If frac{dy}{dx} = 1 + x + y + xy and y(-1) = | vedclass

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(B) Given the differential equation $\frac{dy}{dx} = 1 + x + y + xy$.Factorizing the right side: $\frac{dy}{dx} = (1 + x) + y(1 + x) = (1 + x)(1 + y)$

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$e^{(1 + x)^2/2} - 1$

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(B) Given the differential equation $\frac{dy}{dx} = 1 + x + y + xy$.Factorizing the right side: $\frac{dy}{dx} = (1 + x) + y(1 + x) = (1 + x)(1 + y)$.Separating the variables: $\frac{dy}{1 + y} = (1 + x) dx$.Integrating both sides: $\int \frac{dy}{1 + y} = \int (1 + x) dx$.This gives $\log_e(1 + y) = x + \frac{x^2}{2} + C$.Exponentiating both sides: $1 + y = e^{x + x^2/2 + C} = k e^{x + x^2/2}$,where $k = e^C$.So,$y = k e^{x + x^2/2} - 1$.Using the initial condition $y(-1) = 0$: $0 = k e^{-1 + 1/2} - 1 \implies 0 = k e^{-1/2} - 1 \implies k = e^{1/2}$.Substituting $k$ back: $y = e^{1/2} e^{x + x^2/2} - 1 = e^{(1 + 2x + x^2)/2} - 1 = e^{(1 + x)^2/2} - 1$.

If $\frac{dy}{dx} = 1 + x + y + xy$ and $y(-1) = 0$,then the function $y$ is

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