Question

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

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

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

$$\begin{gathered} y=k_1{e}^{-x}{e}^{ix}+k_2{e}^{-x}{e}^{-ix} \\ y=k_1{e}^{(-1+i)x}+k_2{e}^{(-1-i)x} \end{gathered}$$

where $k_1+k_2=c_1$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, $m_1$and $m_2$are the roots of the required differential equation.

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

$m_1=-1+i,m_2=-1-i$

Using these roots, we can have

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

Finding the differential equation from the auxiliary equation

By multiplying these brackets, we have,

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

This is the auxiliary equation for our required differential equation.

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