The curve satisfying the differential equation ydx − (x + 3y<sup>2</sup>)dy = 0 and passing through the point (1, 1) also passes through the point __________
Rewrite: \(\frac{dx}{dy} = \frac{x}{y} + 3y\), so \(\frac{dx}{dy} - \frac{x}{y} = 3y\). This is linear in x.
The integrating factor is \(\frac{1}{y}\): \(\frac{d}{dy}\left(\frac{x}{y}\right) = 3 \Rightarrow \frac{x}{y} = 3y + C\).
At (1, 1): \(1 = 3 + C \Rightarrow C = -2\). So the curve is \(x = 3y^2 - 2y\).
Check \(\left(-\frac{1}{3}, \frac{1}{3}\right)\): \(3 \cdot \frac{1}{9} - \frac{2}{3} = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}\). It lies on the curve.
Answer: (C)
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