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If x 2 + y 2 =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
x^2 + y^2 = 1, Differentiate w.r.t. x, 2x + 2yy' = 0, Again differentiate, 2 + 2( yy'' +y'^2) = 0, or, yy'' + y'^2 + 1 = 0, ANS is (B).
x^2 + y^2 = 1,
Differentiate w.r.t. x,
2x + 2yy' = 0,
Again differentiate,
2 + 2( yy'' +y'^2) = 0,
or, yy'' + y'^2 + 1 = 0,
ANS is (B).
differnciating x^2 + y^2 w.r.t. x we get 2x + 2yy' = 0 or x + yy' = 0 again differenciating w.r.t x we get 1+y'y' + yy'' = 0 or 1 + (y')^2 + yy'' = 0 so answer is (B)
differnciating x^2 + y^2 w.r.t. x we get
2x + 2yy' = 0 or x + yy' = 0
again differenciating w.r.t x we get
1+y'y' + yy'' = 0 or 1 + (y')^2 + yy'' = 0
so answer is (B)
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