The equation of a curve passing through the p | vedclass

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(B) Given,the slope of the tangent is $\frac{dy}{dx} = x + xy$. Rearranging the terms,we have $\frac{dy}{dx} = x(1+y)$. Separating the variables,we ge

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$y=-1+2 e^{\left(\frac{x^2}{2}\right)}$

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(B) Given,the slope of the tangent is $\frac{dy}{dx} = x + xy$. Rearranging the terms,we have $\frac{dy}{dx} = x(1+y)$. Separating the variables,we get $\frac{dy}{1+y} = x dx$. Integrating both sides,we get $\int \frac{dy}{1+y} = \int x dx$,which gives $\ln(y+1) = \frac{x^2}{2} + C$. Exponentiating both sides,$y+1 = e^{\frac{x^2}{2} + C} = e^C \cdot e^{\frac{x^2}{2}}$. Let $e^C = A$,so $y = A e^{\frac{x^2}{2}} - 1$. Since the curve passes through $(0,1)$,we substitute $x=0$ and $y=1$: $1 = A e^0 - 1 \Rightarrow 1 = A - 1 \Rightarrow A = 2$. Thus,the equation of the curve is $y = 2 e^{\frac{x^2}{2}} - 1$.

The equation of a curve passing through the point $(0,1)$,given that the slope of the tangent to the curve at any point $(x, y)$ is equal to the sum of the $x$-coordinate and the product of $x$ and $y$ coordinates at that point,is

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