How has this result been obtained??

Question

We have the quadratic equation

$$ 2y^2 -(1+x)y + x = 0 $$

The author derives the following result from the above equation -

$$ y' = \frac{y-1}{4y-1-x} = \frac{(x-3)y-x+1}{x^2-6x+1} $$

Now, I understand the first part. It has been obtained by simply differentiating w.r.t to $x$ and then solving for $y'$. But, I am unable to obtain the second part.

I have tried lot of different substitutions, but haven't been able to arrive at the result. Please help!!

Answer 1

$$(x^2-6x+1)(y-1)-(4y-x-1)((x-3)y-x+1)\\ =-2x^2-4xy^2-4xy+6x+12y^2+2yx^2-6y\\ =(6-2x)(2y^2-(1+x)y+x)\\=0.$$

The identity is valid.

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To establish it, solve the initial equation for $y$

$$y=\frac{x+1\pm\sqrt{x^2-6x+1}}4$$

and plug in the fraction, to get

$$y'=\frac{x-3\pm\sqrt{x^2-6x+1}}{\pm4\sqrt{x^2-6x+1}}=\frac{\pm(x-3)\sqrt{x^2-6x+1}+x^2-6x+1}{4(x^2-6x+1)}=\frac{4(x-3)y+x^2-6x+1-(x-3)(x+1)}{4(x^2-6x+1)}.$$

Answer 2

There's a way to simplify the denominator without explicitly solving for $y$. Note that \begin{align} \big(4y-(1+x)\big)^2 &= 16y^2 - 8(1+x)y + (1+x)^2 \\ &= 8\big(2y^2 - (1+x)y\big) + (1+x)^2 \\ &= -8x + (1+x)^2 \\ &= x^2 - 6x + 1 \end{align}

Therefore

\begin{align} y' = \frac{y-1}{4y-(1+x)} &= \frac{(y-1)\big(4y-(1+x)\big)}{\big(4y-(1+x)\big)^2} \\ &= \frac{4y^2 - 4y - (1+x)y+(1+x)}{x^2-6x+1} \\ &= \frac{2\big(2y^2-(1+x)y\big)+(1+x)y - 4y+ (1+x)}{x^2-6x+1} \\ &= \frac{-2x+(1+x)y-4y+1+x}{x^2-6x+1} \\ &= \frac{1-x+(x-3)y}{x^2-6x+1} \end{align}