If x^2+y^2=1,then | vedclass

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Question

(B) Given the equation $x^2+y^2=1$. On differentiating both sides with respect to $x$,we get: $2x + 2y y^{\prime} = 0$ Dividing by $2$,we have $x + y

Vedclass · Accepted Answer

$y y^{\prime \prime}+\left(y^{\prime}\right)^2+1=0$

Vedclass · Answer

(B) Given the equation $x^2+y^2=1$. On differentiating both sides with respect to $x$,we get: $2x + 2y y^{\prime} = 0$ Dividing by $2$,we have $x + y y^{\prime} = 0$. Now,differentiating again with respect to $x$ using the product rule on $y y^{\prime}$: $\frac{d}{dx}(x) + \frac{d}{dx}(y y^{\prime}) = 0$ $1 + (y y^{\prime \prime} + (y^{\prime}) \cdot y^{\prime}) = 0$ $1 + y y^{\prime \prime} + (y^{\prime})^2 = 0$ Thus,the correct differential equation is $y y^{\prime \prime} + (y^{\prime})^2 + 1 = 0$.

If $x^2+y^2=1$,then

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