How can I solve the differential equation $(y'/y)'=ay$? In a physics problem I'm working on, the following differential equation arose.
$$\left(\frac{y'}{y}\right)'=-\alpha y$$
where $y=y(x)$ and $\alpha$ is a positive real number. When I plug it into wolfram alpha, it says the general solution to this second order nonlinear differential equation is
$$y_1(x) = \frac{c_1-c_1 \tanh^2\left(\frac{1}{2} (\sqrt{c_1} c_2-\sqrt{c_1} x)\right)}{2 a}$$
$$y_2(x) = \frac{c_1-c_1 \tanh^2\left(\frac{1}{2} (\sqrt{c_1} c_2+\sqrt{c_1} x)\right)}{2 a}$$
where $c_1$ and $c_2$ are arbitrary real constants. I have no idea how it derived such solutions, but I have checked that they satisfy the differential equation. I've tried to show this for a few hours, but I just can't. Could you help offer some insight?
EDIT: I have almost everything obvious, like writing the LHS of the differential equation as the second derivative of $\log y$, and using the quotient rule on the LHS.
 A: $\DeclareMathOperator{\sech}{sech}$Put $u = \log y$, so that
$$
u'' = -\alpha e^{u}.
\tag{1}
$$
Multiply by $u'$ and integrate, obtaining
$$
\frac{(u')^{2}}{2} = -\alpha e^{u} + \frac{c_{1}}{2},
$$
or
$$
u' = \pm\sqrt{-2\alpha e^{u} + c_{1}}.
\tag{2}
$$
Now let
$$
2\alpha e^{u} = c_{1}\tanh^{2} v,
\tag{3a}
$$
so that
$$
2\alpha e^{u}\, u' = 2c_{1}\tanh v \sech^{2} v\, v'.
\tag{3b}
$$
Dividing (3b) by (3a),
$$
u' = 2\sech^{2} v \coth v\, v' = \frac{2v'}{\cosh v \sinh v},
\tag{4a}
$$
while substituting (3a) in (2) gives
$$
u' = \pm \sqrt{c_{1}} \sech v.
\tag{4b}
$$
Equating (4b) and (4a),
$$
\pm\sqrt{c_{1}} = \frac{2v'}{\sinh v} = \frac{4e^{v}\, v'}{(e^{v})^{2} - 1}.
\tag{5}
$$
Setting $w = e^{v}$ converts (5) into
$$
\pm\sqrt{c_{1}} = \frac{4\, w'}{w^{2} - 1},
$$
which integrates to
$$
\sqrt{c_{1}}(\pm x + c_{2})
  = 2\log\left\lvert\frac{w - 1}{w + 1}\right\rvert
  = 2\log\left\lvert\frac{e^{v} - 1}{e^{v} + 1}\right\rvert
$$
or
$$
e^{\frac{1}{2}\sqrt{c_{1}}(\pm x + c_{2})} = \tanh(\tfrac{v}{2}).
\tag{6}
$$
Now, (3a) reads
$$
\tanh^{2} v = \frac{2\alpha}{c_{1}} e^{u} = \frac{2\alpha}{c_{1}} y,
$$
or
$$
y = \frac{c_{1}}{2\alpha} \tanh^{2}v.
\tag{7}
$$
To massage this into Wolfram's form, note that if $e^{z} = \tanh \frac{v}{2}$, i.e.,
$$
z = \tfrac{1}{2}\sqrt{c_{1}}(\pm x + c_{2}),
\tag{6a}
$$
then
$$
e^{2z} = \tanh^{2}\tfrac{v}{2} = \frac{\cosh v - 1}{\cosh v + 1}.
\tag{8}
$$
Consequently,
\begin{align*}
1 - \tanh^{2} z
  &= (1 - \tanh z)(1 + \tanh z) = \frac{4e^{2z}}{(e^{2z} + 1)^{2}} \\
  &= \frac{4\dfrac{\cosh v - 1}{\cosh v + 1}}{\left(\dfrac{\cosh v - 1}{\cosh v + 1} + 1\right)^{2}}
  = \frac{\cosh v - 1}{\cosh v + 1}\left(\frac{\cosh v + 1}{\cosh v}\right)^{2} = \tanh^{2} v.
\end{align*}
Substituting this into (7) and using (6a),
$$
y = \frac{c_{1}}{2\alpha} (1 - \tanh^{2} z)
  = \frac{c_{1}}{2\alpha} \left[1 - \tanh^{2}\bigl(\tfrac{1}{2}\sqrt{c_{1}}(\pm x + c_{2})\bigr)\right].
$$
A: Set $$z = \frac{y'}{y} \,\,\, \text{ or conversely } \,\,\, y' = y\,z$$ and create the system of ODEs
\begin{align}
y' &= y \, z\\
z' &= -\alpha \, y  
\end{align}
 or written with the $x$ argument variable
\begin{align}
\frac{dy}{dx} &= y \, z\\
\frac{dz}{dx} &= -\alpha \, y  
\end{align}
Eliminating $dx$ gives us the ODE
\begin{align}
\frac{dy}{y \, z} =  dx = - \, \frac{dz}{\alpha \, y } 
\end{align}
$$\frac{dy}{y \, z}  + \, \frac{dz}{\alpha \, y } = 0$$
$$\frac{dy}{ z}  + \, \frac{dz}{\alpha} = 0$$
$$\alpha \,{dy}  + z\, {dz} = 0$$
$$\alpha \, y + \frac{z^2}{2} =
\frac{C}{2}$$
$$2 \, \alpha \, y + {z^2} =
{C}$$ Substituting $z = \frac{y'}{y}$ back
$$2 \, \alpha \, y + \frac{1}{y^2} \, (y')^2 =
{C}$$
\begin{align}
\left(\frac{dy}{dx}\right)^2 = y^2 \, \big(C - 2 \, \alpha \, y\big)
\end{align}
\begin{align}
\frac{dy}{dx} = \varepsilon \, y\, \sqrt{C - 2 \, \alpha \, y}
\end{align}
where $\varepsilon \in \{-1, 1\}$. To integrate the latter equation, I think you should perform the substitute $$u(x) = \frac{1}{y(x)}$$ 
\begin{align}
\end{align}
