Differential Equations Question 6
Question: Solution of the equation $ (1-x^{2})dy+xydx=xy^{2}dx $ is
[DSSE 1989]
Options:
A) $ {{(y-1)}^{2}}(1-x^{2})=0 $
B) $ {{(y-1)}^{2}}{{(1-x)}^{2}}=c^{2}y^{2} $
C) $ {{(y-1)}^{2}}(1+x^{2})=c^{2}y^{2} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ (1-x^{2})dy+xydx=xy^{2}dx $
Therefore $ (1-x^{2})dy=xy(y-1)dx $
Therefore $ \frac{1}{y(y-1)}dy=\frac{x}{(1-x^{2})}dx $
Now on integrating both sides, we get $ \log (y-1)-\log y=-\frac{1}{2}\log (1-x^{2})+\log c $
or $ 2\log (y-1)+\log (1-x^{2})=\log y^{2}c^{2} $
Hence the solution is $ {{(y-1)}^{2}}(1-x^{2})=c^{2}y^{2} $ .
BETA
