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Question

(A) Given the differential equation: $\frac{dy}{dx} = 1 + x + y + xy$Factor the right side of the equation:$\frac{dy}{dx} = (1 + x) + y(1 + x)$$\frac{

Vedclass · Accepted Answer

$\log(1 + y) = x + \frac{x^2}{2} + c$

Vedclass · Answer

(A) Given the differential equation: $\frac{dy}{dx} = 1 + x + y + xy$Factor the right side of the equation:$\frac{dy}{dx} = (1 + x) + y(1 + x)$$\frac{dy}{dx} = (1 + x)(1 + y)$Separate the variables:$\frac{dy}{1 + y} = (1 + x) dx$Integrate both sides:$\int \frac{1}{1 + y} dy = \int (1 + x) dx$Performing the integration:$\log(1 + y) = x + \frac{x^2}{2} + c$Thus,the solution is $\log(1 + y) = x + \frac{x^2}{2} + c$.

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

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