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 = (\tan^{-1} x)^2 $,then $ (x^2 + 1)^2 y_2 + 2x(x^2 + 1) y_1 $ is equal to

If $\frac{d^n y}{d x^n}=y_n$ and $y=e^{\sqrt{x}}+e^{-\sqrt{x}}$,then $4 x y_2+2 y_1=$

$y=e^{a \sin ^{-1} x} \Rightarrow (1-x^2) y_{n+2}-(2 n+1) x y_{n+1}$ is equal to

If $y = \sin(m \sin^{-1} x)$, then $(1-x^2) y_2 - x y_1$ is equal to (Here, $y_n$ denotes $\frac{d^n y}{dx^n}$)

If $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then $\frac{d^2 y}{d x^2}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module