Question

An integrating factor for the differential equation
$1+x^2)dx-(ta{n}^{-1}y-x)dy=0$

[MP PET 1993]

• A. $ta{n}^{-1}y$
• B. $e^{ta{n}^{-1}}y$
• C. $\frac{1}{1+y^2}$
• D. $\frac{1}{x(1+y^2)}$

Answer 1

The correct option is A

$ta{n}^{-1}y$

$(1+y^2)dx-(ta{n}^{-1}y-x)dy=0\Rightarrow \frac{dy}{dx}=\frac{1+y^2}{ta{n}^{-1}y-x}\Rightarrow \frac{dx}{dy}=\frac{ta{n}^{-1}y}{1+y^2}-\frac{y}{1+y^2}\Rightarrow \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{ta{n}^{-1}y}{1+y^2}$
This is equation of the form $\frac{dx}{dy}+px=Q$
So, I.F. $=e^{\int pdy}=e^{\int \frac{1}{1+y^2}}dy=e^{ta{n}^1}y$