Question

Let $\cos ^{-1}\left(\frac{y}{b}\right)=\log \left(\frac{x}{n}\right)^{n}$. Then
( Here $y_{2} \equiv \frac{d^{2} y}{d x^{2}}, y_{1} \equiv \frac{d y}{d x}$ )

• A. $x^{2} y_{2}+x y_{1}+n^{2} y=0$
• B. $x y_{2}-x y_{1}+2 n^{2} y=0$
• C. $x^{2} y_{2}+3 x y_{1}-n^{2} y=0$
• D. $x y_{2}+5 x y_{1}-3 y=0$

Answer 1

Answer: (A)

$\frac{-1}{\sqrt{1-\frac{y^{2}}{b^{2}}}} \cdot \frac{1}{b} \cdot \frac{d y}{d x}=n \cdot \frac{n}{x} \cdot \frac{1}{n} $
$\frac{d y}{d x}\left(\frac{-1}{\sqrt{b^{2}-y^{2}}}\right)=\frac{n}{x} $
$\Rightarrow x y_{1}=-n \sqrt{b^{2}-y^{2}}$
Squaring both sides
$x^{2} y_{1}^{2}=n^{2} b^{2}-n^{2} y^{2}$
Differentiating both sides
$\Rightarrow 2 x y_{1}^{2}+x^{2} \cdot 2 y_{1} y_{2}=-n^{2} \cdot 2 y y_{1}$
$\Rightarrow x y_{1}+x^{2} y_{2}=-n^{2} y $
$\Rightarrow x^{2} y_{2}+x y_{1}+n^{2} y=0$