Problem source

Suppose that $y$ satisfies the differential equation 
\[
y=x\frac{\mathrm{d}y}{\mathrm{d}x}-\cosh\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right).\tag{*}
\]
By differentiating both sides of $(*)$ with respect to $x$, show
that either 
\[
\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=0\qquad\mbox{ or }\qquad x-\sinh\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=0.
\]
Find the general solutions of each of these two equations. Determine the solutions of $(*)$.

Solution source

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

Therefore $\frac{\d^2y}{\d x^2} = 0$ or $ x - \sinh \left ( \frac{\d y}{\d x}\right) = 0$ as required.

\begin{align*}
&& \frac{\d ^2 y}{\d x^2} &= 0 \\
\Rightarrow && y &= ax + b \\
\\
&& 0 &=  x - \sinh \left ( \frac{\d y}{\d x}\right) \\
\Rightarrow && \frac{\d y}{\d x} &= \sinh^{-1} (x) \\
\Rightarrow && y &= x \sinh^{-1} x - \sqrt{x^2+1} + C
\end{align*}

Since it is necessary the solution satisfies one of those equations, we just need to check if either of these types of solutions work for our differential equation, ie

\begin{align*}
&& ax + b &\stackrel{?}{=} ax - \cosh(a) \\
\Rightarrow && b &= -\cosh(a) \\
\Rightarrow && y &= ax -\cosh(a) \\
\\
&& x \sinh^{-1} x - \sqrt{x^2+1} + C &\stackrel{?}{=} x\sinh^{-1} x - \cosh ( \sinh^{-1} x) \\
&&&= \sinh^{-1} x -\sqrt{x^2+1} \\
\Rightarrow && C &= 0
\end{align*}

Therefore the general solutions are, $y = ax - \cosh(a)$ and $y = x \sinh^{-1} x - \sqrt{x^2+1}$