Question

If $\sqrt{y}={\cos }^{-1}x$, then it satisfies the differential equation $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=c$, where $c$ is equal to:

• A. $0$
• B. $3$
• C. $1$
• D. $2$

Answer 1

Answer: (D)

$2$

Given, $\sqrt{y}={\cos }^{-1}x\Rightarrow y={\left({\cos }^{-1}x\right)}^2$
On differentiating both sides with respect to $x$, we get
$\frac{dy}{dx}=2\left({\cos }^{-1}x\right)\times \frac{-1}{\sqrt{1-x^2}}$

Again, differentiating both sides with respect to $x$, we get
$\frac{d^{2}y}{d{x}^2}=-2\left[\frac{\sqrt{1-x^2}\times \frac{-1}{\sqrt{1-x^2}}-{\cos }^{-1}x\times \left(\frac{1}{2}\right)\frac{(-2x)}{{(1-x^2)}^{1/2}}}{{\left(\sqrt{1-x^2}\right)}^2}\right]$

$=-2\left[\frac{-1+\frac{x{\cos }^{-1}x}{{(1-x^2)}^{1/2}}}{(1-x^2)}\right]$

$\frac{d^{2}y}{d{x}^2}=\left[\frac{2-\frac{2x{\cos }^{-1}x}{{(1-x^2)}^{1/2}}}{(1-x^2)}\right]$

$\Rightarrow (1-x^2)\frac{d^{2}y}{d{x}^2}=2+x\frac{dy}{dx}$

$\Rightarrow (1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=2$

But, it is given $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=c$
$\therefore c=2$