how to do this ODE transformation? Could someone please show me how the book did the following transformation? I am not able to reproduce the result shown. I show one attempt, but I tried few others. Here is screen shot of the part of the book page showing the transformation used
The book link at google books is this and the page is page number 3 in the introduction:

\begin{equation}
\frac{d^{2}y}{dx^{2}}\left(  y+x\right)  +\frac{dy}{dx}\left(  \frac{dy}
{dx}-1\right)  =0 \tag{1}
\end{equation}
Since $y=u-v\left(  u\right)  $ then,
\begin{equation}
\frac{dy}{dx}=\frac{du}{dx}-\frac{dv}{du}\frac{du}{dx} \tag{2}
\end{equation}
And%
\begin{align}
\frac{d^{2}y}{dx^{2}} &  =\frac{d^{2}u}{dx^{2}}-\frac{d}{dx}\left(  \frac
{dv}{du}\frac{du}{dx}\right)  \nonumber\\
&  =\frac{d^{2}u}{dx^{2}}-\left(  \frac{d^{2}v}{du^{2}}\left(  \frac{du}
{dx}\right)  ^{2}+\frac{dv}{du}\frac{d^{2}u}{dx^{2}}\right)  \tag{3}
\end{align}
Now Let $\frac{dv}{du}=v^{\prime},\frac{d^{2}v}{du^{2}}=v^{\prime\prime}$,
then (2) and (3) becomes
\begin{align*}
\frac{dy}{dx} &  =\frac{du}{dx}-v^{\prime}\frac{du}{dx}\\
\frac{d^{2}y}{dx^{2}} &  =\frac{d^{2}u}{dx^{2}}-\left(  v^{\prime\prime
}\left(  \frac{du}{dx}\right)  ^{2}+v^{\prime}\frac{d^{2}u}{dx^{2}}\right)
\end{align*}
Substituting the above two equations back into (1), and using the given
$x=u+v,y=u-v$ results in
\begin{align*}
\left[  \frac{d^{2}u}{dx^{2}}-\left(  v^{\prime\prime}\left(  \frac{du}
{dx}\right)  ^{2}+v^{\prime}\frac{d^{2}u}{dx^{2}}\right)  \right]  \left(
u-v+u+v\right)  +\left(  \frac{du}{dx}-v^{\prime}\frac{du}{dx}\right)  \left(
\frac{du}{dx}-v^{\prime}\frac{du}{dx}-1\right)   &  =0\\
\left[  \frac{d^{2}u}{dx^{2}}-\left(  v^{\prime\prime}\left(  \frac{du}
{dx}\right)  ^{2}+v^{\prime}\frac{d^{2}u}{dx^{2}}\right)  \right]  2u+\left(
\frac{du}{dx}\right)  ^{2}-v^{\prime}\left(  \frac{du}{dx}\right)  ^{2}
-\frac{du}{dx}-v^{\prime}\left(  \frac{du}{dx}\right)  ^{2}+v^{\prime2}\left(
\frac{du}{dx}\right)  ^{2}+v^{\prime}\frac{du}{dx} &  =0\\
2u\frac{d^{2}u}{dx^{2}}-2uv^{\prime\prime}\left(  \frac{du}{dx}\right)
^{2}-2uv^{\prime}\frac{d^{2}u}{dx^{2}}+\left(  \frac{du}{dx}\right)
^{2}-v^{\prime}\left(  \frac{du}{dx}\right)  ^{2}-\frac{du}{dx}-v^{\prime
}\left(  \frac{du}{dx}\right)  ^{2}+v^{\prime2}\left(  \frac{du}{dx}\right)
^{2}+v^{\prime}\frac{du}{dx} &  =0
\end{align*}
I do not know how to make the above same as book result.
 A: Let $x = u + v$ and $y = u -v$ so that
$$
x + y = 2u \quad\mbox{and}\quad x - y = 2v \tag{1}
$$
Note that
\begin{eqnarray}
dx &=& du + dv = (1 + dv/du)du = (1 + v')du \\
dy &=& du - dv = (1 - dv/du)du = (1 - v')du
\end{eqnarray}
Therefore
$$
\frac{dy}{dx} = \frac{(1 - v')du}{(1 + v')du} = \frac{1 -v'}{1 + v'} \tag{2}
$$
As for the second derivative
\begin{eqnarray}
dy' &=& d\frac{1-v'}{1+ v'} = -(1 - v')(1 + v')^{-2}dv' - (1+v')dv' \\
&=& -\frac{2}{(1 + v')^2}dv' 
\end{eqnarray}
Using the fact that $dx = (1 + v')du$ we get
$$
y'' = \frac{dy'}{dx} = -\frac{2}{(1 + v')^2}\frac{dv'}{(1 + v')du} = -\frac{2}{(1 + v')^3}v'' \tag{3}
$$
Now, replacing in the original differential equation
\begin{eqnarray}
\left[-\frac{2}{(1 + v')^3}v'' \right]\left[2u \right] + \frac{1-v'}{1+v'}\left[\frac{1-v'}{1+v'} - 1 \right] = 0
\end{eqnarray}
which leads to
$$ \bbox[yellow]
{
2v''u + (1- v'^2)v' = 0
} \tag{4}
$$
IMPORTANT
This last equation is different from the one in the book, but I'm pretty sure there's a mistake in the book! To show that, I numerically integrate the original ODE, the second equation in the book, and equation (4). This is what I got

