The solution of y' = 1 + x + y^2 + xy^2,y(0) | vedclass

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(D) Given the differential equation $\frac{dy}{dx} = 1 + x + y^2 + xy^2$. Factoring the right side,we get $\frac{dy}{dx} = (1 + x) + y^2(1 + x) = (1 +

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$y = 	an \left( x + \frac{x^2}{2} ight)$

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(D) Given the differential equation $\frac{dy}{dx} = 1 + x + y^2 + xy^2$. Factoring the right side,we get $\frac{dy}{dx} = (1 + x) + y^2(1 + x) = (1 + x)(1 + y^2)$. Separating the variables,we have $\frac{dy}{1 + y^2} = (1 + x) dx$. Integrating both sides,$\int \frac{dy}{1 + y^2} = \int (1 + x) dx$. This yields $	an^{-1} y = x + \frac{x^2}{2} + c$. Using the initial condition $y(0) = 0$,we substitute $x = 0$ and $y = 0$: $	an^{-1}(0) = 0 + \frac{0^2}{2} + c$,which gives $c = 0$. Therefore,$	an^{-1} y = x + \frac{x^2}{2}$,which implies $y = 	an \left( x + \frac{x^2}{2} ight)$.

The solution of $y' = 1 + x + y^2 + xy^2$,$y(0) = 0$ is

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