Question

Question: In problems49–58 Find a homogeneous linear differential equation with constant coefficient whose general solution is given.

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

Short answer

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

Step-by-step solution

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

$\begin{array}{rcl}y& =& c_1{e}^0+k_1{e}^{2x}{e}^{5ix}+k_2{e}^{2x}{e}^{-5ix}\\ & =& c_1{e}^0+k_1{e}^{\left(2+5i\right)x}+k_2{e}^{\left(2-5i\right)x}\end{array}$

Where$k_1+k_2=c_{2}and{k}_1-k_2=c_3$which is in the form of$y=c_1{e}^0+k_1{e}^{m_{1}x}+k_{2}x{e}^{m_{2}x}$

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

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

Finding the differential equation from the auxiliary equation

By multiplying these brackets, we have,

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

This is the auxiliary equation for our required differential equation.

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