If y = tan^{-1}(sec x^3 - tan x^3) and frac{p | vedclass

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

Question

(B) Given $y = 	an^{-1}(\sec x^3 - 	an x^3)$.$y = 	an^{-1}\left(\frac{1 - \sin x^3}{\cos x^3}ight) = 	an^{-1}\left(\frac{1 - \cos(\frac{\pi}{2}

Vedclass · Accepted Answer

$x^2 y'' - 6 y + \frac{3\pi}{2} = 0$

Vedclass · Answer

(B) Given $y = 	an^{-1}(\sec x^3 - 	an x^3)$.$y = 	an^{-1}\left(\frac{1 - \sin x^3}{\cos x^3}ight) = 	an^{-1}\left(\frac{1 - \cos(\frac{\pi}{2} - x^3)}{\sin(\frac{\pi}{2} - x^3)}ight)$.Using the half-angle formulas,$1 - \cos 	heta = 2 \sin^2(\frac{	heta}{2})$ and $\sin 	heta = 2 \sin(\frac{	heta}{2}) \cos(\frac{	heta}{2})$,we get:$y = 	an^{-1}\left(	an(\frac{\pi}{4} - \frac{x^3}{2})ight)$.Since $\frac{\pi}{2} Adding $\frac{\pi}{4}$,we get $-\frac{\pi}{2} Thus,$y = \frac{\pi}{4} - \frac{x^3}{2}$.Now,differentiate with respect to $x$:$y' = -\frac{1}{2} \cdot 3x^2 = -\frac{3}{2}x^2$.$y'' = -\frac{3}{2} \cdot 2x = -3x$.From $y = \frac{\pi}{4} - \frac{x^3}{2}$,we have $x^3 = \frac{\pi}{2} - 2y$.Also,$x^2 = -\frac{2}{3} y'$.Substituting $x^2$ into $y'' = -3x$,we get $y'' = -3 \sqrt[3]{-\frac{2}{3} y'}$.Alternatively,using $x^2 = -\frac{2}{3} y'$,we have $x^2 y'' = x^2 (-3x) = -3x^3 = -3(\frac{\pi}{2} - 2y) = -\frac{3\pi}{2} + 6y$.Therefore,$x^2 y'' - 6y + \frac{3\pi}{2} = 0$.

If $y = \tan^{-1}(\sec x^3 - \tan x^3)$ and $\frac{\pi}{2} < x^3 < \frac{3\pi}{2}$,then:

Similar Questions

If $\sin ^{-1}\left(\frac{3}{x}\right)+\sin ^{-1}\left(\frac{4}{x}\right)=\frac{\pi}{2}$, then $x$ is equal to

If ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y + {\tan ^{ - 1}}z = \pi ,$ then $\frac{1}{{xy}} + \frac{1}{{yz}} + \frac{1}{{zx}} = $

The sum of the first $n$ terms of the series $\cot^{-1} 3 + \cot^{-1} 7 + \cot^{-1} 13 + \cot^{-1} 21 + \dots$ is given by:

$2 \pi - \left(\sin ^{-1} \frac{4}{5} + \sin ^{-1} \frac{5}{13} + \sin ^{-1} \frac{16}{65}\right)$ is equal to

The value of $\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module