Serious high school. This is one of the few differential equations I can solve.
The usual particular solution is [tex]u=e^t[/tex] because [tex]e^t[/tex] is its own derivative.
An independent solution is [tex]u=e^{-t}[/tex] which has a negative sign in the first derivative which turns back to positive in the second.
The arbitrary linear combination spans the solution space:
[tex]u= c_1 e^t + c_2 e^{-t}[/tex]
But we only are asked for the basis.
Answer:[tex]\textrm{. } \quad e^t, \quad e^{-t}[/tex]
