Question

In Problems 1–20 solve the given system of differential equations bysystematic elimination.

$$\begin{gathered} \mathbf{(}{\mathit{D}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{)}\mathit{x}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0} \\ \mathbf{(}\mathit{D}\mathbf{-}\mathbf{1}\mathbf{)}\mathit{x}\mathbf{+}\mathit{D}\mathit{y}\mathbf{=}\mathbf{0} \end{gathered}$$

Short answer

The solution is,

$x\left(t\right)=c_1{e}^t+c_2{e^{-\frac{1}{2}}}^t\cos \frac{\sqrt{3}}{2}t+c_3{e^{-\frac{1}{2}}}^t\sin \frac{\sqrt{3}}{2}t$

$y\left(t\right)=(-\frac{3}{2}{c}_2-\frac{\sqrt{3}}{2}{c}_3)e^{\frac{1}{-2t}}\cos \frac{\sqrt{3}}{2}t+(\frac{\sqrt{3}}{2}{c}_2-\frac{3}{2}{c}_3)e^{\frac{{\displaystyle 1}}{{\displaystyle -2t}}}\sin \frac{\sqrt{3}}{2}t$

Step-by-step solution

Step 1:Definition of differential equation.

An equation containing the derivatives of one or more unknown functions (or dependent variables), with respect to one or more independent variables, is said to be a differential equation (DE).

Using Elimination.

We have the system of first order equations

$$\begin{gathered} (D^2-1)x-y=0...........\left(1\right) \\ (D-1)x+Dy=0...........\left(2\right) \end{gathered}$$

We have to solve it using systematic elimination as the following technique:

Operate the first equation by the operator D and the second equation by ,then we obtain

$$\begin{gathered} D(D^2-1)x-Dy=0........\left(1a\right) \\ (D-1)x+Dy=0.............\left(2a\right) \end{gathered}$$

After that, add equation (1a) to equation (2a) and simplify, then we obtain

$$\begin{gathered} \left[D\right(D^2-1)x-Dy]+\left[\right(D-1)x+Dy]=0 \\ \left[D\right(D^2-1)x+(D-1\left)x\right]+[Dy-Dy]=0 \\ D(D^2-1)x+(D-1)x=0 \\ (D^3-D+D-1)x=0 \\ (D^3-1)x=0 \end{gathered}$$

To find the particular solutions.

After assuming that $y=e^{mx}$as a solution for our system, we can have the auxiliary solution for x as , $m^3-1=0$

Now if we substitute with $m=1$into the auxiliary equation ,we find that it achieve it, then we can consider it as a factor $(m-1)=0$, after that we can make a long division, then we can obtain

$(m-1)(m^2+m+1)=0$

Then we have the roots,

$$\begin{gathered} m_1=1, \\ {m_2}_{,3}=\frac{-1\pm \sqrt{1-4\times 1\times 1}}{2} \\ {m_2}_{,3}=-\frac{1}{2}\pm \frac{\sqrt{3}}{2}i \end{gathered}$$

Then we can obtain

$x\left(t\right)=c_1{e}^t+c_2{e^{-\frac{1}{2}}}^t\cos \frac{\sqrt{3}}{2}t+c_3{e^{-\frac{1}{2}}}^t\sin \frac{\sqrt{3}}{2}t..........\left(3\right)$

is the general solution for x.

To find the  solutions.

Now we have to obtain the general solution for y as the following:

We have to expand and rearrange equation (1) as

$$\begin{gathered} D^{2}x-x-y=0 \\ y=D^{2}x-x \end{gathered}$$

After that, differentiate the solution of x shown in equation (3)with respect to t as

$$\begin{gathered} Dx=c_1{e}^t+(-\frac{1}{2}{c}_2+\frac{\sqrt{3}}{2}{c}_3){e^{-\frac{1}{2}}}^t\cos \frac{\sqrt{3}}{2}t+(-\frac{\sqrt{3}}{2}{c}_2-\frac{1}{2}{c}_3){e^{-\frac{1}{2}}}^t\sin \frac{\sqrt{3}}{2}t \\ {D}^{2}x=c_1{e}^t+(-\frac{1}{2}{c}_2-\frac{\sqrt{3}}{2}{c}_3){e^{-\frac{1}{2}}}^t\cos \frac{\sqrt{3}}{2}t+(\frac{\sqrt{3}}{2}{c}_2-\frac{1}{2}{c}_3){e^{-\frac{1}{2}}}^t\sin \frac{\sqrt{3}}{2}t \end{gathered}$$

After that, substitute with equations(3)and (5) into equation (4),then we obtain

$$\begin{gathered} y=(c_1{e}^t+(-\frac{1}{2}{c}_2-\frac{\sqrt{3}}{2}{c}_3){e^{-\frac{1}{2}}}^t\cos \frac{\sqrt{3}}{2}t+(\frac{\sqrt{3}}{2}{c}_2-\frac{1}{2}{c}_3\left){e^{-\frac{1}{2}}}^t\sin \frac{\sqrt{3}}{2}t\right) \\ -(c_1{e}^t+c_2{e^{-\frac{1}{2}}}^t\cos \frac{\sqrt{3}}{2}t+c_3{e^{-\frac{1}{2}}}^t\sin \frac{\sqrt{3}}{2}t) \\ y\left(t\right)=(-\frac{3}{2}{c}_2-\frac{\sqrt{3}}{2}{c}_3){e^{-\frac{{\displaystyle 1}}{{\displaystyle 2}}}}^t\cos \frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2}}t+(\frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2}}{c}_2-\frac{{\displaystyle 1}}{{\displaystyle 2}}{c}_3){e^{-\frac{{\displaystyle 1}}{{\displaystyle 2}}}}^t\sin \frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2}}t) \end{gathered}$$

is the general solution for y