Need serious help Homogeneous Differential Equation

Given (d^2x/dt^2) + 2p(dx/dt) + 169x = 0

which satisfies the conditions t=0, x=0 and dx/dt = 8

Find value of p such that the characteristics equation has repeated roots and determine x in term of t?

answer p=13

Characteristic equation: r^2 + 2pr + 169 = 0
has repeated roots when discriminant = 0

(-2p)^2 - 4(169) = 0
4p^2 = 4(169)
p^2 = 169
p = ± 13

When p = 13, we get characteristic equation
r^2 + 26r + 269 = 0
(r + 13)^2 = 0
r = -13 (double root)

Differential equation: x(t) = A e^(-13t) + B t e^(-13t)

When p = -13, we get characteristic equation
r^2 - 26r + 269 = 0
(r - 13)^2 = 0
r = 13 (double root)

Differential equation: x(t) = A e^(13t) + B t e^(13t)

When x = 0, x(0) = 0
A * 1 + B * 0 = 0 -----> A = 0

x(t) = B t e^(±13t)
x'(t) = ±13B t e^(±13t) + B e^(±13t)

When x = 0, x'(0) = 8

±13 B * 0 + B * 1 = 8 -----> B = 8


Solution: x(t) = 8 t e^(-13t) . . . or . . . x(t) = 8 t e^(13t)


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