2.1.8.6 ✓ Maple. Time used: 0.045 (sec). Leaf size: 14

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

\begin{align*} x = -\ln \left (-b t +{\mathrm e}^{-1}\right ) \end{align*}

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
<- separable successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}x \left (t \right )=b \,{\mathrm e}^{x \left (t \right )}, x \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d t}x \left (t \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}x \left (t \right )=b \,{\mathrm e}^{x \left (t \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d t}x \left (t \right )}{{\mathrm e}^{x \left (t \right )}}=b \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {\frac {d}{d t}x \left (t \right )}{{\mathrm e}^{x \left (t \right )}}d t =\int b d t +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{{\mathrm e}^{x \left (t \right )}}=b t +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \left (t \right ) \\ {} & {} & x \left (t \right )=\ln \left (-\frac {1}{b t +\mathit {C1}}\right )+2 \,\mathrm {I} \pi \_Z1 \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=1 \\ {} & {} & 1=\ln \left (-\frac {1}{\mathit {C1}}\right )+2 \,\mathrm {I} \pi \_Z1 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \\ {} & {} & \mathit {C1} =-\frac {1}{{\mathrm e}^{1-2 \,\mathrm {I} \pi \_Z1}} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \textit {\_C1} =-\frac {1}{{\mathrm e}^{1-2 \,\mathrm {I} \pi \_Z1}}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & x \left (t \right )=\ln \left (\frac {1}{{\mathrm e}^{-1+2 \,\mathrm {I} \pi \_Z1}-b t}\right )+2 \,\mathrm {I} \pi \_Z1 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x \left (t \right )=\ln \left (\frac {1}{{\mathrm e}^{-1+2 \,\mathrm {I} \pi \_Z1}-b t}\right )+2 \,\mathrm {I} \pi \_Z1 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \left \{\_Z1 \right \} \\ {} & {} & x \left (t \right )=\ln \left (\frac {1}{{\mathrm e}^{-1}-b t}\right ) \end {array} \]