The solutions to the linear differential equation d2u/dt2 = u form a vector space (since combinations of solutions are still solutions). find two independent solutions, to give a basis for that solution space.

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DeanR

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]


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