Question

solve the given system of differential equations by systematic elimination.

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

Short answer

$$\begin{gathered} x\left(t\right)=c_1{e}^{2t}+c_{2}cosx+c_{3}sinx+e^t \\ y\left(t\right)=\frac{3}{2}{c}_1{e}^{2t}-c_{2}cosx-c_{3}sinx \\ z\left(t\right)=-2c_1{e}^{2t}+c_{2}sinx-c_{3}cosx \end{gathered}$$

Step-by-step solution

Solution for X

We have the system of first order equations

$$\begin{gathered} Dx+z=e^t\left(1\right) \\ (D-1)x+Dy+Dz=0\left(2\right) \\ x+2y+Dz=e^t\left(3\right) \end{gathered}$$

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

Multiply the equation (2) by -2 and operate equation (3) by the operator D, then we obtain

$$\begin{gathered} -2(D-1)x-2Dy-2Dz=0\left(3\right) \\ Dx+2Dy+D^{2}z=D\left(e^t\right)\left(4\right) \end{gathered}$$

After that, add equation (3) to equation (4) and simplify, then we obtain

localid="1668164518247" $$\begin{gathered} [-2(D-1)x-2Dy-2Dz]+\left[Dx+2Dy+D^{2}z\right]=D\left(e^t\right) \\ [2Dy-2Dy]+\left[Dx-2(D-1)x-2Dz+D^{2}z\right]=e^t \\ Dx-2(D-1)x+\left(D^2-2D\right)z=e^t \\ (D-2D+2)x+\left(D^2-2D\right)z=e^t \\ (-D+2)x+D(D-2)z=e^t \\ (D-2)x-D(D-2)z=-e^t\left(5\right) \end{gathered}$$

Then, operate equation (1) with the operator $D(D-2)$, then we obtain

localid="1668164637215" $$\begin{gathered} D^2(D-2)x+D(D-2)z=D(D-2)e^t \\ {D}^2(D-2)x+D(D-2)z=D\left(e^t\right)+D\left(-2e^t\right) \\ {D}^2(D-2)x+D(D-2)z=e^t-2e^t \\ {D}^2(D-2)x+D(D-2)z=-e^t\left(6\right) \end{gathered}$$

After that, add equation (6) to equation (5), then we obtain

localid="1668164896208" $$\begin{gathered} \left[D^2(D-2)x+D(D-2)z\right]+\left[\right(D-2)x-D(D-2\left)z\right]=-e^t-e^t \\ \left[D\right(D-2)z-D(D-2\left)z\right]+\left[D^2(D-2)x+(D-2)x\right]=-2e^t \\ {D}^2(D-2)x+(D-2)x=-2e^t \\ (D-2)\left(D^2+1\right)x=-2e^t \\ \left(D^3-2D^2-D\right)x=-2e^t\left(7\right) \end{gathered}$$

After that, we have to obtain the solution of the corresponding homogeneous equation $(D-2)\left(D^2+1\right)x=0$as the following:

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

$(m-2)\left(m^2+1\right)=0$

Then we have the roots

$m_1=2and{m}_{2,3}=\pm i$

Then we can obtain the homogeneous solution for x as

$x_c=c_1{e}^{2t}+c_{2}cosx+c_{3}sinx\left(8\right)$

After that, we have to obtain its particular solution by assuming that $x_p=A{e}^t$, then differentiate with respect to t as

$$\begin{gathered} D{x}_p=A{e}^t\left(9\right) \\ {D}^2{x}_p=A{e}^t\left(10\right) \\ {D}^3{x}_p=A{e}^t\left(11\right) \end{gathered}$$

Then substitute with equations (9), (10) and (11) into equation (7), then we obtain

$$\begin{gathered} A{e}^t-2A{e}^t-A{e}^t=-2e^t \\ -2A{e}^t=-2e^t \\ A=1 \end{gathered}$$

Now, we can obtain the particular solution for x as

$x_p=e^t$ (12)

Then from equation (8) and (12), we can obtain the general solution for x as

$x\left(t\right)=x_c+x_p=c_1{e}^{2t}+c_{2}cosx+c_{3}sinx+e^t$$x\left(t\right)=x_c+x_p=c_1{e}^{2t}+c_{2}cosx+c_{3}sinx+e^t$ (13)

Solutions for y and z

Now we can obtain the general solution for z by substituting with the solution of x and z shown in equations (13) and (14) into equation (3), then we obtain

$$\begin{gathered} z\left(t\right)=e^t-Dx \\ z\left(t\right)=e^t-D\left(c_1{e}^{2t}+c_{2}cosx+c_{3}sinx+e^t\right) \\ z\left(t\right)=-2c_1{e}^{2t}+c_{2}sinx-c_{3}cosx\left(14\right) \end{gathered}$$

Also, we can obtain the general solution for y by substituting with the solution of x in equation (13) into equation (1), then we obtain

$$\begin{gathered} y\left(t\right)=\frac{1}{2}\left[e^t-x-Dz\right] \\ y\left(t\right)=\frac{1}{2}{e}^t-\frac{1}{2}\left(c_1{e}^{2t}+c_{2}cosx+c_{3}sinx+e^t\right)-\frac{1}{2}D\left(-2c_1{e}^{2t}+c_{2}sinx-c_{3}cosx\right) \\ y\left(t\right)=\frac{1}{2}{e}^t+\left(-\frac{1}{2}{c}_1{e}^{2t}-\frac{1}{2}{c}_{2}cosx-\frac{1}{2}{c}_{3}sinx-\frac{1}{2}{e}^t\right)-\frac{1}{2}\left(-4c_1{e}^{2t}+c_{2}cosx+c_{3}sinx\right) \\ y\left(t\right)=\left(-\frac{1}{2}{c}_1{e}^{2t}-\frac{1}{2}{c}_{2}cosx-\frac{1}{2}{c}_{3}sinx\right)+\left(2c_1{e}^{2t}-\frac{1}{2}{c}_{2}cosx-\frac{1}{2}{c}_{3}sinx\right) \\ y\left(t\right)=\frac{3}{2}{c}_1{e}^{2t}-c_{2}cosx-c_{3}sinx \end{gathered}$$

Final Answer

$$\begin{gathered} x\left(t\right)=c_1{e}^{2t}+c_{2}cosx+c_{3}sinx+e^t \\ y\left(t\right)=\frac{3}{2}{c}_1{e}^{2t}-c_{2}cosx-c_{3}sinx \\ z\left(t\right)=-2c_1{e}^{2t}+c_{2}sinx-c_{3}cosx \end{gathered}$$