Differential Equations Question 25
Question: The solution of the differential equation $ (1+x^{2})(1+y)dy+(1+x)(1+y^{2})dx=0 $ is
[DSSE 1986]
Options:
A) $ {{\tan }^{-1}}x+\log (1+x^{2})+{{\tan }^{-1}}y+\log (1+y^{2})=c $
B) $ {{\tan }^{-1}}x-\frac{1}{2}\log (1+x^{2})+{{\tan }^{-1}}y-\frac{1}{2}\log (1+y^{2})=c $
C) $ {{\tan }^{-1}}x+\frac{1}{2}\log (1+x^{2})+{{\tan }^{-1}}y+\frac{1}{2}\log (1+y^{2})=c $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Given equation $ (1+x^{2})(1+y)dy+(1+x)(1+y^{2})dx=0 $
Therefore $ \frac{(1+y)}{(1+y^{2})}dy=-\frac{(1+x)}{(1+x^{2})}dx $
Therefore $ \int _{{}}^{{}}{[ \frac{1}{1+y^{2}}+\frac{y}{1+y^{2}} ]}dy+\int _{{}}^{{}}{[ \frac{1}{1+x^{2}}+\frac{x}{1+x^{2}} ]dx+c}=0 $
Therefore $ {{\tan }^{-1}}y+\frac{1}{2}\log (1+y^{2})+{{\tan }^{-1}}x+\frac{1}{2}\log (1+x^{2})=c $ .
BETA
