Example y′ = 1−y2+x2 _____________

3.4.10.20 Example $y^{'} = \frac{1 - y^{2} + x^{2}}{1 + y^{2} - x^{2}}$

\begin{aligned} y^{'} & = \frac{1 - y^{2} + x^{2}}{1 + y^{2} - x^{2}} \\ = ω (x, y) \end{aligned}

Using anstaz’s it is found that

\begin{aligned} ξ & = x - y \\ η & = y - x \end{aligned}

Hence

\begin{aligned} \frac{d x}{ξ} & = \frac{d y}{η} = d S \\ (1) & \frac{d x}{x - y} & = \frac{d y}{y - x} = d S \end{aligned}

The ﬁrst two give

\frac{d y}{d x} = \frac{η}{ξ} = \frac{y - x}{x - y} = - 1

Hence

\begin{matrix} (2) & y = - x + c_{1} \end{matrix}

Therefore

\begin{aligned} R & = c_{1} \\ = y + x \end{aligned}

To ﬁnd $S$ , since both $ξ, η$ depend on both $x, y$ , then $\frac{d y}{η} = d S$ or $\frac{d x}{ξ} = d S$ can be used. Lets try both to show same answer results.

\begin{aligned} \frac{d y}{η} & = d S \\ d S & = \frac{d y}{y - x} \end{aligned}

But from (2), $x = c_{1} - y$ . The above becomes

\begin{aligned} d S & = \frac{d y}{y - (c_{1} - y)} \\ = \frac{d y}{2 y - c_{1}} \end{aligned}

Hence

S = \frac{1}{2} \ln (2 y - c_{1})

But $c_{1} = y + x$ . So the above becomes

\begin{aligned} S & = \frac{1}{2} \ln (2 y - (y + x)) \\ (3) & = \frac{1}{2} \ln (y - x) \end{aligned}

Let us now try the other ode

\begin{aligned} \frac{d x}{ξ} & = d S \\ d S & = \frac{d x}{x - y} \end{aligned}

But from (2) $y = - x + c_{1}$ . The above becomes

\begin{aligned} d S & = \frac{d x}{x - (- x + c_{1})} \\ = \frac{d x}{2 x - c_{1}} \end{aligned}

Therefore

S = \frac{1}{2} \ln (2 x - c_{1})

But $c_{1} = y + x$ . Therefore

\begin{aligned} S & = \frac{1}{2} \ln (2 x - (y + x)) \\ (4) & = \frac{1}{2} \ln (x - y) \end{aligned}

The constant of integration is set to zero when ﬁnding $S$ . What is left is to ﬁnd $\frac{d S}{d R}$ . This is given by

\begin{matrix} (5) & \frac{d S}{d R} = \frac{S_{x} + S_{y} ω}{R_{x} + R_{y} ω} \end{matrix}

But, and using (4) for $S$ we have

\begin{aligned} R_{x} & = 1 \\ R_{y} & = 1 \\ S_{x} & = \frac{- 1}{y - x} \\ S_{y} & = \frac{1}{y - x} \end{aligned}

Hence (2) becomes

\begin{aligned} \frac{d S}{d R} & = \frac{\frac{- 1}{y - x} + \frac{1}{y - x} ω}{1 + ω} \\ = \frac{- \frac{ω - 1}{x - y}}{1 + ω} \\ = \frac{1 - ω}{(1 + ω) (x - y)} \\ = \frac{1 - (\frac{1 - y^{2} + x^{2}}{1 + y^{2} - x^{2}})}{(1 + (\frac{1 - y^{2} + x^{2}}{1 + y^{2} - x^{2}})) (x - y)} \\ = - x - y \\ = - (x + y) \\ = - R \end{aligned}

Hence

\begin{aligned} \frac{d S}{d R} & = - R \\ S & = - \frac{R^{2}}{2} \end{aligned}

Converting back to $x, y$ gives

\ln (y - x) = - \frac{{(y + x)}^{2}}{2}

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3.4.10.20 Example y′=1−y2+x21+y2−x2

3.4.10.20 Example $y^{'} = \frac{1 - y^{2} + x^{2}}{1 + y^{2} - x^{2}}$