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
\[\begin{array}{l}
\therefore \dfrac{d}{{dx}}\left( {{x^2} + {y^2}} \right) = \dfrac{d}{{dx}} \times 1\\
 \Rightarrow \dfrac{d}{{dx}} \times {x^2} + \dfrac{d}{{dx}} \times {y^2} = 0\\
 \Rightarrow 2x + 2y \times \dfrac{{dy}}{{dx}} = 0\\
 \Rightarrow 2\left( {x + y \times \dfrac{{dy}}{{dx}}} \right) = 0\\
 \Rightarrow x + y \times \dfrac{{dy}}{{dx}} = 0
\end{array}\]
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.