Show that the substitution x=varepsilon e^t , y=varepsilon…

Show that the substitution $x=\varepsilon e^t$, $y=\varepsilon e^t\varphi(t)$ ($\varepsilon=\pm1$) reduces $(*)\!: \qquad x^2y'' - y(y')^2 + y = 0$ to the first-order equation $\dfrac{dz}{d\varphi} = \dfrac{\varphi(z^2-1)}{z-\varphi}$, where $z = \dfrac{d\varphi}{dt}+\varphi$.

Step 1: The change of independent variable

Take $\varepsilon=+1$ (the case $\varepsilon=-1$ is structurally identical, since only $\varepsilon^2=1$ ever enters below). With $x=e^t$, $\dfrac{dx}{dt}=e^t=x$, so $\dfrac{dt}{dx}=\dfrac1x$, and for any function of $t$,

\[ \frac{d}{dx} = \frac1x\,\frac{d}{dt}. \tag{1} \]

Step 2: Transforming $y'$

Since $y = e^t\varphi(t) = x\varphi(t(x))$, apply (1) to $y$, writing $\dot\varphi = d\varphi/dt$: \[y' = \frac1x\,\frac{d}{dt}\bigl[e^t\varphi\bigr] = \frac1x\,e^t(\varphi+\dot\varphi) = \varphi+\dot\varphi,\]

using $e^t=x$ to cancel the prefactor exactly. Define

\[ z := \dot\varphi+\varphi, \qquad \text{so that}\qquad y'=z. \tag{2} \]

Step 3: Transforming $y''$

Applying (1) again, now to $y'=z(t)$, \[y'' = \frac1x\,\frac{d}{dt}(z) = \frac{\dot z}{x}. \tag{3}\]

Step 4: Substituting into $(*)$

Insert $y=x\varphi$, $y'=z$ from (2), and $y''=\dot z/x$ from (3) into $(*)$: \[x^2\cdot\frac{\dot z}{x} - x\varphi\cdot z^2 + x\varphi = 0 \;\Longrightarrow\; x\dot z - x\varphi(z^2-1) = 0.\]

Dividing by $x\neq0$, the factor of $x$ cancels completely, leaving a fully autonomous relation in $t$ with no leftover $x$-dependence: \[\dot z = \varphi(z^2-1). \tag{4}\]

Step 5: The autonomous system and its phase-plane reduction

Equation (2) rearranges to $\dot\varphi = z-\varphi$, which together with (4) forms an autonomous first-order system in the pair $(\varphi,z)$: $\dot\varphi = z-\varphi, \qquad \dot z = \varphi(z^2-1)$.

Because neither right-hand side involves $t$ explicitly, $z$ may be regarded as a function of $\varphi$ along trajectories (the classical Poincaré phase-plane reduction), and dividing the two equations eliminates $t$ directly:

\[ \frac{dz}{d\varphi} = \frac{\dot z}{\dot\varphi} = \frac{\varphi(z^2-1)}{z-\varphi}. \qquad \blacksquare \]

Step 6: Verification

Substituting $y=x\varphi(\ln x)$ directly into $x^2y''-y(y')^2+y$, with $\varphi$ treated as an abstract function of $t=\ln x$ and all derivatives expanded by the chain rule, reduces the expression exactly to

\[ x\Bigl[\ddot\varphi+\dot\varphi-(\varphi+\dot\varphi)^2\varphi+\varphi\Bigr] = x\bigl[\dot z - \varphi(z^2-1)\bigr], \]

since $\ddot\varphi+\dot\varphi=\dot z$ by the definition $z=\dot\varphi+\varphi$. The bracket vanishes identically by (4), confirming the reduction holds exactly, with no hidden $x$-dependence and no approximation at any step.\ \checkmark\)

\[\frac{dz}{d\varphi} = \frac{\varphi(z^2-1)}{z-\varphi}, \qquad z=\dot\varphi+\varphi\]