91Ó°ÊÓ

Elimination Procedure for 2 × 2 Systems

To find a general solution for the system

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

1. 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. 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 involvesbut not its derivatives. [Otherwise, go to step (d).] Substitute the found expression forinto this equation to get a formula for. The expressions for, andgive the desired general solution.
4. Eliminate x from the system and solve for. [Solving forgives more constants----in fact, twice as many as needed.]
5. Remove the extra constants by substituting the expressions for x(t) and into one or both of the equations in the system. Write the expressions for and in terms of the remaining constants.

Given that,

$\left(D^2-1\right)\left[u\right]+5v=e^t      ......\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

$2u+\left(D^2+2\right)\left[v\right]=0        ......\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Multiply$\left(D^2+2\right)$ on equation (1).

$$\begin{gathered} \left(D^2+2\right)\left(D^2-1\right)\left[u\right]+5\left(D^2+2\right)v=\left(D^2+2\right)e^t \\ \left(D^2+2\right)\left(D^2-1\right)\left[u\right]+5\left(D^2+2\right)v=e^t+2e^t \\ \left(D^2+2\right)\left(D^2-1\right)\left[u\right]+5\left(D^2+2\right)v=3e^t                   ......\mathbf{\left(}\mathbf{3}\mathbf{\right)} \end{gathered}$$

And multiply 5 on equation (2).

$$\begin{gathered} 2\left(5\right)u+5\left(D^2+2\right)\left[v\right]=0 \\ 10u+5\left(D^2+2\right)\left[v\right]=0        ......\mathbf{\left(}\mathbf{4}\mathbf{\right)} \end{gathered}$$

Then subtract equation (3) and (4) together one gets,

$$\begin{gathered} \left(D^2+2\right)\left(D^2-1\right)\left[u\right]-10u=3e^t \\ \left(D^4-D^2+2D^2-2-10\right)\left[u\right]=3e^t \\ \left(D^4+D^2-12\right)\left[u\right]=3e^t \\ \left(D^2+4\right)\left(D^2-3\right)\left[u\right]=3e^t \end{gathered}$$

$\left(D^2+4\right)\left(D^2-3\right)\left[u\right]=3e^t        ......\mathbf{\left(}\mathbf{5}\mathbf{\right)}$

Since the auxiliary equation to the corresponding homogeneous equation is:

$\left(r^2+4\right)\left(r^2-3\right)=0$

. The roots are$r=Â\pm 2i$ and $r=Â\pm \sqrt{3}$.

Then, the homogeneous solution of u is

$u_h\left(t\right)=c_1{e}^{-\sqrt{3}t}+c_2{e}^{\sqrt{3}t}+c_3\cos \left(2t\right)+c_4\sin \left(2t\right)        ......\mathbf{\left(}\mathbf{6}\mathbf{\right)}$

Let us take the undetermined coefficients and assume that

$u_p\left(t\right)=a{e}^t     ......\mathbf{\left(}\mathbf{7}\mathbf{\right)}$

Now derivate the equation (7)

$$\begin{gathered} D^2\left[u_p\left(t\right)\right]=a{e}^t \\ {D}^4\left[u_p\left(t\right)\right]=a{e}^t \end{gathered}$$

Substitute the derivation in equation (5).

$$\begin{gathered} \left(D^4+D^2-12\right)\left[u\right]=3e^t \\ \left(D^4+D^2-12\right)\left[a{e}^t\right]=3e^t \\ a{e}^t+a{e}^t-12a{e}^t=3e^t \\ -10a{e}^t=3e^t \end{gathered}$$

Now, equalize the like terms.

$$\begin{gathered} -10a=3 \\ a=-\frac{3}{10} \end{gathered}$$

So,$u_p\left(t\right)=-\frac{3}{10}{e}^t       ......\mathbf{\left(}\mathbf{8}\mathbf{\right)}$

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

$$\begin{gathered} u\left(t\right)=u_h\left(t\right)+u_p\left(t\right) \\ u\left(t\right)=c_1{e}^{-\sqrt{3}t}+c_2{e}^{\sqrt{3}t}+c_3\cos \left(2t\right)+c_4\sin \left(2t\right)-\frac{3}{10}{e}^t       ......\mathbf{\left(}\mathbf{9}\mathbf{\right)} \end{gathered}$$

Now, take equation (1).

$$\begin{gathered} \left(D^2-1\right)\left[u\right]+5v=e^t \\ 5v=e^t-\left(D^2-1\right)\left[u\right] \\ v=\frac{e^t-\left(D^2-1\right)\left[u\right]}{5} \end{gathered}$$

Now derivate the value of u to find the value of v.

$$\begin{gathered} u\text{'}\left(t\right)=-\sqrt{3}{c}_1{e}^{-\sqrt{3}t}+\sqrt{3}{c}_2{e}^{\sqrt{3}t}-2c_3\sin \left(2t\right)+2c_4\cos \left(2t\right)-\frac{3}{10}{e}^t \\ u\text{'}\text{'}\left(t\right)=3c_1{e}^{-\sqrt{3}t}+3c_2{e}^{\sqrt{3}t}-4c_3\cos \left(2t\right)-4c_4\sin \left(2t\right)-\frac{3}{10}{e}^t \end{gathered}$$

Then,

$$\begin{gathered} v=\frac{e^t}{5}-\frac{\left(D^2-1\right)\left[u\right]}{5} \\ =\frac{e^t}{5}-\frac{\left(D^2-1\right)\left[c_1{e}^{-\sqrt{3}t}+c_2{e}^{\sqrt{3}t}+c_3\cos \left(2t\right)+c_4\sin \left(2t\right)-\frac{3}{10}{e}^t\right]}{5} \\ =\frac{e^t}{5}-\frac{3c_1{e}^{-\sqrt{3}t}+3c_2{e}^{\sqrt{3}t}-4c_3\cos \left(2t\right)-4c_4\sin \left(2t\right)-\frac{3}{10}{e}^t-\left(c_1{e}^{-\sqrt{3}t}+c_2{e}^{\sqrt{3}t}+c_3\cos \left(2t\right)+c_4\sin \left(2t\right)-\frac{3}{10}{e}^t\right)}{5} \\ =\frac{e^t}{5}-\frac{2c_1{e}^{-\sqrt{3}t}+2c_2{e}^{\sqrt{3}t}-5c_3\cos \left(2t\right)-5c_4\sin \left(2t\right)}{5} \\ =-\frac{2}{5}{c}_1{e}^{-\sqrt{3}t}-\frac{2}{5}{c}_2{e}^{\sqrt{3}t}+c_3\cos \left(2t\right)+c_4\sin \left(2t\right)+\frac{e^t}{5} \end{gathered}$$

So, the solution is founded.