2.2.10.5 ✓ Maple. Time used: 0.019 (sec). Leaf size: 10

ode:=diff(diff(x(t),t),t)+2*diff(x(t),t)+x(t) = 0; 
ic:=[x(0) = 0, D(x)(0) = 1]; 
dsolve([ode,op(ic)],x(t), singsol=all);
 

\begin{align*} x = {\mathrm e}^{-t} t \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} \\ {} & {} & \left [\frac {d^{2}}{d t^{2}}x \left (t \right )+2 \frac {d}{d t}x \left (t \right )+x \left (t \right )=0, x \left (0\right )=0, \left (\frac {d}{d t}x \left (t \right )\right )\bigg | {\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1\right ] \\ \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}+2 r +1=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +1\right )^{2}=0 \\ \bullet & {} & \textrm {Root of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =-1 \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & x_{1}\left (t \right )={\mathrm e}^{-t} \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} x_{1}\left (t \right )\hspace {3pt}\textrm {by}\hspace {3pt} t \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & x_{2}\left (t \right )=t \,{\mathrm e}^{-t} \\ \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}^{-t}+\mathit {C2} t \,{\mathrm e}^{-t} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} x \left (t \right )=\textit {\_C1} {\mathrm e}^{-t}+\textit {\_C2} t {\mathrm e}^{-t} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=0 \\ {} & {} & 0=\textit {\_C1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & \frac {d}{d t}x \left (t \right )=-\textit {\_C1} \,{\mathrm e}^{-t}+\textit {\_C2} \,{\mathrm e}^{-t}-\textit {\_C2} t \,{\mathrm e}^{-t} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} \left (\frac {d}{d t}x \left (t \right )\right )\bigg | {\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-\textit {\_C1} +\textit {\_C2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \hspace {3pt}\textrm {and}\hspace {3pt} \textit {\_C2} \\ {} & {} & \left \{\textit {\_C1} =0, \textit {\_C2} =1\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & x \left (t \right )=t \,{\mathrm e}^{-t} \\ \bullet & {} & \textrm {Solution to the}\hspace {3pt} \hspace {3pt}\textrm {IVP}\hspace {3pt} \\ {} & {} & x \left (t \right )=t \,{\mathrm e}^{-t} \end {array} \]