If cos ^{-1}left(frac{y}{2}right)=log _{e}lef | vedclass

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

Question

(A) Given: $\cos ^{-1}\left(\frac{y}{2}\right)=5 \log _{e}\left(\frac{x}{5}\right)$.Differentiating both sides with respect to $x$:$\frac{-1}{\sqrt{1-

Vedclass · Accepted Answer

$x^{2} y^{\prime \prime}+x y^{\prime}+25 y=0$

Vedclass · Answer

(A) Given: $\cos ^{-1}\left(\frac{y}{2}ight)=5 \log _{e}\left(\frac{x}{5}ight)$.Differentiating both sides with respect to $x$:$\frac{-1}{\sqrt{1-\frac{y^{2}}{4}}} \cdot \frac{y^{\prime}}{2} = 5 \cdot \frac{1}{x/5} \cdot \frac{1}{5} = \frac{5}{x}$.$\frac{-y^{\prime}}{\sqrt{4-y^{2}}} = \frac{5}{x} \implies -x y^{\prime} = 5 \sqrt{4-y^{2}}$.Squaring both sides: $x^{2} (y^{\prime})^{2} = 25(4-y^{2}) = 100 - 25y^{2}$.Differentiating again with respect to $x$:$x^{2} \cdot 2 y^{\prime} y^{\prime \prime} + 2x (y^{\prime})^{2} = -50 y y^{\prime}$.Dividing by $2 y^{\prime}$ (assuming $y^{\prime} 
eq 0$):$x^{2} y^{\prime \prime} + x y^{\prime} = -25 y$.Therefore,$x^{2} y^{\prime \prime} + x y^{\prime} + 25 y = 0$.

If $\cos ^{-1}\left(\frac{y}{2}\right)=\log _{e}\left(\frac{x}{5}\right)^{5}$ and $|y| < 2$,then which of the following is true?

Similar Questions

If $y = \sqrt{x^2 + 8x + 3}$, then the value of $\frac{d^2y}{dx^2}$ at $x = 3$ is

If $y = (\tan^{-1} 2x)^2 + (\cot^{-1} 2x)^2$,then $(1 + 4x^2)^2 y'' - 16 =$

If $y = a \cos(\log x) + b \sin(\log x)$ where $a, b$ are parameters,then ${x^2}y'' + xy' = $

If $f(x)$ and $g(x)$ are twice differentiable functions on $(0,3)$ satisfying $f^{\prime \prime}(x)=g^{\prime \prime}(x)$, $f^{\prime}(1)=4$, $g^{\prime}(1)=6$, $f(2)=3$, and $g(2)=9$, then $f(1)-g(1)$ is

If $x = \sin \theta$ and $y = \cos(p \theta)$,then $(1 - x^2) y_2 =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module