1) The integrating factor of $\frac{xdy}{dx}-y = x^{4}-3x $ is A) x B) log x C) 1/x D) -x Answer: Option CExplanation:Since $\frac{xdy}{dx}-y = x^{4}-3x $ .'. $\frac{dy}{dx}-\frac{y}{x} = x^{3}-3 $ Hence IF = $e^{\int_{}^{}pdx }$ = $e^{-\int_{}^{}\frac{1}{x}dx }$ = $e^{-\log_{}{x} }$ = 1/x
