# Integrating factor of $x\Large \frac{dx}{dy}\normalsize-y=x^4-3x$ is $(A)\;x\quad(B)\;log x\quad(C)\;\frac{1}{x}\quad(D)\;-x$

Toolbox:
• A linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q has the general solution as y e ^{\int pdx}=\int Q.e^{\int pdx}.dx+c, where e^{\int pdx} is the integrating factor (I.F) Given x \large\frac{dy}{dx}$$-y=x^4-3x$
Divide throughout by x
$\large\frac{dy}{dx}-\frac{y}{x}$$=x^3-3 Clearly this is a linear differential equation of the form \large\frac{dy}{dx}$$+Py=Q$
Here $P=\large\frac{-1}{x}$
The Integrating factor $(I.F)=e^{\int pdx}$
$\int pdx=-\int \large \frac{1}{x} $$dx =-\log x \qquad (or)\qquad \log \large\frac{1}{x} Hence I.F=e^{\large[\log ^{\Large\frac{1}{x}}]} But e^{\large\log^{\Large\frac{1}{x}}}$$=\large\frac{1}{x}$
Hence $I.F= \large\frac{1}{x}$
The correct option is $C$