The integrating factor of the differential equation, $x^2\frac{dy}{dx}+xy = \log_ex$ is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $x$
The given equation is
$x^{2}\frac{dy}{dx}+xy=\log x$
Divide by $x^{2}$:
$\frac{dy}{dx}+\frac{1}{x}y=\frac{\log x}{x^{2}}$
This is linear of the form
$\frac{dy}{dx}+P(x)y=Q(x)$ where $P(x)=\frac{1}{x}$
Integrating factor is
$IF=e^{\int \frac{1}{x}dx}$
$=e^{\log x}$
$=x$
Integrating factor is $x$.