Question

In problems 49–58Find a homogeneous linear differential equation with constant coefficient whose general solution is given.

$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{2}\mathbf{x}}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{5}\mathit{x}\mathbf{+}{\mathit{c}}_{\mathbf{3}}{\mathit{e}}^{\mathbf{2}\mathbf{x}}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{5}\mathit{x}$

Short answer

$y\text{'}\text{'}\text{'}-4y\text{'}\text{'}+29y=0$

Step-by-step solution

 Step 1: Finding the roots of the required differential equation

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

$y=c_1+c_2{e}^{2x}\cos 5x+c_3{e}^{2x}\sin 5x$

The given equation equals

$$\begin{gathered} y=c_1{e}^{°}+k_1{e}^{2x}{e}^{5ix}+k_2{e}^{2x}{e}^{-5ix} \\ y=c_1{e}^{°}+k_1{e}^{(2+5i)x}+k_2{e}^{(2-5i)x} \end{gathered}$$

where $$" width="9">k1+k2=c2and $k_1-k_2=c_3$which is in the form of $y=c_1{e}^{°}+k_1{e}^{m_{1}x}+k_{2}x{e}^{m_{2}x}$

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

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

$m_1=0,m_2=2+5i,m_3=2-5i$

Using these roots, we can have

$$\begin{gathered} m\left[m-(2+5i)\right]\left[m-(2-5i)\right]=0 \\ m\left[m+(-2-5i)\right]\left[m+(-2+5i)\right]=0 \end{gathered}$$

Finding the differential equation from the auxiliary equation

By multiplying these brackets, we have,

$$\begin{gathered} m\left(m^2+(-2+5i\right)m+\left(-2-5i\right)m+(-2-5i)(-2+5i)=0 \\ m(m^2-4m+4-25i^2)=0 \\ m(m^2-4m+29)=0 \\ {m}^3-4m^2+29m \end{gathered}$$

This is the auxiliary equation for our required differential equation.

Hence, the homogeneous differential equation corresponds to the above auxiliary equation is${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{\text{'}\text{'}}\mathbf{+}\mathbf{29}\mathit{y}\mathbf{=}\mathbf{0}$.