Question

The integrating factor for the differential equation $x~log_{e}x~dy = (2~log_{e}x - y) dx$ is

• A. $log_{e}x$
• B. $e^{-log_{e}x}$
• C. x
• D. $\frac{1}{log_{e}x}$

Hint:

When calculating an I.F. involving logs and exponentials, remember that $e^{\ln(f(x))} = f(x)$. If you end up with $e^{-\ln(f(x))}$, it becomes $1/f(x)$. Getting comfortable with these logarithmic identities will save you a lot of time.

Answer 1

The Correct Option is A

We need to find the integrating factor by first rearranging the differential equation into the standard linear form: $\frac{dy}{dx} + P(x)y = Q(x)$.

Step 1: \color{redRearrange to Linear Form
$x \ln x \frac{dy}{dx} = 2 \ln x - y$
$x \ln x \frac{dy}{dx} + y = 2 \ln x$.
Divide throughout by $x \ln x$:
$\frac{dy}{dx} + \frac{1}{x \ln x}y = \frac{2}{x}$.
Now the equation is in the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \frac{1}{x \ln x}$.

Step 2: \color{redSet up the Integrating Factor Formula
The integrating factor (I.F.) is given by $e^{\int P(x) dx}$.
$I.F. = e^{\int \frac{1}{x \ln x} dx}$.

Step 3: \color{redEvaluate the Integral
To solve $\int \frac{1}{x \ln x} dx$, let $u = \ln x$. Then $du = \frac{1}{x} dx$.
The integral becomes $\int \frac{1}{u} du = \ln|u|$.
Substitute back $u = \ln x$:
$\int \frac{1}{x \ln x} dx = \ln(\ln x)$.

Step 4: \color{redCalculate the Final I.F.
$I.F. = e^{\ln(\ln x)}$.
Using the property $e^{\ln w} = w$:
$I.F. = \ln x$.
The integrating factor is $\ln x$, which is Option (1).