The general solution of the differential equa | vedclass

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

Vedclass · Accepted Answer

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

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(D) Given the differential equation: $\frac{dy}{dx} = 1 + x + y + xy$ Factorizing the right side: $\frac{dy}{dx} = (1 + x)(1 + y)$ Separating the variables: $\frac{dy}{1 + y} = (1 + x) dx$ Integrating both sides: $\int \frac{dy}{1 + y} = \int (1 + x) dx$ This gives: $\log(1 + y) = x + \frac{x^2}{2} + c$ Taking the exponential of both sides: $1 + y = e^{x + \frac{x^2}{2} + c}$ $1 + y = e^c \cdot e^{x + \frac{x^2}{2}}$ Let $k = e^c$,then $1 + y = k e^{x + \frac{x^2}{2}}$ Therefore,the general solution is: $y = k e^{x + \frac{x^2}{2}} - 1$

The general solution of the differential equation $\frac{dy}{dx} = 1 + x + y + xy$ is

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