If $\cos ^{-1}\left(\frac{y}{b}\right)=2 \log \left(\frac{x}{2}\right)$, where $x0$, then $x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}$ is equal to

Question

If $\cos ^{-1}\left(\frac{y}{b}\right)=2 \log \left(\frac{x}{2}\right)$, where $x>0$, then $x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}$ is equal to

Aicharya · Accepted Answer

$ \cos ^{-1}\left(\frac{y}{b}ight)=2 \log \left(\frac{x}{2}ight), x>0 $ On differentiating w.r.t. $x$ we get $ \begin{array}{cc} & -\frac{1}{\sqrt{1-\frac{y^2}{b^2}}} \cdot \frac{1}{b} \frac{d y}{d x}=2 \cdot \frac{1}{\left(\frac{x}{2}ight)} \cdot \frac{1}{2} \ \Rightarrow \quad & -\frac{1}{\sqrt{b^2-y^2}} \cdot \frac{d y}{d x}=\frac{2}{x} \ \Rightarrow \quad & x \frac{d y}{d x}=-2 \sqrt{b^2-y^2} \end{array} $ Again, differentiating w.r.t. $x$ we get, $ \begin{aligned} & x \frac{d^2 y}{d x^2}+\frac{d y}{d x}=-2 \cdot \frac{1}{2}\left(b^2-y^2ight)^{-1 / 2}(-2 y) \frac{d y}{d x} \ \Rightarrow \quad & \quad x \frac{d^2 y}{d x^2}+\frac{d y}{d x}=\frac{2 y}{\sqrt{b^2-y^2}} \cdot \frac{d y}{d x} \ \Rightarrow \quad & x \frac{d^2 y}{d x^2}+\frac{d y}{d x}=\frac{2 y}{-\frac{x}{2} \cdot \frac{d y}{d x}} \cdot \frac{d y}{d x} \quad 	ext { [from Eq. (i)] } \ \Rightarrow \quad & x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}=-4 y \end{aligned} $

If $\cos ^{-1}\left(\frac{y}{b}\right)=2 \log \left(\frac{x}{2}\right)$, where $x>0$, then $x^2 \frac{d^2…

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