MCQ
If ${x^2} + {y^2} = 1$ then $\left( {y' = \frac{{dy}}{{dx}},y'' = \frac{{{d^2}y}}{{d{x^2}}}} \right)$
- A$yy'' - 2{(y')^2} + 1 = 0$
- ✓$yy'' + {(y')^2} + 1 = 0$
- C$yy'' - {(y')^2} - 1 = 0$
- D$yy'' + 2{(y')^2} + 1 = 0$
✓
Answer
Correct option: B.
$yy'' + {(y')^2} + 1 = 0$
b
(b) Differentiating $w.r.t$. $x,$ $2x + 2yy' = 0$ or $x + yy' = 0$
(b) Differentiating $w.r.t$. $x,$ $2x + 2yy' = 0$ or $x + yy' = 0$
Differentiating again $w.r.t$. $x,$ $1 + {y'^2} + yy'' = 0$.
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