Q37E - 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.

$\mathit{y}\text{'}\text{'}\text{'}\mathbf{+}\mathit{y}\text{'}\text{'}\mathbf{-}\mathbf{6}\mathit{y}\text{'}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{0}$.

Step-by-Step Solution

Answer

$y\left(t\right)=c_1{e}^t+c_2{e}^{(-1-\sqrt{5})t}+c_3{e}^{(-1+\sqrt{5})t}$ is the solution of the given equation $y\text{'}\text{'}\text{'}+y\text{'}\text{'}-6y\text{'}+4y=0$ 

Step 1: Differentiate the value of y

Given differential equation is $y\text{'}\text{'}\text{'}+y\text{'}\text{'}-6y\text{'}+4y=0$

 

Let, $y=e^{rt}$ then $y\text{'}\left(t\right)=r{e}^{rt}$ and $y\left(t\right)=r^2{e}^{rt}$

$y\text{'}\text{'}\text{'}=r^3{e}^{rt}$

Then the auxiliary equation is $r^3{e}^{rt}+r^2{e}^{rt}-6r{e}^{rt}+4e^{rt}=0$

 $$\begin{gathered} \left(r^3+r^2-6r+4\right)e^{rt}=0 \\ r^3+r^2-6r+4=0 \end{gathered}$$

Step 2: Substitute the value for r

Let substitute $r=1$ then one gets $1+1-6+4=0$

Therefore $r=1$ is the solution of the auxiliary equation $r^3+r^2-6r+4=(r-1)\left(r^2+2r-4\right)$

$(r-1)\left(r^2+2r-4\right)=0$

 

Hence, the solutions are;

$r=1$ 

 

And

$$\begin{gathered} r=\frac{-2\pm \sqrt{2^2-4\times 1\times -4}}{2\times 1} \\ r=\frac{-2\pm \sqrt{20}}{2\times 1} \\ r=-1\pm \sqrt{5} \end{gathered}$$ 

 

So, the solutions are $y\left(t\right)=c_1{e}^t+c_2{e}^{(-1-\sqrt{5})t}+c_3{e}^{(-1+\sqrt{5})t}$.