The particular solution of frac{dy}{dx} = 1 + | vedclass

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(D) Given the differential equation: $\frac{dy}{dx} = 1 + x + y^2 + xy^2$ Factorizing the right side: $\frac{dy}{dx} = (1 + x)(1 + y^2)$ Separating th

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

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(D) Given the differential equation: $\frac{dy}{dx} = 1 + x + y^2 + xy^2$ Factorizing the right side: $\frac{dy}{dx} = (1 + x)(1 + y^2)$ Separating the variables: $\int \frac{dy}{1 + y^2} = \int (1 + x) dx$ Integrating both sides: $	an^{-1}(y) = x + \frac{x^2}{2} + C$ Given the initial condition $y(0) = 0$: $	an^{-1}(0) = 0 + \frac{0^2}{2} + C \Rightarrow C = 0$ Thus,the particular solution is: $	an^{-1}(y) = x + \frac{x^2}{2}$ Therefore: $y = 	an \left(x + \frac{x^2}{2}ight)$

The particular solution of $\frac{dy}{dx} = 1 + x + y^2 + xy^2$,when $y(0) = 0$,is

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