where $A$ and $B$ are set by either initial or boundary conditions. Since the $\lambda_\pm$ may be complex, so may $A$ and $B$; it's their combination that should give a real number (as $x(t)$ is real), see problem {numref}`app:solvingde`.\ref{prob:rewritesolutionsecondorderode}.
where $A$ and $B$ are set by either initial or boundary conditions. Since the $\lambda_\pm$ may be complex, so may $A$ and $B$; it's their combination that should give a real number (as $x(t)$ is real), see {numref}`pb:secondorderodesolutions`a.
In the case that equation {eq}`secondorderodesol1` gives only one solution, the corresponding exponential function is still a solution of equation {eq}`secondorderodeconstcoeff`, but it is not the most general one, as we only can put a single undetermined constant in front of it. We therefore need a second, independent solution. To guess one, here's a third useful trick<sup>[^4]</sup>: take the derivative of our known solution, $e^{\lambda t}$, with respect to the parameter $\lambda$. This gives a second Ansatz: $t e^{\lambda t}$, where $\lambda = -b/2a$. Substituting this Ansatz into equation {eq}`secondorderodeconstcoeff` for the case that $c = b^2/2a$, we find:
In the case that equation {eq}`secondorderodesol1` gives only one solution, the corresponding exponential function is still a solution of equation {eq}`secondorderodeconstcoeff`, but it is not the most general one, as we only can put a single undetermined constant in front of it. We therefore need a second, independent solution. To guess one, here's a third useful trick<sup>[^4]</sup>: take the derivative of our known solution, $e^{\lambda t}$, with respect to the parameter $\lambda$. This gives a second Ansatz: $t e^{\lambda t}$, where $\lambda = -b/2a$. Substituting this Ansatz into equation {eq}`secondorderodeconstcoeff` for the case that $c = b^2/2a$, we find:
