Question

I: lf $y=\cos \left(m{\sin }^{-1}x\right)$ then $(1-x^2)y_2-x{y}_1+m^{2}y=0$

II: lf $\log y={\tan }^{-1}x$ then $(1+x^2)\frac{d^{2}y}{d{x}^2}+(2x-1)\frac{dy}{dx}=0$

Which of the following is correct?

• A. only I is true
• B. only II is true
• C. both I and II are true
• D. neither I nor II true

Answer 1

The correct option is C both I and II are true

$i:y^{{}^{\prime }}=-\sin \left(m{\sin }^{-1}x\right)\frac{m}{\sqrt{1-x^2}}$

$y^{{}^{\prime \prime }}=-cos\left(m{\sin }^{-1}x\right)\frac{m^2}{1-x^2}-\sin \left(m{\sin }^{-1}x\right)\frac{mx}{(1-x^2{)}^{3/2}}$

$y^{{}^{\prime \prime }}=-\frac{y{m}^2}{1-x^2}+\frac{x{y}^{{}^{\prime }}}{1-x^2}$

$(1-x^2)y^{{}^{\prime \prime }}+y{m}^2-x{y}^{{}^{\prime }}=0$

$ii:\frac{y^{{}^{\prime }}}{y}=\frac{1}{1+x^2}$

$y^{{}^{\prime \prime }}=\frac{y^{{}^{\prime }}}{1+x^2}-\frac{y2x}{(1+x^2{)}^2}$

$(1+x^2)y^{{}^{\prime \prime }}=y^{{}^{\prime }}-2x{y}^{{}^{\prime }}$

$(1+x^2)y^{{}^{\prime \prime }}0+(2x-1)y^{{}^{\prime }}=0$