Question

Let ${\cos }^{-1}\left(\frac{y}{b}\right)=\log {\left(\frac{x}{n}\right)}^n$ . Then

$(\text{Here}{y}_2\equiv \frac{d^{2}y}{d{x}^2},y_1\equiv \frac{dy}{dx})$

• A. $x^2{y}_2+x{y}_1+n^{2}y=0$
• B. $x{y}_2-x{y}_1+2n^{2}y=0$
• C. $x^2{y}_2+3x{y}_1-n^{2}y=0$
• D. $x{y}_2+5x{y}_1-3y=0$

Answer 1

The correct option is A $x^2{y}_2+x{y}_1+n^{2}y=0$

$\frac{-1}{\sqrt{1-\frac{y^2}{b^2}}}\cdot \frac{1}{b}\cdot \frac{dy}{dx}=n\cdot \frac{n}{x}\cdot \frac{1}{n}\frac{dy}{dx}\left(\frac{-1}{\sqrt{b^2-y^2}}\right)=\frac{n}{x}$
$\Rightarrow x{y}_1=-n\sqrt{b^2-y^2}$

Squaring both sides
$x^2{y}_1^2=n^2{b}^2-n^2{y}^2$

Differentiating both sides
$\Rightarrow 2x{y}_1^2+x^2.2y_1{y}_2=-n^2.2y{y}_1\Rightarrow x{y}_1+x^2{y}_2=-n^{2}y$
$\Rightarrow x^2{y}_2+x{y}_1+n^{2}y=0$