Problem source

Solve the quadratic equation $u^{2}+2u\sinh x-1=0$, giving $u$ in terms of $x$. 
		Find the solution of the differential equation 
		\[
		\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}+2\frac{\mathrm{d}y}{\mathrm{d}x}\sinh x-1=0
		\]
		which satisfies $y=0$ and $y'>0$ at $x=0$. 
		Find the solution of the differential equation 
		\[
		\sinh x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}+2\frac{\mathrm{d}y}{\mathrm{d}x}-\sinh x=0
		\]
		which satisfies $y=0$ at $x=0$.

Solution source

\begin{align*}
&& 0 &= u^2 + 2u \sinh x -1  \\
&&&= u^2 + u(e^x-e^{-x})-e^{x}e^{-x} \\
&&&= (u-e^{-x})(u+e^x) \\
\Rightarrow && u &= e^{-x}, -e^x
\end{align*}

\begin{align*}
&& 0 &= \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}+2\frac{\mathrm{d}y}{\mathrm{d}x}\sinh x-1 \\
\Rightarrow && \frac{\d y}{\d x} &= e^{-x}, -e^x \\
\Rightarrow && y &= -e^{-x}+C, -e^x+C \\
y(0) = 0: && C &= 1\text{ both cases } \\
y'(0) > 0: && y &= 1-e^{-x}
\end{align*}

\begin{align*}
&& 0 &= \sinh x u^2 + 2u -\sinh x \\
\Rightarrow && u &= \frac{-2 \pm \sqrt{4+4\sinh^2 x}}{2\sinh x} \\
&&&= \frac{-1 \pm \cosh x}{\sinh x} = - \textrm{cosech }x \pm \textrm{coth}x \\
\\
&& 0 &= \sinh x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}+2\frac{\mathrm{d}y}{\mathrm{d}x}-\sinh x \\
\Rightarrow && \frac{\d y}{\d x} &=  - \textrm{cosech }x \pm \textrm{coth}x  \\
\Rightarrow && y &= -\ln \left ( \tanh \frac{x}{2} \right)  \pm \ln \sinh x+C 
\end{align*}

For $x \to 0$ to be defined, we need $+$, so

\begin{align*}
&& y &= \ln \left (\frac{\sinh x}{\tanh \frac{x}{2}} \right) + C \\
&& y &= \ln \left (\frac{2\sinh \frac{x}{2} \cosh \frac{x}{2}}{\tanh \frac{x}{2}} \right)+C \\
&&&= \ln \left (2 \cosh^2 x \right) + C \\
y(0) = 0: && 0 &= \ln 2+C \\
\Rightarrow && y &= \ln(2 \cosh^2 x) -\ln 2 \\
&& y &= 2 \ln (\cosh x)
\end{align*}