Question

For $y=\cos \left(m{\sin }^{-1}x\right)$ which of the following is true?

• A. $(1-x^2)y_2+x{y}_1-m^{2}y=0$
• B. $(1-x^2)y_2-x{y}_1-m^{2}y=0$
• C. $(1+x^2)y_2+x{y}_1-m^{2}y=0$
• D. $(1-x^2)y_1+x{y}_1+m^{2}y=0$

Answer 1

The correct option is B $(1-x^2)y_2-x{y}_1-m^{2}y=0$

Given : $y=\cos \left(m{\sin }^{-1}x\right)$

$⟹y_1=\frac{dy}{dx}=\cos \left(m{\sin }^{-1}x\right)\cdot \frac{m}{\sqrt{1-x^2}}$ .... $\left(i\right)$

$y_2=\frac{d^{2}y}{d{x}^2}=\frac{m}{\sqrt{1-x^2}}\cos \left(m{\sin }^{-1}x\right)\cdot \frac{m}{\sqrt{1-x^2}}+m\cos \left(m{\sin }^{-1}x\right)(-\frac{1}{2}\cdot \frac{-2x}{(1-x^2{)}^{\frac{3}{2}}})$

$=\frac{m^2}{(1-x^2)}\cos \left(m{\sin }^{-1}x\right)+\frac{mx}{(1-x^2)\sqrt{1-x^2}}\cos \left(m{\sin }^{-1}x\right)$

Now, consider

$(1-x^2)y_2-x{y}_1-m^{2}y$

$=(1-x^2)(\frac{m^2\cos \left(m{\sin }^{-1}x\right)}{(1-x^2)}+\frac{mx\cos \left(m{\sin }^{-1}x\right)}{(1-x^2)\sqrt{1-x^2}})-x\left(\frac{m\cos \left(m{\sin }^{-1}x\right)}{\sqrt{1-x^2}}\right)-m^2\cos \left(m{\sin }^{-1}x\right)$

$=(m^2\cos \left(m{\sin }^{-1}x\right)+\frac{mx\cos \left(m{\sin }^{-1}x\right)}{\sqrt{1-x^2}})-\frac{mx\cos \left(m{\sin }^{-1}x\right)}{\sqrt{1-x^2}}-m^2\cos \left(m{\sin }^{-1}x\right)$

$=0$