If y = x log left( frac{x}{a + bx} right),the | vedclass

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

Question

(B) Given $y = x \log \left( \frac{x}{a + bx} \right)$.Dividing by $x$,we get $\frac{y}{x} = \log x - \log(a + bx)$.Differentiating with respect to $x

Vedclass · Accepted Answer

$\left( x \frac{dy}{dx} - y \right)^2$

Vedclass · Answer

(B) Given $y = x \log \left( \frac{x}{a + bx} \right)$.Dividing by $x$,we get $\frac{y}{x} = \log x - \log(a + bx)$.Differentiating with respect to $x$:$\frac{x \frac{dy}{dx} - y}{x^2} = \frac{1}{x} - \frac{b}{a + bx} = \frac{a + bx - bx}{x(a + bx)} = \frac{a}{x(a + bx)}$.Thus,$x \frac{dy}{dx} - y = \frac{ax}{a + bx}$ ..... $(i)$.Differentiating equation $(i)$ with respect to $x$:$x \frac{d^2y}{dx^2} + \frac{dy}{dx} - \frac{dy}{dx} = \frac{(a + bx)a - ax(b)}{(a + bx)^2}$.$x \frac{d^2y}{dx^2} = \frac{a^2}{(a + bx)^2}$.Multiplying both sides by $x^2$:$x^3 \frac{d^2y}{dx^2} = \frac{a^2 x^2}{(a + bx)^2} = \left( \frac{ax}{a + bx} \right)^2$.From equation $(i)$,$\frac{ax}{a + bx} = x \frac{dy}{dx} - y$.Therefore,$x^3 \frac{d^2y}{dx^2} = \left( x \frac{dy}{dx} - y \right)^2$.

If $y = x \log \left( \frac{x}{a + bx} \right)$,then $x^3 \frac{d^2y}{dx^2} = $

Similar Questions

If $y=ae^x+be^{-x}+c$, where $a, b, c$ are parameters, then $y^{\prime \prime \prime}$ is equal to

If $y = \cos^{-1} \left\{ \frac{a \cos x - b \sin x}{\sqrt{a^2 + b^2}} \right\}$,then $\frac{d^2 y}{d x^2}$ is equal to

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?

If $a \neq b, x \neq n \pi, n \in Z$ and $y^2 = a^2 \cos^2 x + b^2 \sin^2 x$,then $\frac{d^2 y}{dx^2} + y =$

If $f(x) = \frac{x^2}{x+a}$,then $f^{\prime \prime}(a)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module