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

Step 1: Understanding the Concept:
A first-order linear differential equation is a specific class of ordinary differential equations that can be expressed in the form \( \frac{dy}{dx} + P(x)y = Q(x) \).
The primary challenge in solving these equations lies in the fact that the left-hand side is not initially a direct derivative of a product of functions.
To overcome this, we employ a mathematical tool known as the Integrating Factor (I.F.).
The concept of the Integrating Factor was historically developed by mathematicians like Leibniz and Euler to transform an inexact differential equation into an exact one.
When the entire differential equation is multiplied by the I.F., denoted as \( \mu(x) \), the left-hand side becomes the derivative of the product \( \mu(x) \cdot y \) according to the product rule of differentiation.
This transformation allows us to solve the equation by simple integration of both sides with respect to \( x \).
Step 2: Key Formula or Approach:
For a linear differential equation written in the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \), the formula to calculate the Integrating Factor is:
\[ \text{I.F.} = e^{\int P(x) dx} \]
This formula is derived by requiring that \( \frac{d}{dx}(\mu y) = \mu \frac{dy}{dx} + \mu P(x) y \), which leads to the separable equation \( \frac{d\mu}{dx} = \mu P(x) \).
Step 3: Detailed Explanation:
The given differential equation is provided as:
\[ \frac{dy}{dx} + \left( \frac{2x}{1 + x^2} \right) y = \frac{1}{(1 + x^2)^2} \]
First, we compare this given equation to the standard linear layout \( \frac{dy}{dx} + P(x)y = Q(x) \).
By direct comparison, we identify the coefficient function of \( y \), which is \( P(x) \):
\[ P(x) = \frac{2x}{1 + x^2} \]
Now, we must integrate this function with respect to \( x \) to find the exponent of our Integrating Factor:
\[ \int P(x) dx = \int \frac{2x}{1 + x^2} dx \]
To evaluate this integral, we use the method of substitution (u-substitution).
Let \( u = 1 + x^2 \).
Differentiating both sides with respect to \( x \), we obtain:
\[ \frac{du}{dx} = 2x \implies du = 2x dx \]
Substituting these into the integral, we get a standard form:
\[ \int \frac{1}{u} du = \ln|u| \]
Substituting back the value of \( u \), we have:
\[ \int P(x) dx = \ln(1 + x^2) \]
(Note: We do not need to add the constant of integration \( C \) when finding the I.F., as it would eventually cancel out in the final solution).
Finally, we substitute this result into the I.F. formula:
\[ \text{I.F.} = e^{\ln(1 + x^2)} \]
According to the fundamental identity of logarithms where \( e^{\ln(f(x))} = f(x) \), the expression simplifies perfectly:
\[ \text{I.F.} = 1 + x^2 \]
Step 4: Final Answer:
The Integrating Factor (I.F.) for the given linear differential equation is \( 1 + x^2 \).
This result matches Option (A).