Differential Equations Question 314
Question: The solution of the differential equation $ \frac{dy}{dx}=\frac{1+y^{2}}{1+x^{2}} $ is
[SCRA 1986]
Options:
A) $ 1+xy+c(y+x)=0 $
B) $ x+y=c(1-xy) $
C) $ y-x=c(1+xy) $
D) $ 1+xy=c(x+y) $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{dy}{dx}=\frac{1+y^{2}}{1+x^{2}}\Rightarrow \frac{1}{1+y^{2}}dy=\frac{1}{1+x^{2}}dx $
Now on integrating both sides, we get $ {{\tan }^{-1}}y={{\tan }^{-1}}x+{{\tan }^{-1}}c $
Therefore $ {{\tan }^{-1}}y={{\tan }^{-1}}( \frac{x+c}{1-cx} ) $
Therefore $ y=\frac{x+c}{1-cx} $
Therefore $ y-x=c(1+xy) $ .
BETA
