Differential Equations Question 165
Question: An integrating factor for the differential equation $ (1+y^{2})dx-({{\tan }^{-1}}y-x)dy=0 $
[MP PET 1993]
Options:
A) $ {{\tan }^{-1}}y $
B) $ {e^{{{\tan }^{-1}}y}} $
C) $ \frac{1}{1+y^{2}} $
D) $ \frac{1}{x(1+y^{2})} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ (1+y^{2})dx-({{\tan }^{-1}}y-x)dy=0 $
Therefore $ \frac{dy}{dx}=\frac{1+y^{2}}{{{\tan }^{-1}}y-x} $
Therefore $ \frac{dx}{dy}=\frac{{{\tan }^{-1}}y}{1+y^{2}}-\frac{x}{1+y^{2}} $
Therefore $ \frac{dx}{dy}+\frac{x}{1+y^{2}}=\frac{{{\tan }^{-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^{{{\tan }^{-1}}y}} $ .
BETA
