In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \mathbf{2}\mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{,} \\ \mathbf{x}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{t}} \end{gathered}$$

Elimination Procedure for 2 × 2 Systems.

To find a general solution for the system

$$\begin{gathered} {\mathrm{L}}_1\left[\mathrm{x}\right]+{\mathrm{L}}_2\left[\mathrm{y}\right]={\mathrm{f}}_1, \\ {\mathrm{L}}_3\left[\mathrm{x}\right]+{\mathrm{L}}_4\left[\mathrm{y}\right]={\mathrm{f}}_2, \end{gathered}$$

Where${\mathbf{L}}_{\mathbf{1}}\mathbf{,}{\mathbf{L}}_{\mathbf{2}}\mathbf{,}{\mathbf{L}}_{\mathbf{3}}\mathbf{,}$ and${\mathbf{L}}_{\mathbf{4}}$ are polynomials in$D=\frac{d}{dt}$

Given that,

$2x\text{'}+y\text{'}-x-y=e^{-t} ......\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

$x\text{'}+y\text{'}+2x+y=e^t ......\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Let us rewrite this system of operators in operator form:

$$\begin{gathered} \left(2D-1\right)\left[x\right]+\left(D-1\right)\left[y\right]=e^{-t} ......\mathbf{\left(}\mathbf{3}\mathbf{\right)} \\ \left(D+2\right)\left[x\right]+\left(D+1\right)\left[y\right]=e^t ......\mathbf{\left(}\mathbf{4}\mathbf{\right)} \end{gathered}$$

Multiply (D+1) on equation (3).

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

And multiply (D-1) on equation (2).

$$\begin{gathered} \left(D-1\right)\left(D+2\right)\left[x\right]+\left(D-1\right)\left(D+1\right)\left[y\right]=\left(D-1\right)e^t \\ \left(D^2+2D-D-2\right)\left[x\right]+\left(D^2-1\right)\left[y\right]=e^t-e^t \\ \left(D^2+D-2\right)\left[x\right]+\left(D^2-1\right)\left[y\right]=0 ......\mathbf{\left(}\mathbf{6}\mathbf{\right)} \end{gathered}$$

Then subtract equation (5) and (6) together one gets,

$$\begin{gathered} \left(2D^2+D-1\right)\left[x\right]+\left(D^2-1\right)\left[y\right]-\left(\left(D^2+D-2\right)\left[x\right]+\left(D^2-1\right)\left[y\right]\right)=0 \\ \left(2D^2+D-1-D^2-D+2\right)\left[x\right]=0 \\ \left(D^2+1\right)\left[x\right]=0 \\ \left(D^2+1\right)\left[x\right]=0 ......\mathbf{\left(}\mathbf{7}\mathbf{\right)} \end{gathered}$$

Now solve equation (7).

$$\begin{gathered} \left(D^2+1\right)\left[x\right]=0 \\ x\text{'}\text{'}+x=0 \\ x=A\sin t+B\cos t \end{gathered}$$

Substitute the value of x in equation (3)

$$\begin{gathered} \left(2D-1\right)\left[x\right]+\left(D-1\right)\left[y\right]=e^{-t} \\ \left(2D-1\right)\left[A\sin t+B\cos t\right]+\left(D-1\right)\left[y\right]=e^{-t} \\ 2A\cos t-2B\sin t-A\sin t-B\cos t+\left(D-1\right)\left[y\right]=e^{-t} \\ \left(2A-B\right)\cos t-\left(A+2B\right)\sin t+\left(D-1\right)\left[y\right]=e^{-t} \end{gathered}$$

$\left(D-1\right)\left[y\right]=e^{-t}+\left(A+2B\right)\sin t-\left(2A-B\right)\cos t ......\mathbf{\left(}\mathbf{8}\mathbf{\right)}$

Since the auxiliary equation to the corresponding homogeneous equation is $r-1=0$.

The root is r=1.

Then, the homogeneous solution of u is$y_h\left(t\right)=c_1{e}^t ......\mathbf{\left(}\mathbf{9}\mathbf{\right)}$

Let us take the undetermined coefficients and assume that

$y_p\left(t\right)=a\sin t+b\cos t+c_2{e}^{-t} ......\mathbf{\left(}\mathbf{10}\mathbf{\right)}$

Now derivate the equation (7)

$D\left[y_p\left(t\right)\right]=a\cos t-b\sin t-c_2{e}^{-t} ......\mathbf{\left(}\mathbf{11}\mathbf{\right)}$

Subtract the equations (11) and (10).

$$\begin{gathered} D\left[y_p\left(t\right)\right]-y_p\left(t\right)=a\cos t-b\sin t-c_2{e}^{-t}-a\sin t-b\cos t-c_2{e}^{-t} \\ =\left(a-b\right)\cos t-\left(a+b\right)\sin t-2c_2{e}^{-t} \end{gathered}$$

Now, equalize the like terms of equation (8).

$$\begin{gathered} A+2B=-\left(a+b\right) \\ -2A+B=\left(a-b\right) \\ 1=-2c_2 \\ {c}_2=-\frac{1}{2} \end{gathered}$$

Solve the equations to find the value of A and B.

$$\begin{gathered} 3B-3A=-2b \\ b=\frac{3\left(A-B\right)}{2} \end{gathered}$$

Then,

$$\begin{gathered} A+B=-2a \\ a=-\frac{A+B}{2} \end{gathered}$$

So,$y_p\left(t\right)=-\frac{A+B}{2}\sin t+\frac{3\left(A-B\right)}{2}\cos t-\frac{1}{2}{e}^{-t} ......\mathbf{\left(}\mathbf{12}\mathbf{\right)}$

Use equations (9) and (12) to get,

$$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ y\left(t\right)=c_1{e}^t-\frac{A+B}{2}\sin t+\frac{3\left(A-B\right)}{2}\cos t-\frac{1}{2}{e}^{-t} \end{gathered}$$

So, the solution is founded.