Question

Question: In problems49–58 find a homogeneous linear differential equation with

constant coefficients whose general solution is given.

$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{5}\mathbf{x}}$

Short answer

$y-6y\text{'}+5y=0$

Step-by-step solution

Solve the given general solution

We have a general solution for a homogeneous second order differential equation as

$y=c_1{e}^x+c_2{e}^{5x}$

as the form, $y=c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x}$

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

From this general solution we can see that we have the roots

$m_1=1$and$m_2=5$

Find the homogeneous linear differential equation

After that, using these roots we can have

$\left(m-1\right)\left(m-5\right)=0$

Multiply the two, then we can obtain

$$\begin{gathered} m^2-5m-m+5=0 \\ {m}^2-6m+5=0 \end{gathered}$$

is the auxiliary equation of our required differential equation.

After that we can obtain the homogeneous differential equation that corresponds to this

Auxiliary equation as

$y-6y\text{'}+5y=0$

Therefore,thehomogeneous linear differential equation with constant coefficients using the given general solution is found to be $y-6y\text{'}+5y=0$.