The standard method of solution is as follows. Differentiate , $${\dot x} = {\dot x} + t\,{\ddot x}  + \psi^\prime({\dot x})\,{\ddot x}\,,$$ and simplify the result to obtain $${\ddot x}\,\left(t + \psi^\prime({\dot x})\right) = 0\,.$$ This offers two possibilities: \[eq:ddot\] [x]{} = 0 and/or \[eq:other\] t + \^([x]{}) = 0. Solve each of the equations with the initial condition $x(t_0) = x_0$. The first equation implies \[sol:ddot\] x(t) = x\_0 + c(t - t\_0) with a constant $c$. Substitute this into the original equation : $$x_0 + c\,(t - t_0) = t\,c + \psi(c)\,.$$ This yields the equation for the possible value(s) of $c$: \[sol:c\] x\_0 = c t\_0 + (c). Assuming $c(t_0, x_0)$ is a solution of (there may be many solutions), we can re-write the solution as follows: \[sol:ddot-2\] x(t) = (c(t\_0, x\_0)) + c(t\_0, x\_0)t. The second equation requires some kind of invertibility of the function $\psi^\prime$ (the simplest case is when $\psi^{\prime\prime} \neq 0$, i.e., $\psi$ is convex or concave). Assuming the inverse function $(\psi^\prime)^{-1}$ makes sense, we obtain \[eq:other-2\] [x]{} = (\^)\^[-1]{}(-t). After integration, \[sol:other\] x(t) = x\_0 + \_[t\_0]{}\^t (\^)\^[-1]{}(-s)ds. Substitute this expression into : \[eq:temp-11\] x\_0 + \_[t\_0]{}\^t (\^)\^[-1]{}(-s)ds = t (\^)\^[-1]{}(-t) + 