Question

In Problems 65–68 use a computer either as an aid in solving the auxiliary equation or as a means of directly obtaining the general solution of the given differential equation. If you use a CAS to obtain the general solution, simplify the output and, if necessary, write the solution in terms of real functions.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{ }$

Short answer

$y =k_1{e}^{5.612 x}+k_2{e}^{ 0.6587 x}+k_3{e}^{-0.2705 x}$

Step-by-step solution

General solution

The general solution of the DE is

Solving the third order differential equation

We have the third order differential equation

$y\text{'}\text{'}\text{'}-6y\text{'}\text{'}+2y\text{'}+y=0······ \left(1\right)$

Consider$y=e^{mx}$as the solution of the differential equation,

Substitute$y=e^{mx},y\text{'}=m{e}^{mx}$ and$y\text{'}\text{'}=m^2{e}^{mx}andy\text{'}\text{'}\text{'}=m^3{e}^{mx}$ into$y\text{'}\text{'}\text{'}-6y\text{'}\text{'}+2y\text{'}+y=0$ to obtain the auxiliary equation.

$\begin{array}{c}{m}^3 e^{mx}-6m^2 e^{mx}+2m e^{mx}+e^{mx}=0\\ e^{mx}\left(m^3-6m^2+2m+1\right)=0\end{array}$

For any realrole="math" localid="1663914297647" $x,e^{mx}\ne 0$, we have

$m^3-6m^2+2m+1=0$

This auxiliary equation cannot be solved easily, then using Computer Algebra System we can solve it and obtain the roots as

$m_1=5.612 , m_2=0.6587\mathrm{and}{m}_3=-0.2705$

which are real and distinct.

Then the general solution of this homogeneous equation$y\text{'}\text{'}\text{'}-6y\text{'}\text{'}+2y\text{'}+y=0$ can be given as

$\begin{array}{l}y=k_1{e}^{m_1 x}+k_2{e}^{m_2 x}+k_3{e}^{m_3 x}\\ =k_1{e}^{5.612 x}+k_2{e}^{ 0.6587 x}+k_3{e}^{-0.2705 x}\end{array}$

Where$k_1,k_2$and$k_3$are arbitrary constants.

Hence, the final solution is$y =k_1{e}^{5.612 x}+k_2{e}^{ 0.6587 x}+k_3{e}^{-0.2705 x}$