Differential Equations Question 284
Question: The differential equation of the curve $ \frac{x}{c-1}+\frac{y}{c+1}=1 $ is given by
Options:
A) $ ( \frac{dy}{dx}-1 )( y+x\frac{dy}{dx} )=2\frac{dy}{dx} $
B) $ ( \frac{dy}{dx}+1 )( y-x\frac{dy}{dx} )=\frac{dy}{dx} $
C) $ ( \frac{dy}{dx}+1 )( y-x\frac{dy}{dx} )=2\frac{dy}{dx} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \frac{x}{c-1}+\frac{y}{c+1}=1 $
-
(i)
$ \Rightarrow \frac{1}{c-1}+\frac{y’}{c+1}=0 $ -
(ii)
$ \Rightarrow \frac{y’}{1}=\frac{c+1}{1-c}\Rightarrow \frac{y’-1}{y’+1}=c $ Put value of c in (1)
$ \Rightarrow \frac{x}{\frac{y’-1}{y’+1}-1}+\frac{y}{\frac{y’-1}{y’+1}+1}=1 $
$ \Rightarrow \frac{x(y’)+1}{-2}+\frac{y(y’+1)}{2y’}=1 $
$ \Rightarrow \frac{(y’+1)}{2}( \frac{y}{y’}-x )=1 $
$ \Rightarrow ( 1+\frac{dy}{dx} )( y-x\frac{dy}{dx} )=2\frac{dy}{dx} $
BETA
