If the general solution of frac{dy}{dx} = fra | vedclass

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(D) Given the differential equation $\frac{dy}{dx} = \frac{y^2}{xy - y^2 - x^2}$.Substitute $y = mx$,then $\frac{dy}{dx} = m + x \frac{dm}{dx}$.$m + x

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$3$

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(D) Given the differential equation $\frac{dy}{dx} = \frac{y^2}{xy - y^2 - x^2}$.Substitute $y = mx$,then $\frac{dy}{dx} = m + x \frac{dm}{dx}$.$m + x \frac{dm}{dx} = \frac{m^2 x^2}{x(mx) - m^2 x^2 - x^2} = \frac{m^2}{m - m^2 - 1}$.$x \frac{dm}{dx} = \frac{m^2}{m - m^2 - 1} - m = \frac{m^2 - m^2 + m^3 + m}{m - m^2 - 1} = \frac{m^3 + m}{m - m^2 - 1}$.Separating variables: $\int \frac{m - m^2 - 1}{m^3 + m} dm = \int \frac{dx}{x}$.$\int \left( \frac{m}{m(m^2 + 1)} - \frac{m^2 + 1}{m(m^2 + 1)} ight) dm = \ln|x| + C$.$\int \left( \frac{1}{m^2 + 1} - \frac{1}{m} ight) dm = \ln|x| + C$.$	an^{-1}(m) - \ln|m| = \ln|x| + C$.Substitute $m = \frac{y}{x}$: $	an^{-1}\left(\frac{y}{x}ight) - \ln\left|\frac{y}{x}ight| = \ln|x| + C$.$	an^{-1}\left(\frac{y}{x}ight) - (\ln|y| - \ln|x|) = \ln|x| + C$.$	an^{-1}\left(\frac{y}{x}ight) = \ln|y| + C$.Comparing with $	an^{-1}\left(\frac{y}{x}ight) = f(y) + C$,we get $f(y) = \ln|y|$.Therefore,$f(e^3) = \ln|e^3| = 3$.

If the general solution of $\frac{dy}{dx} = \frac{y^2}{xy - y^2 - x^2}$ is $\tan^{-1}\left(\frac{y}{x}\right) = f(y) + C$,then $f(e^3) = $

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