Question

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

$$\begin{gathered} \mathbf{19}\mathbf{.}\mathbf{}\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{6}\mathit{y} \\ \frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{x}\mathbf{+}\mathit{z} \\ \frac{\mathbf{d}\mathbf{z}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{x}\mathbf{+}\mathit{y} \end{gathered}$$

Short answer

$$\begin{gathered} x\left(t\right)=c_1{e}^{-t}+c_2{e}^{-2t}+c_3{e}^{3t} \\ y\left(t\right)=-\frac{1}{6}{c}_1{e}^{-t}-\frac{1}{3}{c}_2{e}^{-2t}+\frac{1}{2}{c}_3{e}^{3t} \\ z\left(t\right)=-\frac{5}{6}{c}_1{e}^{-t}-\frac{1}{3}{c}_2{e}^{-2t}+\frac{1}{2}{c}_3{e}^{3t} \end{gathered}$$

Step-by-step solution

To solve it using systematic elimination as the following technique:

We have the system of differential equations

$$\begin{gathered} \frac{dx}{dt}=6y \\ \frac{dy}{dt}=x+z \\ \frac{dz}{dt}=x+y \end{gathered}$$

Which by using differential operating are equivalent to

$$\begin{gathered} Dx=6y \\ Dy=x+z \\ Dz=x+y \end{gathered}$$ ….. (1), (2) & (3)

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

Operate the equation by the operator D and multiply equation (2) by 6, then we obtain

$$\begin{gathered} D^{2}x-6Dy=0 \\ 6Dy=6x+6z \end{gathered}$$ …. (4) & (5)

After that, add equation (4) to equation (5) and simplify, then we obtain

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

Then, operate equation by the operator and multiply equation (3) by 6, then we obtain

$$\begin{gathered} D^{3}x-6Dz=6Dx \\ 6Dz=6x+6y \end{gathered}$$ …. (7) & (8)

After that, add equation (8) to equation (7) and simplify, then we obtain

$$\begin{gathered} \left(D^{3}x-6Dz\right)+\left(6Dz\right)=6Dx+6x+6y \\ {D}^{3}x=6Dx+6x+6y \end{gathered}$$ …… (9)

After that, by solving the two equations (1) and (9), we can obtain

$$\begin{gathered} D^{3}x=6Dx+6x+Dx \\ {D}^{3}x-7Dx-6x=0 \\ \left(D^3-7D-6\right)x=0 \end{gathered}$$

Final proof

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

$m^3-7m-6=0$

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

$$\begin{gathered} (m+1)\left(m^2-m-6\right)=0 \\ (m+1)(m+2)(m-3)=0 \end{gathered}$$

which has the roots

$m_1=-1,m_2=-2\text{and}{m}_3=3$

Then we obtain

$x\left(t\right)=c_1{e}^{-t}+c_2{e}^{-2t}+c_3{e}^{3t}$ …… (10)

is the solution for x.

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

$$\begin{gathered} y\left(t\right)=\frac{1}{6}Dx \\ =\frac{1}{6}D\left(c_1{e}^{-t}+c_2{e}^{-2t}+c_3{e}^{3t}\right) \\ =\frac{1}{6}\left(-c_1{e}^{-t}-2c_2{e}^{-2t}+3c_3{e}^{3t}\right) \\ =-\frac{1}{6}{c}_1{e}^{-t}-\frac{1}{3}{c}_2{e}^{-2t}+\frac{1}{2}{c}_3{e}^{3t} \end{gathered}$$ ......(11)

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

$$\begin{gathered} z\left(t\right)=Dy-x \\ =D\left(-\frac{1}{6}{c}_1{e}^{-t}-\frac{1}{3}{c}_2{e}^{-2t}+\frac{1}{2}{c}_3{e}^{3t}\right)-\left(c_1{e}^{-t}+c_2{e}^{-2t}+c_3{e}^{3t}\right) \\ =\left(\frac{1}{6}{c}_1{e}^{-t}+\frac{2}{3}{c}_2{e}^{-2t}+\frac{3}{2}{c}_3{e}^{3t}\right)-\left(c_1{e}^{-t}+c_2{e}^{-2t}+c_3{e}^{3t}\right) \\ =-\frac{5}{6}{c}_1{e}^{-t}-\frac{1}{3}{c}_2{e}^{-2t}+\frac{1}{2}{c}_3{e}^{3t} \end{gathered}$$

The result is

$$\begin{gathered} x\left(t\right)=c_1{e}^{-t}+c_2{e}^{-2t}+c_3{e}^{3t} \\ y\left(t\right)=-\frac{1}{6}{c}_1{e}^{-t}-\frac{1}{3}{c}_2{e}^{-2t}+\frac{1}{2}{c}_3{e}^{3t} \\ z\left(t\right)=-\frac{5}{6}{c}_1{e}^{-t}-\frac{1}{3}{c}_2{e}^{-2t}+\frac{1}{2}{c}_3{e}^{3t} \end{gathered}$$