Q39E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find three linearly independent solutions (see Problem 35) of the given third-order differential equation and write a general solution as an arbitrary linear combination of it.  $\mathit{z}\text{'}\text{'}\text{'}\mathbf{+}\mathbf{2}\mathit{z}\text{'}\text{'}\mathbf{-}\mathbf{4}\mathit{z}\text{'}\mathbf{-}\mathbf{8}\mathit{z}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

$z\left(t\right)=c_1{e}^{-2t}+c_{2}t{e}^{-2t}+c_3{e}^{2t}$ is the solution of the given equation $z\text{'}\text{'}\text{'}+2z\text{'}\text{'}-4z\text{'}-8z=0$

Step 1: Differentiate the value of z

Given differential equation is $z\text{'}\text{'}\text{'}+2z\text{'}\text{'}-4z\text{'}-8z=0$  

Let $z=e^{rt}$, then $z\text{'}\left(t\right)=r{e}^{rt}$ and

$$\begin{gathered} z\text{'}\text{'}\left(t\right)=r^2{e}^{rt} \\ z\text{'}\text{'}\text{'}=r^3{e}^{rt} \end{gathered}$$ 

 

Then the auxiliary equation is $r^3{e}^{rt}+2r^2{e}^{rt}-4r{e}^{rt}-8e^{rt}=0$

$$\begin{gathered} \left(r^3+2r^2-4r-8\right)e^{rt}=0 \\ r^3+2r^2-4r-8=0 \end{gathered}$$ 

Step 2: Substitute the value for r

Let substitute $r=1$ 

$1^3+2(1{)}^2-4-8\ne 0$ 

 

Now substitute $r=2$  

$2^3+2(2{)}^2-4\left(2\right)-8=0$ 

 

Therefore $r=2$ is the solution of the auxiliary equation $r^3+2r^2-4r-8=(r-2)\left(r^2+4r+4\right)$ 

$(r-2)\left(r^2+4r+4\right)=0$ 

 

Therefore, the solutions are $r=2$ and ${(r+2)}^2=0$ 

$$\begin{gathered} r=-2,-2,2 \\ z\left(t\right)=c_1{e}^{-2t}+c_{2}t{e}^{-2t}+c_3{e}^{2t} \end{gathered}$$