Q41E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find three linearly independent solutions (see Problem35) of the given third-order differential equation and write a general solution as an arbitrary linear combination of it.

y"' + 3y" - 4y' - 12y = 0.

Step-by-Step Solution

Answer

The three independent solutions of the given differential equation are: 

y₁(t) = e^2t,y₂(t) = e^-2t and y₃(t) = e^-3t

The general solution is y(t) = c₁e^2t + c₂e^-2t + c₃e^-3t

Step 1: Finding the roots of the auxiliary equation.

The given differential equation is y"' + 3y" - 4y' - 12y = 0 

The auxiliary equation for the given differential equation is r³ + 3r² - 4r - 12 = 0

Finding the roots of the auxiliary equation.

$\begin{array}{rcl}{r}^3+3r^2-4r-12& =& 0\\ r^2(r+3)-4(r+3)& =& 0\\ (r-2)(r+2)(r+3)& =& 0\\ r& =& 2,-2,-3\end{array}$

Hence, the roots for the auxiliary equation are 2,-2 and -3.

Step 2: Substitute the root values

Therefore the three independent solutions of the given differential equation are: 

 

y₁(t) = e^2t,y₂(t) = e^-2t and y₃(t) = e^-3t

 

Therefore, the general solution is y(t) = c₁e^2t + c₂e^-2t + c₃e^-3t.