Question

In Problems 1-20 solve the given system of differential equations by systematic elimination.

20.

$$\begin{gathered} \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathit{x}\mathbf{+}\mathit{z} \\ \frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{y}\mathbf{+}\mathit{z} \\ \frac{\mathbf{d}\mathbf{z}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathit{x}\mathbf{+}\mathit{y} \end{gathered}$$

Short answer

$$\begin{gathered} x\left(t\right)=c_1+c_2{e}^{-t}+c_{3}t{e}^{-t} \\ y\left(t\right)=c_1+\left(c_2-c_3\right)e^{-t}+c_{3}t{e}^{-t} \\ z\left(t\right)=c_1+c_3{e}^{-t} \end{gathered}$$

Step-by-step solution

To solve it using systematic elimination as the following technique:

We have the system of differential equations

$$\begin{gathered} \frac{dx}{dt}=-x+z \\ \frac{dy}{dt}=-y+z \\ \frac{dz}{dt}=-x+y \end{gathered}$$

which by using differential operating are equivalent to

$$\begin{gathered} Dx+x-z=0 \\ Dy+y-z=0 \\ Dz+x-y=0 \end{gathered}$$ …. (1), (2) & (3)

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

Rearrange equation (2) and operate equation (3) by the operator (D+1) and rearrange it, then we obtain

$$\begin{gathered} (D+1)y-z=0 \\ D(D+1)z+(D+1)x-(D+1)y=0 \end{gathered}$$ ….. (2) & (4)

After that, add equation (4) to equation (2) and simplify, then we obtain

$$\begin{gathered} \left[D\right(D+1)z+(D+1)x-(D+1\left)y\right]+\left[\right(D+1)y-z]=0 \\ D(D+1)z+(D+1)x-z=0 \\ \left(D^2+D-1\right)z+(D+1)x=0 \end{gathered}$$ ……. (5)

Then, operate equation by the operator $\left(D^2+D-1\right)$and rearrange it, then we obtain

$$\begin{gathered} D\left(D^2+D-1\right)x+\left(D^2+D-1\right)x-\left(D^2+D-1\right)z=0 \\ (D+1)\left(D^2+D-1\right)x-\left(D^2+D-1\right)z=0 \end{gathered}$$ …….. (6)

After that, add equation to equation and simplify, then we obtain

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

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

Final proof

After assuming that$x=e^{mt}$as a solution for our system, we can have the auxiliary solution for x as

$$\begin{gathered} m(m+1)(m+1)=0 \\ {m}_1=0\text{and}{m}_{2,3}=-1 \end{gathered}$$

Then we obtain

$x\left(t\right)=c_1+c_2{e}^{-t}+c_{3}t{e}^{-t}$ …… (7)

is the solution for.

Now, we can obtain the solution for z by substituting with the solution for x in equation (7) into equation (1), then we obtain

$$\begin{gathered} z\left(t\right)=Dx+x \\ =D\left(c_1+c_2{e}^{-t}+c_{3}t{e}^{-t}\right)+\left(c_1+c_2{e}^{-t}+c_{3}t{e}^{-t}\right) \\ =\left(-c_2{e}^{-t}-c_{3}t{e}^{-t}+c_3{e}^{-t}\right)+\left(c_1+c_2{e}^{-t}+c_{3}t{e}^{-t}\right) \\ =c_1+c_3{e}^{-t} \end{gathered}$$

Also, we can obtain the solution for y by substituting with the solution for x in equation (10) and the solution for (y) in equation (11) into equation (2), then we obtain

$$\begin{gathered} y\left(t\right)=Dz+x \\ =D\left(c_1+c_3{e}^{-t}\right)+\left(c_1+c_2{e}^{-t}+c_{3}t{e}^{-t}\right) \\ =\left(-c_3{e}^{-t}\right)+\left(c_1+c_2{e}^{-t}+c_{3}t{e}^{-t}\right) \\ =c_1+c_2{e}^{-t}-c_3{e}^{-t}+c_{3}t{e}^{-t} \\ =c_1+\left(c_2-c_3\right)e^{-t}+c_{3}t{e}^{-t} \end{gathered}$$

The result is

$$\begin{gathered} x\left(t\right)=c_1+c_2{e}^{-t}+c_{3}t{e}^{-t} \\ y\left(t\right)=c_1+\left(c_2-c_3\right)e^{-t}+c_{3}t{e}^{-t} \\ z\left(t\right)=c_1+c_3{e}^{-t} \end{gathered}$$