Elimination of arbitrary constant

Question

using elimination of arbitrary constant, determine differential equations satisfied by the following functions: v=c1e^-w + c2 we^-w + c3 e^5w

a. D³v - 9D²v - 3Dv - 5v = 0

b. D³v - 3D²v - 9Dv + 5v = 0

c. D³v - 5D²v - 3Dv -5v = 0

d. D³v - 3D²v - 9Dv -5v = 0

Answer 1

The solution is a linear combination of e^(-w), we^(-w), and e^(5w). This shows that the characteristic equation has a root of -1 with multiplicity 2, and root of 5. That is

(w +1)²(w -5) = (w² +2w +1)(w -5) = w³ -3w² -9w -5

Which shows that the solution is D.