Question

In problems49–58Find a homogeneous linear differential equation with constant coefficient whose general solution is given.$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{x}\mathbf{+}{\mathit{c}}_{\mathbf{3}}{\mathit{e}}^{\mathbf{8}\mathbf{x}}$

Short answer

$y\text{'}\text{'}\text{'}-8y\text{'}\text{'}=0$

Step-by-step solution

Finding the roots of the required differential equation

A general solution for a homogeneous third order differential equation is given as,

$y=c_1+c_{2}x+c_3{e}^{8x}$

which is in the form of $\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{0}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{x}{\mathit{e}}^{\mathbf{0}}\mathbf{+}{\mathit{c}}_{\mathbf{3}}{\mathit{e}}^{{\mathbf{m}}_{\mathbf{3}}\mathbf{x}}$

Here, $m_1,m_2,m_3$are the roots of the required differential equation.

Form the given general solution, we can see that the roots,

$m_{1,2}=0,m_3=8$

Using these roots, we can have

$\begin{array}{c}{\left(m-0\right)}^2\left(m-8\right)=0\\ m^2\left(m-8\right)=0\end{array}$

Finding the differential equation from the auxiliary equation

By multiplying these brackets, we have,

$\begin{array}{c}{m}^2\left(m-8\right)=0\\ m^3-8m^2=0\end{array}$

This is the auxiliary equation for our required differential equation.

Hence, the homogeneous differential equation corresponds to the above auxiliary equation is $y\text{'}\text{'}\text{'}-8y\text{'}\text{'}=0$.