Question

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

$\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{10}\mathbf{x}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{xe}}^{\mathbf{10}\mathbf{x}}$

Short answer

$y\text{'}\text{'}+20y\text{'}+100y=0$

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, $c_1{e}^{10x}+c_{2}x{e}^{10x}$which is $y=c_1{e}^{m_{1}x}+c_{2}x{e}^{m_{2}x}$

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

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

$m_1=m_2=10$

Using these roots, we can have

$(m-10)(m-10)=0$

Finding the differential equation from the auxiliary equation

By multiplying these brackets, we have,

$$\begin{gathered} m^2-10m+10m+100=0 \\ {m}^2-20m+100=0 \end{gathered}$$

This is the auxiliary equation for our required differential equation.

Hence, the homogeneous differential equation corresponds to the above auxiliary equation is$\mathit{y}\mathbf{\text{'}\text{'}}\mathbf{+}\mathbf{20}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{100}\mathit{y}\mathbf{=}\mathbf{0}$