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in Differential equations by (36.3k points)

If x2 + y2 = 1, then 

(a) yy'' – 2(y')2 + 1 = 0 

(b) yy'' + (y')2 + 1 = 0 

(c) yy'' + (y')2 – 1 = 0 

(d) yy'' + 2(y')2 + 1 = 0.

1 Answer

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Best answer

Answer is (b) yy'' + (y')2 + 1 = 0

Explanation:

The given expression is x2 + y2 = 1

Differentiating w.r.t. x, we get 

2x + 2y(dy/dx) = 0

x + y(dy/dx) = 0

Again differentiating, w.r.t. x, we get 

1 + (dy/dx)2 + y(d2y/dx2) = 0

1 + y'2 + y.y'' = 0

y.y'' + (y')2 + 1 = 0

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