The general solution of the differential equation \(\displaystyle\frac{dy}{dx} = \frac{y^2}{1 - xy}\), \(y \ne 0\), is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(xy = \log |y| + C\)
The equation is not linear in y, but taking the reciprocal makes it linear in x:
\( \displaystyle \frac{dx}{dy} = \frac{1 - xy}{y^2} = \frac{1}{y^2} - \frac{x}{y} \)
\( \displaystyle \frac{dx}{dy} + \frac{1}{y}\,x = \frac{1}{y^2} \)
The integrating factor is
\( e^{\int \frac{1}{y}\,dy} = e^{\log |y|} = |y| \)
and y itself may be used, since a constant sign does not matter. So
\( \displaystyle xy = \int \frac{1}{y^2} \cdot y\,dy = \int \frac{1}{y}\,dy \)
\( xy = \log |y| + C \)