Do the functions y1x e2x y2x 1 form a basis for the soluti

Do the functions y_1(x) = e^2x, y_2(x) = 1 form a basis for the solution space of the DE y\"\'(x) - 2y\"(x) = 0? Justify your answer. Solution: Quick answer: this is a third-order linear DE, so there must be three LI solution functions in order to span the solution space, but only two are presented, so there must be one more that is missing. Long answer: r^3 - 2r^2 = r^2(r - 2) = 0 r = 0, 0, 2 There is a repeated root so span{1, t, e^2t} span the solution space.

Solution

We have a third order ode so we expect three fundamental solutoins

Substituting y=exp(rx) gives the characteristic

r^3-2r^2=0

r=2, r=0

r=0 is repeated root

So, general solutin is

y=Ae^{2t}+B +Ct

So basis for solution is

{1,t,e^{2t}}

t is a solution because r=0 was a repeated root

When we have r=k as a repeated root

Then, exp(kt) and t exp(kt) are both solutions