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(D) Given the differential equation: $ydx - xdy = xy\,dx$Divide both sides by $xy$ (assuming $x, y 
eq 0$):$\frac{ydx - xdy}{xy} = dx$We recognize th

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$y{e^x} = cx$

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(D) Given the differential equation: $ydx - xdy = xy\,dx$Divide both sides by $xy$ (assuming $x, y 
eq 0$):$\frac{ydx - xdy}{xy} = dx$We recognize that $\frac{ydx - xdy}{x^2} = d(\frac{y}{x})$,but here we have $xy$ in the denominator. Let us rearrange:$\frac{ydx - xdy}{xy} = dx$$\frac{1}{x}dx - \frac{1}{y}dy = dx$Integrating both sides:$\int \frac{1}{x}dx - \int \frac{1}{y}dy = \int dx$$\ln|x| - \ln|y| = x + C$$\ln|\frac{x}{y}| = x + C$Taking the exponential of both sides:$\frac{x}{y} = e^{x+C} = e^C \cdot e^x$Let $e^C = \frac{1}{c}$ (where $c$ is an arbitrary constant):$\frac{x}{y} = \frac{1}{c} e^x$$cx = y e^x$Thus,the general solution is $y e^x = cx$.

If $c$ is any arbitrary constant,then the general solution of the differential equation $ydx - xdy = xy\,dx$ is given by

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