Question

Given the third-order differential equation z′′′+2z′′−4z′−8z=0, find the general solution. Enter your solution as z(x)=... . In your answer, use c1 and c2 to denote arbitrary constants and x the independent variable. Enter c1 as c1, c2 as c2 and c3 as c3.

Answer 1

Answer: $z(x) = c_{1} e^{2x} +c_{2} e^{-2x} + c_{3} xe^{-2x}$

Step-by-step explanation:

This is a homogeneous linear third-order differential equation

z′′′+2z′′−4z′−8z=0 so the

FIRST STEP is to find the characteristic equation and its roots

$m^{3} +2m^{2} -4m -8 = 0$

Using the method of finding roots of a polynomial (using m = 2) would provide the solution below

(m-2)(m+2)(m+2)=0;  m = 2 and m = -2 (repeated twice)

SECOND STEP is to write the general solution with the arbitrary constants.

The general solution based on the roots and the using x as the independent variable would provide

$z(x) = c_{1} e^{2x} +c_{2} e^{-2x} + c_{3} xe^{-2x}$

*It should be noted that when the characteristic equation has a repeated root, the general equation form becomes similar to the last part of the answer*