Question

$\boxed{\begin{array}{cc}\bf \: if \: \: \rm \: ( {x - y)}^{2} = \frac{x}{y} , \: \: then \: prove \: that : \\ \\ \\ \displaystyle \int \rm \: \frac{dx}{(x - 3y) {}^{2} } = \frac{1}{2} \ln [( {x - y)}^{2} - 1] + c\end{array}}$
​

Answer 1

Correct Statement :-

If

$\rm :\longmapsto\: {(x - y)}^{2} = \dfrac{x}{y}$

Prove that,

$\rm :\longmapsto\:\dfrac{1}{2}log[ {(x - y)}^{2} - 1] = \displaystyle \int \frac{dx}{x - 3y}$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\: {(x - y)}^{2} = \dfrac{x}{y} - - - - (1)$

Let assume that,

$\rm :\longmapsto\:I = \displaystyle \int \: \frac{dx}{x - 3y} - - - (2)$

Let further assume that

$\rm :\longmapsto\:M = \dfrac{1}{2}log[ {(x - y)}^{2} - 1] - - - (3)$

On differentiating both sides w. r. t. x

$\rm :\longmapsto\:\dfrac{d}{dx}M = \dfrac{1}{2}\dfrac{d}{dx}log[ {(x - y)}^{2} - 1]$

$\rm :\longmapsto\:\dfrac{dM}{dx} = \dfrac{1}{2[ {(x - y)}^{2} - 1] }\dfrac{d}{dx}[ {(x - y)}^{2} - 1]$

$\rm :\longmapsto\:\dfrac{dM}{dx} = \dfrac{1}{2[ {(x - y)}^{2} - 1] }[2{(x - y)}]\dfrac{d}{dx}(x - y)$

$\rm :\longmapsto\:\dfrac{dM}{dx} = \dfrac{1}{[ {(x - y)}^{2} - 1] }[{(x - y)}]\bigg[1 - \dfrac{dy}{dx} \bigg]$

$\rm :\longmapsto\:\dfrac{dM}{dx} = \dfrac{x - y}{[ {(x - y)}^{2} - 1] }\bigg[1 - \dfrac{dy}{dx} \bigg] - - - (4)$

Now, from equation (1),

$\rm :\longmapsto\: {(x - y)}^{2} = \dfrac{x}{y}$

$\rm :\longmapsto\:y{(x - y)}^{2} = x$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y{(x - y)}^{2} = \dfrac{d}{dx}x$

$\rm :\longmapsto\:y\dfrac{d}{dx}{(x - y)}^{2} + {(x - y)}^{2}\dfrac{d}{dx}y = 1$

$\rm :\longmapsto\:2y(x - y)\dfrac{d}{dx}{(x - y)} + {(x - y)}^{2}\dfrac{dy}{dx} = 1$

$\rm :\longmapsto\:2y(x - y)\bigg(1 - \dfrac{dy}{dx}\bigg) + {(x - y)}^{2}\dfrac{dy}{dx} = 1$

$\rm :\longmapsto\:2y(x - y) - 2y(x - y)\dfrac{dy}{dx} + {(x - y)}^{2}\dfrac{dy}{dx} = 1$

$\rm :\longmapsto\:\dfrac{dy}{dx}[ {(x - y)}^{2} - 2y(x - y)] = 1 - 2y(x - y)$

$\rm :\longmapsto\:\dfrac{dy}{dx}(x - y)[x - y - 2y] = 1 - 2y(x - y)$

$\rm :\longmapsto\:\dfrac{dy}{dx}(x - y)[x - 3y] = 1 - 2y(x - y)$

$\rm \implies\:\boxed{ \tt{ \: \dfrac{dy}{dx} = \dfrac{1 - 2y(x - y)}{(x - y)(x - 3y)} }} - - - (5)$

On substituting equation (5) in equation (4), we get

$\rm :\longmapsto\:\dfrac{dM}{dx} = \dfrac{x - y}{[ {(x - y)}^{2} - 1] }\bigg[1 - \dfrac{1 - 2y(x - y)}{(x - y)(x - 3y)} \bigg]$

$\rm :\longmapsto\:\dfrac{dM}{dx} = \dfrac{x - y}{[ {(x - y)}^{2} - 1] }\bigg[ \dfrac{(x - y)(x - 3y) - 1 + 2y(x - y)}{(x - y)(x - 3y)} \bigg]$

$\rm :\longmapsto\:\dfrac{dM}{dx} = \dfrac{1}{[ {(x - y)}^{2} - 1] }\bigg[ \dfrac{(x - y)[x - 3y +2y] - 1}{(x - 3y)} \bigg]$

