91Ó°ÊÓ

Elimination Procedure for 2 × 2 Systems

To find a general solution for the system

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

Where${\mathrm{L}}_1,{\mathrm{L}}_2,{\mathrm{L}}_3$ and L₄are polynomials in$\mathbf{D}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}$

Given that,

${\mathbf{D}}^{\mathbf{2}}\left[\mathbf{u}\right]\mathbf{+}\mathbf{D}\left[\mathbf{v}\right]\mathbf{=}\mathbf{2}                  …\left(1\right)$

$4\mathrm{u}+\mathrm{D}\left[\mathrm{v}\right]=6                 …\left(2\right)$

Then subtract equation (1) and (2) together one gets,

$$\begin{gathered} {\mathrm{D}}^2\left[\mathrm{u}\right]+\mathrm{D}\left[\mathrm{v}\right]-\left(4\mathrm{u}+\mathrm{D}\left[\mathrm{v}\right]\right)=2-6 \\ \left({\mathrm{D}}^2-4\right)\left[\mathrm{u}\right]=-4 \\ \left({\mathrm{D}}^2-4\right)\left[\mathrm{u}\right]=-4                      …\left(3\right) \end{gathered}$$

Since the auxiliary equation to the corresponding homogeneous equation is ${\mathrm{r}}^2-4=0$. The roots are r=-2 and r=2.

Then, the homogeneous solution of u is;

${\mathrm{u}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{-2\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{2\mathrm{t}}                    …\left(4\right)$

Let us take the undetermined coefficients and assume that,

${\mathrm{u}}_{\mathrm{p}}\left(\mathrm{t}\right)=1               …\left(5\right)$

Use equations (4) and (5) to get,

$$\begin{gathered} \mathrm{u}\left(\mathrm{t}\right)={\mathrm{u}}_{\mathrm{h}}\left(\mathrm{t}\right)+{\mathrm{u}}_{\mathrm{p}}\left(\mathrm{t}\right) \\ \mathrm{u}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{-2\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{2\mathrm{t}}+1                 …\left(6\right) \end{gathered}$$

Now, take equation (2).

$$\begin{gathered} 4\mathrm{u}+\mathrm{D}\left[\mathrm{v}\right]=6 \\ \mathrm{D}\left[\mathrm{v}\right]=6-4\left[{\mathrm{c}}_1{\mathrm{e}}^{-2\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{2\mathrm{t}}+1\right] \\ \mathrm{D}\left[\mathrm{v}\right]=-4{\mathrm{c}}_1{\mathrm{e}}^{-2\mathrm{t}}-4{\mathrm{c}}_2{\mathrm{e}}^{2\mathrm{t}}+2 \\ \mathrm{v}\left(\mathrm{t}\right)=2{\mathrm{c}}_1{\mathrm{e}}^{-2\mathrm{t}}-2{\mathrm{c}}_2{\mathrm{e}}^{2\mathrm{t}}+2\mathrm{t}+{\mathrm{c}}_3 \end{gathered}$$

Thus, the solutions for the given linear system are$\mathbf{u}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}2\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{2\mathbf{t}}\mathbf{+}1$ and $\mathbf{v}\left(\mathbf{t}\right)\mathbf{=}\mathbf{2}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}\mathbf{+}\mathbf{2}\mathbf{t}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}$.