The solution of the differential equation ydx | vedclass

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(C) Given the differential equation: $ydx - xdy = x^2 ydx$.Rearranging the terms,we get: $-xdy = x^2 ydx - ydx$.$-xdy = y(x^2 - 1)dx$.Separating the v

Vedclass · Accepted Answer

$y^2 e^{x^2} = c x^2$

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(C) Given the differential equation: $ydx - xdy = x^2 ydx$.Rearranging the terms,we get: $-xdy = x^2 ydx - ydx$.$-xdy = y(x^2 - 1)dx$.Separating the variables,we have: $\frac{dy}{y} = \frac{x^2 - 1}{-x} dx = (\frac{1}{x} - x) dx$.Integrating both sides: $\int \frac{dy}{y} = \int (\frac{1}{x} - x) dx$.$\ln|y| = \ln|x| - \frac{x^2}{2} + C$.Multiplying by $2$: $2 \ln|y| = 2 \ln|x| - x^2 + 2C$.$\ln(y^2) = \ln(x^2) - x^2 + K$ (where $K = 2C$).$\ln(y^2) - \ln(x^2) = -x^2 + K$.$\ln(\frac{y^2}{x^2}) = -x^2 + K$.Taking the exponential of both sides: $\frac{y^2}{x^2} = e^{-x^2 + K} = e^K e^{-x^2}$.Let $e^K = \frac{1}{c}$,then $y^2 = \frac{x^2}{c} e^{-x^2}$,which implies $c y^2 = x^2 e^{-x^2}$ or $y^2 e^{x^2} = c x^2$ (adjusting the constant $c$ appropriately).Thus,the correct option is $(c)$.

The solution of the differential equation $ydx - xdy = x^2 ydx$ is

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