Differential Equations Question 301
Question: If integrating factor of $ x(1-x^{2})dy+(2x^{2}y-y-ax^{3})dx=0 $ is $ {e^{\int _{{}}^{{}}{Pdx}}}, $ then P is equal to
[MP PET 1999]
Options:
A) $ \frac{2x^{2}-ax^{3}}{x(1-x^{2})} $
B) $ (2x^{2}-1) $
C) $ \frac{2x^{2}-1}{ax^{3}} $
D) $ \frac{(2x^{2}-1)}{x(1-x^{2})} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ x(1-x^{2})dy+(2x^{2}y-y-ax^{3})dx=0 $
$ \frac{dy}{dx}+\frac{(2x^{2}-1)}{x(1-x^{2})}y=\frac{ax^{2}}{(1-x^{2})} $ ,
$ \therefore P=\frac{2x^{2}-1}{x(1-x^{2})} $ .
BETA
