that point. If the curve passes through $(0, 1)$, then the equation of the curve is

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