Browse Questions

# The integrating factor of the differential equation $\Large \frac{dy}{dx}\normalsize+y=\Large \frac{1+y}{x}$ is

$(A)\;\frac{x}{e^x} \quad (B)\;\frac{e^x}{x} \quad (C)\;xe^x \quad (D)\;e^x$

Toolbox:
• If the linear differential equation is of the form $\large\frac{dy}{dx}$$+Py=Q, then its general solution is given by ye^{\int pdx}=\int Q. e ^{\int pdx}.dx+c • Here e ^{\int pdx} is the integrating factor (I.F) Given \large\frac{dy}{dx}$$+y=\large\frac{1+y}{x}$
This can be written as
$\large\frac{dy}{dx}$$+y=\large\frac{y}{x}+\frac{1}{x} =>\large\frac{dy}{dx}$$+y\bigg(1-\large\frac{1}{x}\bigg)$$=\large\frac{1}{x} =>\large\frac{dy}{dx}$$+\bigg(\large\frac{1-x}{x}\bigg)$$y=\large\frac{1}{x} Clearly this is a linear differential equation of the form \large\frac{dy}{dx}$$+Py=Q$
Where $P=\large\frac{1-x}{x}$ and $Q=\large\frac{1}{x}$
The integrating factor is $e^{\int pdx}$