Show that $Z^{\prime }\left( s\right) -2\sqrt{-1}H\left( s\right) Z\left( s\right) -1=0 $

Question

Consider the equations

$2H\left( s\right) y\left( s\right) -x^{\prime }\left( s\right) -x^{\prime \prime }\left( s\right) y\left( s\right) y^{\prime }\left( s\right) +x^{\prime }\left( s\right) y\left( s\right) y^{\prime \prime }\left( s\right) =0$

$x^{\prime }\left( s\right) ^{2}+y^{\prime }\left( s\right) ^{2}=1$

$Z\left( s\right) =y\left( s\right) y^{\prime }\left( s\right) +\sqrt{-1}% y\left( s\right) x^{\prime }\left( s\right) $

where $H,x,y:I\rightarrow %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion $ are functions real of one variable

Show that $Z^{\prime }\left( s\right) -2\sqrt{-1}H\left( s\right) Z\left( s\right) -1=0 $, $s\in I\subset \mathbb{R}$

I have opened the accounts anyway, but can not show the equality of the differential equation.

Answer 1

From your equations you get $$ 1=(y'+ix')(y'-ix')\\ 0=x'x''+y'y'' $$ So for the function under consideration, $Z=y(y'+ix')$, we get \begin{align} \frac{Z'}{Z}&=\frac{y'}{y}+\frac{y''+ix''}{y'+ix'} \\& =\frac{y'}y+(y''+ix'')(y'-ix') =\frac{y'}y+i(x''y'-y''x') \\& =\frac1y\left[y'+i(2Hy-x')\right] \\ \implies Z'& =\frac{Z}{y(y'+ix')}+2iHZ=1+2iHZ \end{align} which is indeed the given differential equation.