91Ó°ÊÓ

Elimination Procedure for 2 × 2 Systems:

To find a general solution for the system

$$\begin{gathered} {\mathbf{L}}_{\mathbf{1}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{1}}\mathbf{,} \\ {\mathbf{L}}_{\mathbf{3}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{4}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{2}}\mathbf{,} \end{gathered}$$

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

1. Make sure that the system is written in operator form.
2. Eliminate one of the variables, say, y, and solve the resulting equation for x(t). If the system is degenerating, stop! A separate analysis is required to determine whether or not there are solutions.
3. (Shortcut) If possible, use the system to derive an equation that involves y(t) but not its derivatives. [Otherwise, go to step (d).] Substitute the found expression for x(t) into this equation to get a formula for y(t). The expressions for x(t), and y(t) give the desired general solution.
4. Eliminate x from the system and solve for y(t). [Solving for y(t) gives more constants----in fact, twice as many as needed.]

Remove the extra constants by substituting the expressions for x(t) and y(t) into one or both of the equations in the system. Write the expressions for x(t)

Given that, $$\begin{gathered} x\text{'}\text{'}+y\text{'}\text{'}-x\text{'}=2t        ......\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ x\text{'}\text{'}+y\text{'}-x+y=-1         ......\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

Let us rewrite the system in operator form,

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

Multiply (D+1) on equation (3) and -D on equation (4) then add them together to get,

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

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

Since the corresponding auxiliary equation is $r\left(r+1\right)\left(r-1\right)=0$. The roots are$r=0,r=-1$ and $r=1$.

Then, the homogeneous solution of y is:

$y_h\left(t\right)=c_1+c_2{e}^{-t}+c_3{e}^t         ......\mathbf{\left(}\mathbf{6}\mathbf{\right)}$

Let us take the undetermined coefficients and assume that

$y_p\left(t\right)=A{t}^2+Bt+C         ......\mathbf{\left(}\mathbf{7}\mathbf{\right)}$

Now derivate the equation (7)

$$\begin{gathered} D\left[y_p\left(t\right)\right]=2At+B \\ {D}^2\left[y_p\left(t\right)\right]=2A \end{gathered}$$

Substitute the derivation in equation (5).

$$\begin{gathered} D\left(D-1\right)\left(D+1\right)\left[y\right]=2t+2 \\ \left(D^3-D\right)\left[A{t}^2+Bt+C\right]=2t+2 \\ 0-2At-B=2t+2 \\ -2At-B=2t+2 \end{gathered}$$

Now, equalize the like terms.

$$\begin{gathered} -2A=2 \\ A=-1 \\ -B=2 \\ B=-2 \end{gathered}$$

So,$y_p\left(t\right)=-t^2-2t         ......\mathbf{\left(}\mathbf{8}\mathbf{\right)}$

Use equations (6) and (8) to get,

$$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ y\left(t\right)=c_1+c_2{e}^{-t}+c_3{e}^t-t^2-2t         ......\mathbf{\left(}\mathbf{9}\mathbf{\right)} \end{gathered}$$

Now, take equation (3).

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

Now derivate the y value.

$$\begin{gathered} D\left[y\left(t\right)\right]=-c_2{e}^{-t}+c_3{e}^t-2t-2 \\ {D}^2\left[y\left(t\right)\right]=c_2{e}^{-t}+c_3{e}^t-2 \end{gathered}$$

Substitute in the above equation.

$\left(D^2-D\right)\left[x\right]=2t-c_2{e}^{-t}-c_3{e}^t+2         ......\mathbf{\left(}\mathbf{10}\mathbf{\right)}$

Since the corresponding auxiliary equation is $r^2-r=0$. The roots are r=0 and r=1

.

Then, the homogeneous solution of x is

$x_h\left(t\right)=d_1+d_2{e}^t        ......\mathbf{\left(}\mathbf{11}\mathbf{\right)}$

Let us take the undetermined coefficients and assume that

$x_p\left(t\right)=a{t}^2+bt+c+m{e}^{-t}+n{e}^t+kt{e}^t         ......\mathbf{\left(}\mathbf{12}\mathbf{\right)}$

Now derivate the equation (12)

$$\begin{gathered} D\left[x_p\left(t\right)\right]=2at+b-m{e}^{-t}+n{e}^t+k{e}^t+kt{e}^t \\ =2at+b-m{e}^{-t}+\left(n+k\right)e^t+kt{e}^t \\ {D}^2\left[x_p\left(t\right)\right]=2a+m{e}^{-t}+\left(n+k\right)e^t+k{e}^t+kt{e}^t \\ =2a+m{e}^{-t}+\left(n+2k\right)e^t+kt{e}^t \end{gathered}$$

Substitute in equation (10)

$$\begin{gathered} \left(D^2-D\right)\left[x\right]=2t-c_2{e}^{-t}-c_3{e}^t+2 \\ \left(D^2-D\right)\left[a{t}^2+bt+c+m{e}^{-t}+n{e}^t+kt{e}^t\right]=2t-c_2{e}^{-t}-c_3{e}^t+2 \\ -2at+\left(2a-b\right)+2m{e}^t+t{e}^t=2t-c_2{e}^{-t}-c_3{e}^t+2 \end{gathered}$$

Now, equalize the like terms.

$$\begin{gathered} -2a=2 \\ a=-1 \\ 2a-b=2 \\ b=-4 \\ 2m=-c_2 \\ m=-\frac{c_2}{2} \\ k=-c_3,c,n \end{gathered}$$

$x_p\left(t\right)=-t^2-4t+c-\frac{c_2}{2}{e}^{-t}+n{e}^t-c_{3}t{e}^t         ......\mathbf{\left(}\mathbf{13}\mathbf{\right)}$

$$\begin{gathered} x\left(t\right)=x_h\left(t\right)+x_p\left(t\right) \\ x\left(t\right)=d_1+d_2{e}^t-t^2-4t+c-\frac{c_2}{2}{e}^{-t}+n{e}^t-c_{3}t{e}^t \end{gathered}$$

So, the solution is founded.