Question

In problems49–58 find a homogeneous linear differential equation with

constant coefficients whose general solution is given.

$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{-}\mathbf{4}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{-}\mathbf{3}\mathbf{x}}$

Short answer

$y\text{'}\text{'}+7y\text{'}+12y=0$

Step-by-step solution

Solving the given general solution

We have a general solution for a homogeneous second order differential equation as$y=c_1{e}^{-4x}+c_2{e}^{-3x}$

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

$m_1=-4and{m}_2=-3$

Finding homogeneous linear differential equation

After that, using these roots we can have

$$\begin{gathered} \left(m-(-4)\right)\left(m-(-3)\right)=0 \\ (m+4)(m+3)=0 \end{gathered}$$

Multiply the two, then we can obtain

$$\begin{gathered} m^2+4\mathrm{m}+3m+12=0 \\ {m}^2+7m+12=0 \end{gathered}$$

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

$y\text{'}\text{'}+7y\text{'}+12y=0$

Therefore, the homogeneous linear differential equation with constant coefficients using the given general solution is found to be $\mathit{y}\mathbf{\text{'}\text{'}}\mathbf{+}\mathbf{7}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{12}\mathit{y}\mathbf{=}\mathbf{0}$