Question

The integrating factor of the differential equation \(\frac{dy}{dx} + \frac{2y}{20 - x} = 10\) is

• A. \(\frac{1}{(20 - x)^2}\)
• B. \(\frac{1}{20 - x}\)
• C. \(20 - x\)
• D. \(\log_e |20 - x|\)

Hint:

Always be careful with the negative sign when integrating terms like \(1/(a-x)\). A missing negative sign in the exponent is the most frequent error in I.F. calculations.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A first-order linear differential equation in standard form is \(\frac{dy}{dx} + P(x)y = Q(x)\).
The integrating factor (I.F.) is defined as \(e^{\int P(x) dx}\).

Step 2: Key Formula or Approach:
1. Identify \(P(x)\) from the given equation.
2. Calculate \(\int P(x) dx\).
3. Apply I.F. \(= e^{\int P(x) dx}\).

Step 3: Detailed Explanation:
The given equation is \(\frac{dy}{dx} + \frac{2}{20 - x}y = 10\).
Here, \(P(x) = \frac{2}{20 - x}\).
Calculate the integral of \(P(x)\):
\[ \int P(x) dx = \int \frac{2}{20 - x} dx \]
Using the substitution \(u = 20 - x \implies du = -dx\):
\[ = -2 \int \frac{1}{u} du = -2 \log_e |u| = -2 \log_e |20 - x| \]
Using logarithmic properties \(a \log b = \log b^a\):
\[ = \log_e (20 - x)^{-2} = \log_e \left( \frac{1}{(20 - x)^2} \right) \]
Now, calculate the Integrating Factor:
\[ \text{I.F.} = e^{\log_e \left( \frac{1}{(20 - x)^2} \right)} \]
Since \(e^{\log z} = z\):
\[ \text{I.F.} = \frac{1}{(20 - x)^2} \]

Step 4: Final Answer:
The integrating factor is \(\frac{1}{(20 - x)^2}\).