Discussion Forum

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

Answer:

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