The differential equation of y = sec(tan^{-1} | vedclass

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

Question

(C) Given the equation: $y = \sec(	an^{-1}x)$.Differentiating both sides with respect to $x$ using the chain rule:$\frac{dy}{dx} = \frac{d}{dx}[\sec(

Vedclass · Accepted Answer

$(1 + x^2)\frac{dy}{dx} = xy$

Vedclass · Answer

(C) Given the equation: $y = \sec(	an^{-1}x)$.Differentiating both sides with respect to $x$ using the chain rule:$\frac{dy}{dx} = \frac{d}{dx}[\sec(	an^{-1}x)]$Using the derivative formula $\frac{d}{dx}[\sec(u)] = \sec(u)	an(u)\frac{du}{dx}$ and $\frac{d}{dx}[	an^{-1}x] = \frac{1}{1+x^2}$:$\frac{dy}{dx} = \sec(	an^{-1}x) \cdot 	an(	an^{-1}x) \cdot \frac{1}{1+x^2}$Since $	an(	an^{-1}x) = x$ and $y = \sec(	an^{-1}x)$:$\frac{dy}{dx} = y \cdot x \cdot \frac{1}{1+x^2}$Rearranging the terms:$(1 + x^2)\frac{dy}{dx} = xy$.

The differential equation of $y = \sec(\tan^{-1}x)$ is

Similar Questions

The differential equation of all lines perpendicular to the line $5x + 2y + 7 = 0$ is

If the family of curves $y = a e^{4x} + b e^{-x}$, where $a, b$ are arbitrary constants, represents the general solution of the differential equation $f(x, y, \frac{dy}{dx}, \frac{d^2y}{dx^2}) = 0$, then find $\frac{df}{dx}$.

The differential equation whose solution is $y=c^2+\frac{c}{x}$,where $c$ is a constant,is

The differential equation of the family of curves,$x^{2}=4 b(y+b), b \in R,$ is

Verify that the given function $y = Ax$ is a solution of the differential equation $xy' = y$ $(x \neq 0)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module