The solution of the differential equation (x | vedclass

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(A) Given differential equation is $(x - y^2x)dx = (y - x^2y)dy$.Factorizing both sides,we get $x(1 - y^2)dx = y(1 - x^2)dy$.Separating the variables,

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

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(A) Given differential equation is $(x - y^2x)dx = (y - x^2y)dy$.Factorizing both sides,we get $x(1 - y^2)dx = y(1 - x^2)dy$.Separating the variables,we have $\frac{x}{1 - x^2}dx = \frac{y}{1 - y^2}dy$.Integrating both sides,we get $\int \frac{x}{1 - x^2}dx = \int \frac{y}{1 - y^2}dy$.Let $u = 1 - x^2$,then $du = -2xdx$,so $xdx = -\frac{1}{2}du$. Similarly,$ydy = -\frac{1}{2}dv$ where $v = 1 - y^2$.Substituting these,we get $-\frac{1}{2} \int \frac{1}{u}du = -\frac{1}{2} \int \frac{1}{v}dv + C$.$-\frac{1}{2} \ln|1 - x^2| = -\frac{1}{2} \ln|1 - y^2| + \ln c$.Multiplying by $-2$,we get $\ln|1 - x^2| = \ln|1 - y^2| - 2\ln c$.$\ln|1 - x^2| - \ln|1 - y^2| = \ln(c^{-2})$.$\ln|\frac{1 - x^2}{1 - y^2}| = \ln(c^{-2})$.$\frac{1 - x^2}{1 - y^2} = \frac{1}{c^2}$.Therefore,$(1 - y^2) = c^2(1 - x^2)$.

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

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