2.6.3.3 ✓ Maple. Time used: 0.282 (sec). Leaf size: 38

ode:=diff(diff(y(x),x),x) = m/k*(1+diff(y(x),x)^2)^(1/2); 
dsolve(ode,y(x), singsol=all);
 

\begin{gather*} \begin {aligned} y &= -i x +c_{1}\\ y &= i x +c_{1}\\ y &= \frac {k \cosh \left (\frac {m \left (x +c_{1} \right )}{k}\right )}{m}+c_{2}\\ \end {aligned} \end{gather*}

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying 2nd order Liouville 
trying 2nd order WeierstrassP 
trying 2nd order JacobiSN 
differential order: 2; trying a linearization to 3rd order 
trying 2nd order ODE linearizable_by_differentiation 
-> Calling odsolve with the ODE, diff(diff(diff(y(x),x),x),x)-m^2/k^2*diff(y(x),x), y(x) 
   *** Sublevel 2 *** 
   Methods for third order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   checking if the LODE has constant coefficients 
   <- constant coefficients successful 
trying 2nd order, 2 integrating factors of the form mu(x,y) 
trying differential order: 2; missing variables 
   -> Computing symmetries using: way = 3 
-> Calling odsolve with the ODE, diff(_b(_a),_a) = m/k*(1+_b(_a)^2)^(1/2), _b(_a), HINT = [[1, 0]] 
   *** Sublevel 2 *** 
   symmetry methods on request 
   1st order, trying reduction of order with given symmetries: 
[1, 0] 
   1st order, trying the canonical coordinates of the invariance group 
   <- 1st order, canonical coordinates successful 
<- differential order: 2; canonical coordinates successful 
<- differential order 2; missing variables successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=\frac {m \sqrt {1+\left (\frac {d}{d x}y \left (x \right )\right )^{2}}}{k} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d x}u \left (x \right )=\frac {m \sqrt {1+u \left (x \right )^{2}}}{k} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}u \left (x \right )=\frac {m \sqrt {1+u \left (x \right )^{2}}}{k} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}u \left (x \right )}{\sqrt {1+u \left (x \right )^{2}}}=\frac {m}{k} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}u \left (x \right )}{\sqrt {1+u \left (x \right )^{2}}}d x =\int \frac {m}{k}d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \mathrm {arcsinh}\left (u \left (x \right )\right )=\frac {m x}{k}+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\sinh \left (\frac {\mathit {C1} k +m x}{k}\right ) \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & u \left (x \right )=\sinh \left (\frac {m x}{k}+\mathit {C1} \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\sinh \left (\frac {m x}{k}+\mathit {C1} \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=\sinh \left (\frac {m x}{k}+\mathit {C1} \right ) \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right )d x =\int \sinh \left (\frac {m x}{k}+\mathit {C1} \right )d x +\mathit {C2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {k \cosh \left (\frac {m x}{k}+\mathit {C1} \right )}{m}+\mathit {C2} \end {array} \]