Differential Equations Question 266
Question: What is the differential equation for $ y^{2}=4a(x-a) $ -
Options:
A) $ yy’-2xyy’+y^{2}=0 $
B) $ yy’(yy’+2x)+y^{2}=0 $
C) $ yy’(yy’-2x)+y^{2}=0 $
D) $ yy’-2xyy’+y=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Given curve is $ y^{2}=4a(x-a) $
- (i) On differentiating w.r.t. x we get $ 2yy’=4a $
$ \Rightarrow a=\frac{yy’}{2} $ O putting the value of a in Eq. (i) we get $ y^{2}=4( \frac{yy’}{2} )( x-\frac{yy’}{2} )=yy’(2x-yy’) $
$ \Rightarrow yy’(yy’-2x)+y^{2}=0 $
BETA
