The general solution of {y^2},dx + ({x^2} - x | vedclass

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(A) Given the differential equation: ${y^2}\,dx + ({x^2} - xy + {y^2})\,dy = 0$Rearranging the terms,we get: $\frac{{dx}}{{dy}} = -\frac{{{x^2} - xy +

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${	an ^{ - 1}}\left( {\frac{x}{y}} ight) + \log y + c = 0$

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(A) Given the differential equation: ${y^2}\,dx + ({x^2} - xy + {y^2})\,dy = 0$Rearranging the terms,we get: $\frac{{dx}}{{dy}} = -\frac{{{x^2} - xy + {y^2}}}{{{y^2}}}$This is a homogeneous differential equation. Let $x = vy$,then $\frac{{dx}}{{dy}} = v + y\frac{{dv}}{{dy}}$.Substituting these into the equation: $v + y\frac{{dv}}{{dy}} = -\left( {\frac{{{v^2}{y^2} - vy^2 + {y^2}}}{{{y^2}}}} ight)$$v + y\frac{{dv}}{{dy}} = -({v^2} - v + 1)$$y\frac{{dv}}{{dy}} = -{v^2} + v - 1 - v = -({v^2} + 1)$Separating the variables: $\frac{{dv}}{{{v^2} + 1}} = -\frac{{dy}}{y}$Integrating both sides: $\int \frac{{dv}}{{{v^2} + 1}} = -\int \frac{{dy}}{y}$${	an ^{ - 1}}(v) = -\log |y| + C$Substituting $v = \frac{x}{y}$ back: ${	an ^{ - 1}}\left( {\frac{x}{y}} ight) + \log |y| = C$Thus,the general solution is ${	an ^{ - 1}}\left( {\frac{x}{y}} ight) + \log y + c = 0$.

The general solution of ${y^2}\,dx + ({x^2} - xy + {y^2})\,dy = 0$ is

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