Question

The integrating factor of the differential equation $(1+x^2)\frac{dy}{dx}+y=e^{{\tan }^{-1}x}$ is:

• A. ${\tan }^{-1}x$
• B. $1+x^2$
• C. $e^{{\tan }^{-1}x}$
• D. ${\log }_e(1+x^2)$

Answer 1

The correct option is C $e^{{\tan }^{-1}x}$

Given equation is
$(1+x^2)\frac{dy}{dx}+y=e^{{\tan }^{-1}x}$
or it can be written as
$\frac{dy}{dx}+\frac{1}{(1+x^2)}y=\frac{e^{{\tan }^{-1}x}}{1+x^2}$

It is a linear differential equation of the form of
$\frac{dy}{dx}+Py=Q$
where,

$P=\frac{1}{1+x^2}$ and $Q=\frac{e^{{\tan }^{-1}x}}{1+x^2}$$\therefore$ Integrating factor,

$IF=e^{\int Pdx}$
$=e^{\int \frac{1}{1+x^2}dx}=e^{{\tan }^{-1}x}$