Question:
The solution curve of the differential equation, \((1 + e^{-x})(1 + y^2)\frac{dy}{dx} = y^2\), which passes through the point \((0, 1)\), is :
The solution curve of the differential equation, \((1 + e^{-x})(1 + y^2)\frac{dy}{dx} = y^2\), which passes through the point \((0, 1)\), is :
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Separate variables: \(\frac{1 + y^2}{y^2}dy = \frac{e^x}{1 + e^x}dx\), integrate, and use the point \((0, 1)\) to fix the constant.
Updated On: Nov 5, 2026
- \(y^2 + 1 = y\left(\log_e\left(\frac{1 + e^{-x}}{2}\right) + 2\right)\)
- \(y^2 + 1 = y\left(\log_e\left(\frac{1 + e^{x}}{2}\right) + 2\right)\)
- \(y^2 = 1 + y\log_e\left(\frac{1 + e^{-x}}{2}\right)\)
- \(y^2 = 1 + y\log_e\left(\frac{1 + e^{x}}{2}\right)\)
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The Correct Option is D
Solution and Explanation
\(\frac{1 + y^2}{y^2}\,dy = \frac{dx}{1 + e^{-x}} = \frac{e^x}{1 + e^x}\,dx\).
Integrating: \(\int \left(y^{-2} + 1\right)dy = \int \frac{e^x}{1 + e^x}dx\), so \(-\frac{1}{y} + y = \log_e(1 + e^x) + C\).
At \((0, 1)\): \(-1 + 1 = \log_e 2 + C\), so \(C = -\log_e 2\).
\(y - \frac{1}{y} = \log_e\left(\frac{1 + e^x}{2}\right)\).
Multiplying by y: \(y^2 - 1 = y\log_e\left(\frac{1 + e^x}{2}\right)\), i.e. \(y^2 = 1 + y\log_e\left(\frac{1 + e^x}{2}\right)\).
Answer: (4) \(y^2 = 1 + y\log_e\left(\frac{1 + e^{x}}{2}\right)\)
Integrating: \(\int \left(y^{-2} + 1\right)dy = \int \frac{e^x}{1 + e^x}dx\), so \(-\frac{1}{y} + y = \log_e(1 + e^x) + C\).
At \((0, 1)\): \(-1 + 1 = \log_e 2 + C\), so \(C = -\log_e 2\).
\(y - \frac{1}{y} = \log_e\left(\frac{1 + e^x}{2}\right)\).
Multiplying by y: \(y^2 - 1 = y\log_e\left(\frac{1 + e^x}{2}\right)\), i.e. \(y^2 = 1 + y\log_e\left(\frac{1 + e^x}{2}\right)\).
Answer: (4) \(y^2 = 1 + y\log_e\left(\frac{1 + e^{x}}{2}\right)\)
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