Question

Find the integrating factor (I.F.) for the linear differential equation: \[ \frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{1}{(1+x^2)^2} \]

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

Hint:

Whenever the numerator is the exact derivative of the denominator inside an exponent, the integrating factor simplifies directly to the denominator function itself: \( e^{\int \frac{f'(x)}{f(x)}\,dx} = e^{\ln|f(x)|} = f(x) \).

Answer 1

The Correct Option is A

Concept: A first-order linear differential equation written in the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \) can be solved by multiplying through by an Integrating Factor (I.F.), which is defined as: \[ \text{I.F.} = e^{\int P(x)\,dx} \]

Step 1: Identify the function \( P(x) \) from the given differential equation. By comparing our equation directly to the standard linear layout, we find the coefficient function of \( y \): \[ P(x) = \frac{2x}{1+x^2} \]

Step 2: Integrate \( P(x) \) with respect to \( x \). To evaluate \( \int \frac{2x}{1+x^2}\,dx \), we use integration by substitution. Let \( u = 1+x^2 \), which means its differential is \( du = 2x\,dx \): \[ \int \frac{2x}{1+x^2}\,dx = \int \frac{du}{u} = \ln|u| = \ln(1+x^2) \]

Step 3: Raise the integrated function as a power of base \( e \). Substitute the result back into the exponential format to finalize our integrating factor: \[ \text{I.F.} = e^{\ln(1+x^2)} \] Using the fundamental logarithmic identity \( e^{\ln(f(x))} = f(x) \), the expression simplifies perfectly: \[ \text{I.F.} = 1+x^2 \]