Question

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

• A. $0$
• B. $3$
• C. $1$
• D. $2$

Answer 1

Answer: (D)

Given, $\sqrt{y}={\cos }^{-1}x$
$\Rightarrow y={\left({\cos }^{-1}x\right)}^2$
On differentiating both sides w.r.t. $x$, we get
$\frac{dy}{dx}=2\left({\cos }^{-1}x\right)\times \frac{-1}{\sqrt{1-x^2}}$
Again, differentiating both sides w.r.t. $x$, we get
$\frac{d^{2}y}{d{x}^2}=-2\frac{\sqrt{1-x^2}\times \frac{-1}{\sqrt{1-x^2}}-{\cos }^{-1}x\times \left(\frac{1}{2}\right)\frac{(-2x)}{{\left(1-x^2\right)}^{1/2}}}{{\left(\sqrt{1-x^2}\right)}^2}$
$=-2\left[\frac{-1+\frac{x{\cos }^{-1}x}{{\left(1-x^2\right)}^{1/2}}}{\left(1-x^2\right)}\right]$
$\frac{d^{2}y}{d{x}^2}=\left[\frac{2-\frac{2x{\cos }^{-1}x}{{\left(1-x^2\right)}^{1/2}}}{\left(1-x^2\right)}\right]$
$\Rightarrow \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}=2+x\frac{dy}{dx}$
$\Rightarrow \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=2$
But, it is given
$\left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=c$
$\therefore c=2$