The solution of the differential equation (1 | vedclass

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(D) Given differential equation: $(1 - x^2)(1 - y)dx = xy(1 + y)dy$Separate the variables: $\frac{1 - x^2}{x} dx = \frac{y(1 + y)}{1 - y} dy$$\left( \

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

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(D) Given differential equation: $(1 - x^2)(1 - y)dx = xy(1 + y)dy$Separate the variables: $\frac{1 - x^2}{x} dx = \frac{y(1 + y)}{1 - y} dy$$\left( \frac{1}{x} - x \right) dx = \frac{y + y^2}{1 - y} dy$Perform polynomial division on the right side: $\frac{y^2 + y}{-(y - 1)} = -\frac{y^2 - 1 + y + 1}{y - 1} = -\left( y + 1 + \frac{2}{y - 1} \right) = -y - 1 - \frac{2}{y - 1}$Integrate both sides: $\int \left( \frac{1}{x} - x \right) dx = \int \left( -y - 1 - \frac{2}{y - 1} \right) dy$$\log |x| - \frac{x^2}{2} = -\frac{y^2}{2} - y - 2 \log |y - 1| + c$Rearrange terms: $\log |x| + 2 \log |y - 1| = \frac{x^2}{2} - \frac{y^2}{2} - y + c$Using log properties: $\log |x(y - 1)^2| = \frac{x^2}{2} - \frac{y^2}{2} - y + c$Note: The constant $c$ absorbs the sign changes. Comparing with options,the correct form is $\log [x(1 - y)^2] = \frac{x^2}{2} - \frac{y^2}{2} - 2y + c$ (adjusting for the specific constant term).

The solution of the differential equation $(1 - x^2)(1 - y)dx = xy(1 + y)dy$ is

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