Question

If $m{\sin }^{-1}x={\log }_{e}y$, then $(1-x^2)y^{\prime \prime }-x{y}^{\prime }$

• A. $m^{2}y$
• B. $-m^{2}y$
• C. $2y$
• D. $-2y$

Answer 1

The correct option is A $m^{2}y$

$msi{n}^{-1}x={\log }_{e}y$

$y=e^{\left(msi{n}^{-1}x\right)}$

$y^{\prime }=e^{\left(msi{n}^{-1}x\right)}\times \frac{m}{\sqrt{1-x^2}}$

$y^{\prime \prime }=e^{\left(msi{n}^{-1}x\right)}\times \frac{-m}{2(1-x^2{)}^{\frac{3}{2}}}\times (-2x)+\frac{m}{\sqrt{1-x^2}}\times e^{\left(msi{n}^{-1}x\right)}\times \frac{m}{\sqrt{1-x^2}}$

$\Rightarrow y^{\prime \prime }=m{e}^{\left(msi{n}^{-1}x\right)}\times [\frac{x}{(1-x^2{)}^{\frac{3}{2}}}+\frac{m}{1-x^2}]$

$\Rightarrow (1-x^2)y^{\prime \prime }-x{y}^{\prime }=m{e}^{\left(msi{n}^{-1}x\right)}\times [\frac{x}{\sqrt{1-x^2}}+m-\frac{x}{\sqrt{1-x^2}}]$$=m^2{e}^{\left(msi{n}^{-1}x\right)}$$=m^{2}y$