Question

The integrating factor of $\frac{xdy}{dx}-y=x^4-3x$ is

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

Answer 1

Answer: (C)

Since $x\frac{dy}{dx}-y=x^4-3x$
$\therefore \frac{dy}{dx}-\frac{y}{x}=x^3-3$
Hence
$IF=e^{\int Pdx}=e^{-\int \frac{1}{x}dx}=e^{-logx}=\frac{1}{x}$