Question

In problems,49–58 find a homogeneous linear differential equation with

constant coefficients whose general solution is given.

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

Short answer

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

Step-by-step solution

Solving the given general solution

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

$y=c_{1}e°+c_2{e}^{2x}$

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

$m_1=0$and$m_2=2$

Finding homogeneous linear differential equation

After that, using these roots we can have

$$\begin{gathered} (m-0)(m-2)=0 \\ m(m-2)=0 \end{gathered}$$

Multiply the two, then we can obtain

$m^2-2m=0$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

role="math" localid="1664267347125" $y\text{'}\text{'}-2y\text{'}=0$

Therefore, the homogeneous linear differential equation with constant coefficients using the given general solution is found to be $\mathit{y}\mathbf{\text{'}\text{'}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0}$.