Q38E - 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. $\mathit{y}\text{'}\text{'}\text{'}\mathbf{-}\mathbf{6}\mathit{y}\text{'}\text{'}\mathbf{-}\mathit{y}\text{'}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{=}\mathbf{0}$.

Step-by-Step Solution

Answer

$y\left(t\right)=c_1{e}^t+c_2{e}^{-t}+c_3{e}^{6t}$ is the solution of the given equation $y\text{'}\text{'}\text{'}-6y\text{'}\text{'}-y\text{'}+6y=0$.

Step 1: Differentiate the value of y

Given differential equation is $y\text{'}\text{'}\text{'}-6y\text{'}\text{'}-y\text{'}+6y=0$

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

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

 

Then the auxiliary equation is:

$$\begin{gathered} r^3{e}^{rt}-6r^2{e}^{rt}-r{e}^{rt}+6e^{rt}=0 \\ \left(r^3-6r^2-r+6\right)e^{rt}=0 \\ r^3-6r^2-r+6=0 \end{gathered}$$ 

Step 2: Substitute the value for r

Let substitute $r=1$

$1-6-1+6=0$ 

 

Therefore $r=1$ is the solution of the auxiliary equation $r^3-6r^2-r+6=(r-1)\left(r^2-5r-6\right)$

$(r-1)\left(r^2-5r-6\right)=0$

Hence, the solutions are $r=1$ and $\left(r+1\right)\left(r-6\right)=0$

$$\begin{gathered} r=1,-1,6 \\ y\left(t\right)=c_1{e}^t+c_2{e}^{-t}+c_3{e}^{6t} \end{gathered}$$