Question:
The curve satisfying the differential equation, \(y\,dx - (x + 3y^2)\,dy = 0\) and passing through the point \((1, 1)\), also passes through the point :
The curve satisfying the differential equation, \(y\,dx - (x + 3y^2)\,dy = 0\) and passing through the point \((1, 1)\), also passes through the point :
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Treat \(x\) as a function of \(y\): the equation becomes linear, \(\frac{dx}{dy} - \frac{x}{y} = 3y\). Solve and use \((1,1)\) to find the constant.
Updated On: Nov 11, 2026
- \(\left(\frac{1}{4}, -\frac{1}{2}\right)\)
- \(\left(-\frac{1}{3}, \frac{1}{3}\right)\)
- \(\left(\frac{1}{3}, -\frac{1}{3}\right)\)
- \(\left(\frac{1}{4}, \frac{1}{2}\right)\)
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The Correct Option is B
Solution and Explanation
\(\frac{dx}{dy} = \frac{x + 3y^2}{y}\), so \(\frac{dx}{dy} - \frac{1}{y}x = 3y\), linear in \(x\).
Integrating factor \(= e^{-\int dy/y} = \frac{1}{y}\).
\(\frac{d}{dy}\left(\frac{x}{y}\right) = 3\), so \(\frac{x}{y} = 3y + c\).
Through \((1, 1)\): \(1 = 3 + c\), \(c = -2\). So \(x = 3y^2 - 2y\).
Check options: \(\left(\frac{1}{4}, -\frac{1}{2}\right)\): \(3\cdot\frac{1}{4} + 1 = \frac{7}{4} \neq \frac{1}{4}\). \(\left(-\frac{1}{3}, \frac{1}{3}\right)\): \(\frac{1}{3} - \frac{2}{3} = -\frac{1}{3}\), yes. \(\left(\frac{1}{3}, -\frac{1}{3}\right)\): \(\frac{1}{3} + \frac{2}{3} = 1 \neq \frac{1}{3}\). \(\left(\frac{1}{4}, \frac{1}{2}\right)\): \(\frac{3}{4} - 1 = -\frac{1}{4} \neq \frac{1}{4}\).
Answer: (2) \(\left(-\frac{1}{3}, \frac{1}{3}\right)\)
Integrating factor \(= e^{-\int dy/y} = \frac{1}{y}\).
\(\frac{d}{dy}\left(\frac{x}{y}\right) = 3\), so \(\frac{x}{y} = 3y + c\).
Through \((1, 1)\): \(1 = 3 + c\), \(c = -2\). So \(x = 3y^2 - 2y\).
Check options: \(\left(\frac{1}{4}, -\frac{1}{2}\right)\): \(3\cdot\frac{1}{4} + 1 = \frac{7}{4} \neq \frac{1}{4}\). \(\left(-\frac{1}{3}, \frac{1}{3}\right)\): \(\frac{1}{3} - \frac{2}{3} = -\frac{1}{3}\), yes. \(\left(\frac{1}{3}, -\frac{1}{3}\right)\): \(\frac{1}{3} + \frac{2}{3} = 1 \neq \frac{1}{3}\). \(\left(\frac{1}{4}, \frac{1}{2}\right)\): \(\frac{3}{4} - 1 = -\frac{1}{4} \neq \frac{1}{4}\).
Answer: (2) \(\left(-\frac{1}{3}, \frac{1}{3}\right)\)
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