Question

The integrating factor of the differential equation \((1 + x^{2})dy = (1 - 2xy)dx\) is

• A. \(x^{2} + 1\)
• B. \(\log_{e}(x^{2} + 1)\)
• C. \(\frac{x}{x^{2} + 1}\)
• D. \(x(x^{2} + 1)\)
• E. \(\log_{e}|x|\)

Hint:

For linear DE \(\frac{dy}{dx} + P(x)y = Q(x)\), IF = \(e^{\int P(x)dx}\).

Answer 1

The Correct Option is A

Step 1: Concept:
• Rewrite the equation in standard linear form: \[ \frac{dy}{dx} + P(x)y = Q(x) \]

Step 2: Detailed Explanation:
• Given: \[ (1+x^2)dy = (1-2xy)dx \]
• Divide by \(1+x^2\): \[ (1+x^2)\frac{dy}{dx} = 1 - 2xy \]
• Rewrite: \[ \frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{1}{1+x^2} \]
• Identify: \[ P(x) = \frac{2x}{1+x^2} \]
• Integrating Factor (IF): \[ IF = e^{\int P(x)\,dx} = e^{\int \frac{2x}{1+x^2}\,dx} \]
• Compute: \[ = e^{\ln(1+x^2)} = 1 + x^2 \]

Step 3: Final Answer:
• The integrating factor is: \[ x^2 + 1 \]