Question

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

• A. $\log_e x$
• B. x
• C. $\frac{1}{x}$
• D. $\frac{1}{\log_e x}$

Hint:

When finding the integrating factor, always ensure the differential equation is in the standard form with the coefficient of $\frac{dy}{dx}$ being 1. The integral in the exponent often requires a substitution, especially when logarithmic or trigonometric functions are involved.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The given differential equation is a linear first-order differential equation. To solve it, we first need to convert it to the standard form $\frac{dy}{dx} + P(x)y = Q(x)$ and then find the integrating factor (I.F.).
Step 2: Key Formula or Approach:
The standard form of a linear differential equation is $\frac{dy}{dx} + P(x)y = Q(x)$.
The integrating factor is given by the formula: I.F. = $e^{\int P(x)dx}$.
Step 3: Detailed Explanation:
The given differential equation is:
\[ (x \log_e x) \frac{dy}{dx} + y = 2\log_e x \] To convert this to the standard form, divide the entire equation by $(x \log_e x)$:
\[ \frac{dy}{dx} + \frac{1}{x \log_e x} y = \frac{2\log_e x}{x \log_e x} \] \[ \frac{dy}{dx} + \left(\frac{1}{x \log_e x}\right)y = \frac{2}{x} \] Comparing this with the standard form, we have:
\[ P(x) = \frac{1}{x \log_e x} \] Now, we calculate the integrating factor:
\[ \text{I.F.} = e^{\int P(x)dx} = e^{\int \frac{1}{x \log_e x} dx} \] To evaluate the integral $\int \frac{1}{x \log_e x} dx$, we use the substitution method.
Let $t = \log_e x$. Then, $dt = \frac{1}{x} dx$.
The integral becomes:
\[ \int \frac{1}{t} dt = \ln|t| = \ln(\log_e x) \] Now, substitute this back into the formula for the integrating factor:
\[ \text{I.F.} = e^{\ln(\log_e x)} \] Using the property $e^{\ln u} = u$, we get:
\[ \text{I.F.} = \log_e x \] Step 4: Final Answer:
The integrating factor of the given differential equation is $\log_e x$.