Question:
The general solution of the differential equation \((1 + y)dx - (1 - x)dy = 0\) is
The general solution of the differential equation \((1 + y)dx - (1 - x)dy = 0\) is
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For \((a+y)dx = (b-x)dy\), separate to \(\frac{dy}{a+y} = \frac{dx}{b-x}\).
Updated On: Apr 24, 2026
- \(x^{2} + y^{2} + x - y = C\)
- \(x + y - xy = C\)
- \(x - y + xy = C\)
- \(x - y - xy = C\)
- \(x^{2} - y^{2} + x + y = C\)
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The Correct Option is C
Solution and Explanation
Step 1: Concept:
• Separate the variables.
Step 2: Detailed Explanation:
• Given: \[ (1+y)dx = (1-x)dy \]
• Rearrange: \[ \frac{dy}{1+y} = \frac{dx}{1-x} \]
• Integrate both sides: \[ \int \frac{dy}{1+y} = \int \frac{dx}{1-x} \]
• This gives: \[ \ln|1+y| = -\ln|1-x| + \ln C \]
• Simplify: \[ \ln|1+y| = \ln\left|\frac{C}{1-x}\right| \]
• Remove logarithm: \[ 1+y = \frac{C}{1-x} \]
• Multiply: \[ (1+y)(1-x) = C \]
• Expand: \[ 1 - x + y - xy = C \]
• Rearranging: \[ x - y + xy = 1 - C \]
• Let \(C_1 = 1 - C\), then: \[ x - y + xy = C_1 \]
Step 3: Final Answer:
• General solution: \[ x - y + xy = C \]
• Separate the variables.
Step 2: Detailed Explanation:
• Given: \[ (1+y)dx = (1-x)dy \]
• Rearrange: \[ \frac{dy}{1+y} = \frac{dx}{1-x} \]
• Integrate both sides: \[ \int \frac{dy}{1+y} = \int \frac{dx}{1-x} \]
• This gives: \[ \ln|1+y| = -\ln|1-x| + \ln C \]
• Simplify: \[ \ln|1+y| = \ln\left|\frac{C}{1-x}\right| \]
• Remove logarithm: \[ 1+y = \frac{C}{1-x} \]
• Multiply: \[ (1+y)(1-x) = C \]
• Expand: \[ 1 - x + y - xy = C \]
• Rearranging: \[ x - y + xy = 1 - C \]
• Let \(C_1 = 1 - C\), then: \[ x - y + xy = C_1 \]
Step 3: Final Answer:
• General solution: \[ x - y + xy = C \]
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