If sqrt{y}=cos ^{-1} x, then it satisfies the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Given, $\sqrt{y}=\cos ^{-1} x \Rightarrow y=(\cos ^{-1} x)^{2}$.On differentiating both sides with respect to $x$, we get:$\frac{dy}{dx} = 2(\cos ^{-1} x) 	imes \left(\frac{-1}{\sqrt{1-x^{2}}}ight)$.Multiplying both sides by $\sqrt{1-x^{2}}$, we get:$\sqrt{1-x^{2}} \frac{dy}{dx} = -2 \cos ^{-1} x$.Again, differentiating both sides with respect to $x$ using the product rule:$\sqrt{1-x^{2}} \frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} 	imes \left(\frac{-2x}{2\sqrt{1-x^{2}}}ight) = -2 	imes \left(\frac{-1}{\sqrt{1-x^{2}}}ight)$.$\sqrt{1-x^{2}} \frac{d^{2}y}{dx^{2}} - \frac{x}{\sqrt{1-x^{2}}} \frac{dy}{dx} = \frac{2}{\sqrt{1-x^{2}}}$.Multiplying the entire equation by $\sqrt{1-x^{2}}$, we get:$(1-x^{2}) \frac{d^{2}y}{dx^{2}} - x \frac{dy}{dx} = 2$.Comparing this with the given equation $(1-x^{2}) \frac{d^{2}y}{dx^{2}} - x \frac{dy}{dx} = c$, we find $c = 2$.

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

Similar Questions

The differential equation of the family of all straight lines passing through the origin is

The differential equation of $y=ae^{bx}$ (where $a$ and $b$ are parameters) is:

The differential equation formed by eliminating $a$ and $b$ from the equation $y=a e^{2 x}+b x e^{2 x}$ is

The differential equation of all parabolas whose axis is the $y$-axis is

The differential equation of the family of curves $r^2 = a^2 \cos 2\theta$,where '$a$' is an arbitrary constant,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module