Question

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

$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathbf{}\mathbf{7}\mathit{x}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathbf{}\mathbf{7}\mathit{x}$

Short answer

$y\text{'}\text{'}-49y=0$

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,

$y=c_1\cosh 7x+c_2\sinh 7x$

Since we have$\cosh 7x=\frac{e^{7x}+e^{-7x}}{2}\text{and}\sinh 7x=\frac{e^{7x}-e^{-7x}}{2}$

We can obtain the general solution as

$\begin{array}{rcl}y& =& c_1\frac{e^{7x}+e^{-7x}}{2}+c_2\frac{e^{7x}-e^{-7x}}{2}\\ & =& \left(c_1+c_2\right)\frac{e^{7x}}{2}+\left(c_1-c_2\right)\frac{e^{-7x}}{2}\\ & =& \frac{c_1+c_2}{2}{e}^{7x}+\frac{c_1-c_2}{2}{e}^{-7x}\\ & =& k_1{e}^{7x}+k_2{e}^{-7x}\end{array}$

Where$k_1=\frac{c_1+c_2}{2}\text{and}{k}_2=\frac{c_1-c_2}{2}$ is in the form of$y=c_1{e}^{m_{1}x}+c_{2}x{e}^{m_{2}x}$

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

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

$m_1=7,m_2=-7$

Using these roots, we can have

$$\begin{gathered} \left(m-7\right)\left(m-\left(-7\right)\right)=0 \\ \left(m-7\right)\left(m+7\right)=0 \end{gathered}$$

Finding the differential equation from the auxiliary equation

By multiplying these brackets, we have,

$$\begin{gathered} m^2+7m-7m-49=0 \\ {m}^2-49=0 \end{gathered}$$

This is the auxiliary equation for our required differential equation.

Hence, the homogeneous differential equation corresponds to the above auxiliary equation is$y\text{'}\text{'}-49y=0$