Question

Find the general solution of $\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{34}\mathit{y}\mathbf{=}\mathbf{0}$if it is known that${\mathit{y}}_{\mathbf{1}}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{4}\mathbf{x}}\mathit{c}\mathit{o}\mathit{s}\mathit{x}$ is one solution.

Short answer

$y=c_1{e}^{-4x}\cos x+c_2{e}^{-4x}\sin x+c_3{e}^{2x}$

Step-by-step solution

Finding the auxiliary equation using the given differential equation

We have the third order differential equation$y\text{'}\text{'}\text{'}+6y\text{'}\text{'}+y\text{'}-34y=0$ with one solution $y_1=e^{-4x}\cos x$, and we need to find the general solution.

Consider$y=e^{mx}$as the solution of the differential equation,

Substitute $y=e^{mx},y\text{'}=m{e}^{mx}$and $y\text{'}\text{'}=m^2{e}^{mx}andy\text{'}\text{'}\text{'}=m^3{e}^{mx}$into $y\text{'}\text{'}\text{'}+6y\text{'}\text{'}+y\text{'}-34y=0$to obtain the auxiliary equation.

$\begin{array}{c}{m}^3{e}^{mx}+6m^2{e}^{mx}+m{e}^{mx}-34e^{mx}=0\\ e^{mx}\left(m^3+6m^2+m-34\right)=0\end{array}$

For any real$x,e^{mx}\ne 0$, we have

role="math" localid="1663910185237" $m^3+6m^2+m-34=0······\left(1\right)$

This is the auxiliary equation of the given differential equation.

Finding the other roots of the auxiliary equation and the general solution

Secondly, we have one solution as $y_1=e^{-4x}\cos x$, we can obtain its root as

$m_1=-4+i$

And then the second root of this two complex pairs will be

$m_2=-4-i$

After that we can obtain an auxiliary equation for these two complex roots as,

$\begin{array}{l}\left(m-\left(-4+i\right)\right)\left(m-\left(-4-i\right)\right)=0\\ \left(m+\left(4-i\right)\right)\left(m+\left(4+i\right)\right)=0\\ m^2+8m+\left(16-i^2\right)=0\\ m^2+8m+17=0······\left(2\right)\end{array}$

This is the auxiliary equation of the roots

From the auxiliary equation (1) and (2) , we can make a long division to obtain the second parameter contains the third root of the given differential equation as,

$\begin{array}{c}\mathbf{SecondParameter}=\frac{m^3+6m^2+m-34}{m^2+8m+17}\\ =m-2\end{array}$

Then the roots are,

$m_{1,2}=-4\pm i,m_3=2$

Which$c_1,c_2,c_3$ are real and repeated roots.

Finally now, we obtain the general solution of our differential equation,$y\text{'}\text{'}\text{'}+6y\text{'}\text{'}+y\text{'}-34y=0$ as

$\begin{array}{c}y=c_1{e}^{-4x}\cos x+c_2{e}^{-4x}\sin x+c_3{e}^{m_{3}x}\\ =c_1{e}^{-4x}\cos x+c_2{e}^{-4x}\sin x+c_3{e}^{2x}\end{array}$

Where $c_1,c_2,c_3$are arbitrary constants.

Thus the general solution of the given differential equation is $y=c_1{e}^{-4x}\cos x+c_2{e}^{-4x}\sin x+c_3{e}^{2x}$.