Question

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