Question

Two roots of a cubic auxiliary equation with real coefficients are ${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}$ and ${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{+}\mathit{i}$. What is the corresponding homogeneous linear differential equation? Discuss: Is your answer unique?

Short answer

$2y\text{'}\text{'}\text{'}-11y\text{'}\text{'}+14y+10=0$

Step-by-step solution

Find the required differential equation

As we know, if an equation has a complex root, then it also has another complex conjugate root.

Hence the roots of the cubic auxiliary equation is$m_1=-\frac{1}{2},m_2=3+i,m_3=3-i$.

Use the given roots of the cubic auxiliary equation

Using these roots, we can have

$\begin{array}{l}m+\frac{1}{2}\left(m-\left(3+i\right)\right)\left(m-\left(3-i\right)=0\right)\\ \frac{2m+1}{2}\left(m-3-i\right)\left(m-3+i\right)=0\\ \left(2m+1\right)\left({\left(m-3\right)}^2-i^2\right)=0\\ \left(2m+1\right)\left(m^2+9-6m+1\right)=0\end{array}$

Solve further as:

$\begin{array}{l}\left(2m+1\right)\left(m^2-6m+10\right)=0\\ 2m^3+m^2-12m^2-6m+20m+10=0\\ 2m^3-11m^2+14m+10=0\end{array}$

This is the auxiliary equation for our required differential equation.

Hence, the homogeneous differential equation corresponds to the above auxiliary equation is $2y\text{'}\text{'}\text{'}-11y\text{'}\text{'}+14y+10=0$.