Question

The integrating factor of $\frac{dy}{dx}+y=\frac{1+y}{x}$ is

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

Answer 1

Answer: (C)

Given $\frac{dy}{dx}+y=\frac{1+y}{x}$
$\Rightarrow \frac{dy}{dx}+y=\frac{1}{x}+\frac{y}{x}$
$\Rightarrow \frac{dy}{dx}+\left(1-\frac{1}{x}\right)y=\frac{1}{x}$
$\therefore IF-e^{\int \left(1-\frac{1}{x}\right)dx}=e^{x-\log x}=\frac{e^x}{x}$