Question

Find the general solution of y''' + 6y'' + y' − 34y = 0 if it is known that y1 = e−4x cos(x) is one solution.

Answer 1

Answer:

Step-by-step explanation:

Given is a differential equation of III order,

$y''' + 6y'' + y' - 34y = 0$

The characteristic equation would be cubic as

$m^3+6m^2+m-34=0$

By trial and error, we find that

$f(2) = 2^3+6(2^2)+2-34 =0$

Thus m=2 is one solution

Since given that $e^{-4x} cos x$is one solution we get

m = -4+i and hence other root is conjugate $m=-4-i$

Hence general solution would be

$y=Ae^{2x} +e^{-4x} (Bcosx +C sinx)$