Question

If y=xlog x/a+bx,prove x³y₂=(xy₁-y)²

Answer 1

Answer:

It is proved that

$y=xlog(\dfrac{x}{a+bx})\\then x^3y_2=(xy_1-y)^2$

Step-by-step explanation:

Given:

$y=xlog(\dfrac{x}{a+bx})\\\dfrac{y}{x}=log(\dfrac{x}{a+bx})$

Here we are using the formula of:

$log(a)-log(b)=log(\dfrac{a}{b})$

Now differentiating with respect to x.

Using formula of $(\dfrac{u}{v})$ method:

$\dfrac{v\frac{d}{dx}(u)-u\frac{d}{dx}(v)}{v^2}$

Therefore:

$\dfrac{x\frac{d}{dx}(y)-y\frac{d}{dx}(x)}{x^2}=\dfrac{d}{dx}(logx)-\dfrac{d}{dx}(log(a+bx)}\\\dfrac{x\frac{dy}{dx}-y}{x^2}=\dfrac{1}{x}-\dfrac{1}{a+bx}\cdot b\\x\dfrac{dy}{dx}-y=x^2(\dfrac{1}{x}-\dfrac{b}{a-bx})\\x\dfrac{dy}{dx}-y=x-\dfrac{bx^2}{a+bx}\\x\dfrac{dy}{dx}-y=\dfrac{x(a+bx)-bx^2}{a+bx}\\x\dfrac{dy}{dx}-y=\dfrac{ax+bx^2-bx^2}{a+bx}$

Taking $\dfrac{dy}{dx}=y_1$.

So the equation will be:

$xy_1-y=\dfrac{ax}{a+bx}\,\,\,\,\,eqn(1)$

Again differentiating with respect to x.

$x\dfrac{d}{dx}(\dfrac{dy}{dx})+{\dfrac{dy}{dx}(\dfrac{d}{dx}(x))-\dfrac{d}{dx}(y)=\dfrac{(a+bx)\dfrac{d}{dx}(ax)-ax\dfrac{d}{dx}(a+bx)}{(a+bx)^2}$

$x\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}-\dfrac{dy}{dx}=\dfrac{(a+bx)a-(ax)\cdot b}{(a+bx)^2}\\\\x\dfrac{d^2y}{dx^2}=\dfrac{a^2+abx-abx}{(a+bx)^2}$

Taking $\dfrac{d^2y}{dx^2}=y_2$

So the equation will be:

$xy_2=\dfrac{a^2}{(a+bx)^2}$

Multiplying both side of the equation with $x^2$ we get:

$x^3y_2=\dfrac{a^2x^2}{(a+bx)^2}\\x^3y_2=(\dfrac{ax}{a+bx})^2$

Form eqn(1) we can write as:

$x^3y_2=(xy_1-y)^2$