Question

If $y=co{s}^{-1}x$ then $(1-x^2)y^{\prime \prime }-x{y}^{\prime }=?$ where $y^{\prime }=\frac{dy}{dx}$

• A. $1$
• B. $0$
• C. $2x$
• D. $2y$

Answer 1

The correct option is B $0$

$\begin{array}{c}y={\cos }^{-1}x\\ \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}\\ \left(1-x^2\right)y_1^2-1=0\\ y_1=\frac{dy}{dx},y_2=\frac{d^{2}y}{d{x}^2}\\ -2x{y}_1^2+\left(1-x^2\right)2y_1{y}_2=0\\ -x{y}_1+\left(1-x^2\right)y_2=0\\ \left(1-x^2\right)y_2-x{y}_1=0\\ \left(1-x^2\right)y^{\prime \prime }-x{y}^{\prime }=0\\ OptionBiscorrectanswer.\end{array}$