Differential Equations Question 133
Question: If $ x^{2}+y^{2}=1 $ then $ ( {y}’=\frac{dy}{dx},{y}’’=\frac{d^{2}y}{dx^{2}} ) $
[IIT Screening 2000]
Options:
A) $ y{y}’’-2{{({y}’)}^{2}}+1=0 $
B) $ y{y}’’+{{({y}’)}^{2}}+1=0 $
C) $ y{y}’’-{{({y}’)}^{2}}-1=0 $
D) $ y{y}’’+2{{({y}’)}^{2}}+1=0 $
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Answer:
Correct Answer: B
Solution:
Differentiating w.r.t. $ x, $
$ \frac{dy}{dx}=1+x+y+xy $ or $ x+y{y}’=0 $
Differentiating again w.r.t. $ x, $
$ 1+{{{y}’}^{2}}+y{y}’’=0 $ .
BETA
