Each of these ODEs is linear and homogeneous with constant coefficients, so we only need to find the roots to their respective characteristic equations.
(a) The characteristic equation for
[tex]y'' - 2y' - 8y = 0[/tex]
is
[tex]r^2 - 2r - 8 = (r - 4) (r + 2) = 0[/tex]
which arises from the ansatz [tex]y = e^{rx}[/tex].
The characteristic roots are [tex]r=4[/tex] and [tex]r=-2[/tex]. Then the general solution is
[tex]\boxed{y = C_1 e^{4x} + C_2 e^{-2x}}[/tex]
where [tex]C_1,C_2[/tex] are arbitrary constants.
(b) The characteristic equation here is
[tex]25r^2 - 20r + 4 = (5r - 2)^2 = 0[/tex]
with a root at [tex]r=\frac25[/tex] of multiplicity 2. Then the general solution is
[tex]\boxed{y = C_1 e^{2/5\,x} + C_2 x e^{2/5\,x}}[/tex]
(c) The characteristic equation is
[tex]r^2 + 2r + 2 = (r + 1)^2 + 1 = 0[/tex]
with roots at [tex]r = -1 \pm i[/tex], where [tex]i=\sqrt{-1}[/tex]. Then the general solution is
[tex]y = C_1 e^{(-1+i)x} + C_2 e^{(-1-i)x}[/tex]
Recall Euler's identity,
[tex]e^{ix} = \cos(x) + i \sin(x)[/tex]
Then we can rewrite the solution as
[tex]y = C_1 e^{-x} (\cos(x) + i \sin(x)) + C_2 e^{-x} (\cos(x) - i \sin(x))[/tex]
or even more simply as
[tex]\boxed{y = C_1 e^{-x} \cos(x) + C_2 e^{-x} \sin(x)}[/tex]