$\rm :\longmapsto\:\dfrac{dM}{dx} = \dfrac{1}{[ {(x - y)}^{2} - 1] }\bigg[ \dfrac{(x - y)(x - y) - 1}{(x - 3y)} \bigg]$

$\rm :\longmapsto\:\dfrac{dM}{dx} = \dfrac{1}{[ {(x - y)}^{2} - 1] }\bigg[ \dfrac{ {(x - y)}^{2} - 1}{(x - 3y)} \bigg]$

$\bf\implies \:\boxed{ \tt{ \: \dfrac{dM}{dx} = \frac{1}{x - 3y} \: }}$

So, on integrating both sides, we get

$\rm :\longmapsto\:\displaystyle \int\: \dfrac{dM}{dx} \: dx= \displaystyle \int\dfrac{1}{x - 3y} \:dx$

$\rm :\longmapsto\:M= \displaystyle \int\dfrac{1}{x - 3y} \:dx - - - (6)$

Hence, From equation (3) and equation (6), we have

$\rm :\longmapsto\:\dfrac{1}{2}log[ {(x - y)}^{2} - 1] = \displaystyle \int \frac{dx}{x - 3y}$

Answer 2

Step-by-step explanation:

$\color{red} \text{ \tt \: Let \( \tt P= \displaystyle \tt\int \dfrac{d x}{(x-3 y)}=\frac{1}{2} \ln \left\{(x-y)^{2}-1\right\} \)}$

$\color{blue} \displaystyle \because \: \: \: \tt \: P=\int \frac{d x}{(x-3 y)}$

$\color{navy} \begin{aligned} \\& \tt \Rightarrow \quad \frac{d P}{d x}=\frac{1}{(x-3 y)} \\ \\ & \tt Also, \quad P=\frac{1}{2} \ln \left\{(x-y)^{2}-1\right\} \\ \\ & \tt\therefore \quad \frac{d P}{d x}=\frac{2(x-y)\left(1-\frac{d y}{d x}\right)}{2\left\{(x-y)^{2}-1\right\}}=\frac{(x-y)\left(1-\frac{d y}{d x}\right)}{(x-y)^{2}-1} \\ \\ & \tt \text{Also,} \quad P=\frac{1}{2} \ln \left\{(x-y)^{2}-1\right\} \\ \\ & \tt \therefore \quad \frac{d P}{d x}=\frac{2(x-y)\left(1-\frac{d y}{d x}\right)}{2\left\{(x-y)^{2}-1\right\}}=\frac{(x-y)\left(1-\frac{d y}{d x}\right)}{(x-y)^{2}-1} \\ \\ & \tt \text{ Given, } \quad y(x-y)^{2}=x \\ \\ & \tt \Rightarrow \quad \frac{\ln y+2 \ln (x-y)=\ln x}{2} \frac{d y}{d x}+\frac{2}{(x-y)}\left(1-\frac{d y}{d x}\right)=\frac{1}{x} \\ \\ & \tt \Rightarrow \quad \frac{d y}{d x}\left(\frac{1}{y}-\frac{2}{x-y}\right)=\frac{1}{x}-\frac{2}{x-y}=\frac{x-y-2 x}{x(x-y)} \\ \\ & \tt \Rightarrow \quad \frac{d y}{d x}\left(\frac{x-3 y}{y(x-y)}\right)=-\frac{(x+y)}{x(x-y)} \\ \\ & \tt \text{ or }\quad \frac{d y}{d x}=-\frac{y(x+y)}{x(x-3 y)} \end{aligned}$

$\color{darkviolet}\[ \begin{aligned} \text{Now, from Eq. (ii), }\\ \\ & \tt\frac{d P}{d x} =\frac{(x-y)\left\{1+\frac{y(x+y)}{x(x-3 y)}\right\}}{(x-y)^{2}-1} \\ \\ & \tt=\frac{(x-y)\left\{\frac{x^{2}-2 x y+y^{2}}{x(x-3 y)}\right\}}{\left(\frac{x}{y}-1\right)} \\ \\ & \tt=\frac{y(x-y)^{2}}{x(x-3 y)}=\frac{1}{x-3 y} \\ \\ & \text { It is true from (i) } Hence, \int \frac{d x}{x-3 y}=\frac{1}{2} \ln \left\{(x-y)^{2}-1\right\} \\ \\ & \tt \because \quad \text{y is variable} \\ \\ & \tt\therefore \quad \int \frac{d x}{x-2 y} \neq \ln (x-3 y) \end{aligned} \]$