Answer:
Step-by-step explanation:
Given is a differential equation of III order,
[tex]y''' + 6y'' + y' - 34y = 0[/tex]
The characteristic equation would be cubic as
[tex]m^3+6m^2+m-34=0[/tex]
By trial and error, we find that
[tex]f(2) = 2^3+6(2^2)+2-34 =0\\[/tex]
Thus m=2 is one solution
Since given that [tex]e^{-4x} cos x[/tex]is one solution we get
m = -4+i and hence other root is conjugate [tex]m=-4-i[/tex]
Hence general solution would be
[tex]y=Ae^{2x} +e^{-4x} (Bcosx +C sinx)[/tex]
