Discussion Forum



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 C

Explanation:

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