Question

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

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

Short answer

$y\text{'}\text{'}+2y\text{'}+2y=0$

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,

$y=c_1{e}^{-x}\cos x+c_2{e}^{-x}\sin x$

The given equation equals

$\begin{array}{rcl}y& =& k_1{e}^{-x}{e}^{ix}+k_2{e}^{-x}{e}^{-ix}\\ & =& k_1{e}^{\left(-1+i\right)x}+k_2{e}^{\left(-1-i\right)x}\end{array}$

Where$k_1+k_2=c_1\text{and}{k}_1-k_2=c_2$which is in the form of$y=k_1{e}^{m_{1}x}+k_{2}x{e}^{m_{2}x}$

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

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

$m_1=-1+i,m_2=-1-i$width="170">m1=-1+i,m2=-1-i

Using these roots, we can have

$$\begin{gathered} \left(m-\left(-1+i\right)\right)\left(m-\left(-1-i\right)\right)=0 \\ \left(m+\left(1-i\right)\right)\left(m+\left(1+i\right)\right)=0 \end{gathered}$$

Finding the differential equation from the auxiliary equation

By multiplying these brackets, we have,

$$\begin{gathered} m^2+\left(1+i\right)m+\left(1-i\right)m+\left(1-i\right)\left(1+i\right)=0 \\ {m}^2+2m+1-i^2=0 \\ {m}^2+2m+2=0 \end{gathered}$$

This is the auxiliary equation for our required differential equation.

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