Question

If $y=\log \left(m{\cos }^{-1}x\right)$ is a solution of the differential equation $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=k{e}^{-2y}$ then $k=$

• A. $m^2$
• B. $2m^2$
• C. $-m^2$
• D. $-2m^2$

Answer 1

The correct option is C $-m^2$

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

Differentiating the given relation, $\frac{dy}{dx}=\frac{1}{m{\cos }^{-1}x}.\frac{-m}{\sqrt{1-x^2}}$ Square
$\therefore (1-x^2){\left(\frac{dy}{dx}\right)}^2=\frac{1}{{\left({\cos }^{-1}x\right)}^2}\cdots \left(1\right)$
Now $e^y=m{\cos }^{-1}x$

$\therefore \frac{1}{{\left({\cos }^{-1}x\right)}^2}=m^2{e}^{-2y}$
Putting in (1), $(1-x^2){\left(\frac{dy}{dx}\right)}^2=m^2{e}^{-2y}$
Differentiating again $(1-x^2).2\left(\frac{dy}{dx}\right)\frac{d^{2}y}{d{x}^2}-2x{\left(\frac{dy}{dx}\right)}^2=-2m^2{e}^{-2y}\frac{dy}{dx}$
Cancel $2\left(\frac{dy}{dx}\right)$ or $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=-m^2{e}^{-2y}=k{e}^{-2y}$

Hence $k=-m^2$