Question

If ${x^2} + {y^2} = 1$ then
A) $yy'' - 2{\left( {y'} \right)^2} + 1 = 0$
B) $yy'' + {\left( {y'} \right)^2} + 1 = 0$
C) $yy'' - {\left( {y'} \right)^2} - 1 = 0$
D) $yy'' + 2{\left( {y'} \right)^2} + 1 = 0$

Answer 1

Hint: We will used chain rule of derivative to solve this question and also $y' = \dfrac{{dy}}{{dx}}\& y'' = \dfrac{{{d^2}y}}{{d{x^2}}}$ so do the derivative on both the sides at once.
Complete Step by Step Solution:
We are given that ${x^2} + {y^2} = 1$
Differentiating both the sides we will get it as

\therefore \dfrac{d}{{dx}}\left( {{x^2} + {y^2}} \right) = \dfrac{d}{{dx}} \times 1\\\ \Rightarrow \dfrac{d}{{dx}} \times {x^2} + \dfrac{d}{{dx}} \times {y^2} = 0\\\ \Rightarrow 2x + 2y \times \dfrac{{dy}}{{dx}} = 0\\\ \Rightarrow 2\left( {x + y \times \dfrac{{dy}}{{dx}}} \right) = 0\\\ \Rightarrow x + y \times \dfrac{{dy}}{{dx}} = 0 \end{array}$$ So this is what we are getting after doing the first derivative Let us differentiate both the sides one more time $$\begin{array}{l} \therefore \dfrac{{dx}}{{dx}} + \dfrac{d}{{dx}}\left( {y \times \dfrac{{dy}}{{dx}}} \right) = 0\\\ \Rightarrow 1 + y \times \dfrac{{d\dfrac{{dy}}{{dx}}}}{{dx}} + \dfrac{{dy}}{{dx}} \times \dfrac{{dy}}{{dx}} = 0\\\ \Rightarrow 1 + y \times \dfrac{{{d^2}y}}{{d{x^2}}} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = 0\\\ \Rightarrow y\left( {y''} \right) + {\left( {y'} \right)^2} + 1 = 0 \end{array}$$ Clearly option B is the correct option. Note: I have used $$\dfrac{d}{{dx}}(uv) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u$$ this formula while differentiating $$\dfrac{{dy}}{{dx}} \times y$$ by taking $$u = \dfrac{{dy}}{{dx}}\& v = y$$ as u and v are functions of x.