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$\begingroup$

$$y'=\frac{x+y-2}{x-y}$$

$$\begin{vmatrix} 1 & 1\\ 1 & -1 \end{vmatrix}\neq0 $$

Substation: $x=u+\alpha$ , $y=v+\beta$

$\frac{dy}{dx}=\frac{dv}{du}$

$$\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}\begin{pmatrix} \alpha \\ \beta \end{pmatrix} =\begin{pmatrix} 2 \\ 0 \end{pmatrix}\Rightarrow \alpha=1\text{ , }\beta=1 $$

So:

$x=u+1$

$y=v+1$

$$y'=\frac{dv}{du}=\frac{u+1+v+1-2}{u+1-(v+1)}$$

$$\frac{dv}{du}=\frac{u+v}{u-v}$$

$$\frac{dv}{du}=\frac{u+v}{u-v}$$

$$\frac{dv}{du}=\frac{1+\frac{v}{u}}{1-\frac{v}{u}}$$

$z=\frac{v}{u}\Rightarrow v=uz$

$$\frac{dv}{du}=z+u\frac{dz}{du}=\frac{1+z}{1-z}$$

How should I continue?

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    this is a equation of Abel-type – 2017-01-22
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    A small drawing indicates that the graph of the solution (in $u$ and $v$) hits the lines through the origin in the $(u,v)$ coordinate system all under the same angle (45°). This is a fundamental property of the logarithmic spirals. So can try a parametrized representation of the logarithmic spirals. – 2017-01-22

1 Answers 1

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$u\frac{dz}{du}=\frac{1+z}{1-z}-z=\frac{z^2+1}{1-z}$ thus $$\frac{1-z}{1+z^2}dz=\frac{1}{u}du$$ or $$\left(\frac{1}{1+z^2}-\frac{z}{1+z^2}\right)dz=\frac{1}{u}du$$ as a result $$\tan^{-1}(z)-\frac12\ln(1+z^2)=C+\ln(u)$$