The solution of the differential equation (1 | vedclass

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(B) Given equation: $(1 - x^2)dy + xydx = xy^2dx$Rearranging the terms: $(1 - x^2)dy = xy^2dx - xydx = xy(y - 1)dx$Separating the variables: $\frac{1}

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$(y - 1)^2(1 - x^2) = c^2y^2$

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(B) Given equation: $(1 - x^2)dy + xydx = xy^2dx$Rearranging the terms: $(1 - x^2)dy = xy^2dx - xydx = xy(y - 1)dx$Separating the variables: $\frac{1}{y(y - 1)}dy = \frac{x}{1 - x^2}dx$Using partial fractions on the left side: $(\frac{1}{y - 1} - \frac{1}{y})dy = \frac{x}{1 - x^2}dx$Integrating both sides: $\int (\frac{1}{y - 1} - \frac{1}{y})dy = \int \frac{x}{1 - x^2}dx$$\log|y - 1| - \log|y| = -\frac{1}{2}\log|1 - x^2| + \log|c|$Multiply by $2$: $2\log|\frac{y - 1}{y}| = -\log|1 - x^2| + 2\log|c|$$2\log|\frac{y - 1}{y}| + \log|1 - x^2| = \log|c^2|$$\log|(\frac{y - 1}{y})^2(1 - x^2)| = \log|c^2|$Taking exponential on both sides: $\frac{(y - 1)^2}{y^2}(1 - x^2) = c^2$Thus,the solution is $(y - 1)^2(1 - x^2) = c^2y^2$.

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

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