2.2.6.3 ✓ Maple. Time used: 0.002 (sec). Leaf size: 18

ode:=diff(diff(x(t),t),t)-4*diff(x(t),t)+5*x(t) = 0; 
dsolve(ode,x(t), singsol=all);
 

\begin{align*} x = {\mathrm e}^{2 t} \left (c_1 \sin \left (t \right )+c_2 \cos \left (t \right )\right ) \end{align*}

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}x \left (t \right )-4 \frac {d}{d t}x \left (t \right )+5 x \left (t \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d t^{2}}x \left (t \right ) \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-4 r +5=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {4\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (2-\mathrm {I}, 2+\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & x_{1}\left (t \right )={\mathrm e}^{2 t} \cos \left (t \right ) \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & x_{2}\left (t \right )={\mathrm e}^{2 t} \sin \left (t \right ) \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & x \left (t \right )=\mathit {C1} x_{1}\left (t \right )+\mathit {C2} x_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & x \left (t \right )=\mathit {C1} \,{\mathrm e}^{2 t} \cos \left (t \right )+\mathit {C2} \,{\mathrm e}^{2 t} \sin \left (t \right ) \end {array} \]