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Question

Mathematics Question on Differentiability

If x2+y2=1 x^2 + y^2 = 1, then

A

yy"−2(y′)2+1=0yy" - 2 (y' )^2 + 1 = 0

B

yy"+(y′)2+1=0yy" + (y' )^2 + 1 = 0

C

yy"+(y′)2−1=0yy" + (y' )^2 - 1 = 0

D

yy"+2(y′)2+1=0yy" + 2 (y' )^2 + 1 = 0

Answer

yy"+(y′)2+1=0yy" + (y' )^2 + 1 = 0

Explanation

Solution

The correct answer is:(y′)2+yy′′+1=0(y')^2+yy''+1=0
Given that;
x2+y2=1x^2+y^2=1
Now differentiate w.r.t ‘x’
∴2x+2ydydx=0\therefore 2x+2y\frac{dy}{dx}=0
again differentiate w.r.t ‘x’
∴2+2(yd2ydx2+(dydx)2)=0\therefore 2+2(y\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2)=0
⇒1+yf′′(x)+(y′)2=01+yf''(x)+(y')^2=0
⇒(y′)2+yy′′+1=0(y')^2+yy''+1=0
differentiate