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.

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

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${\mathbf{L}}_{\mathbf{1}}\mathbf{,}{\mathbf{L}}_{\mathbf{2}}\mathbf{,}{\mathbf{L}}_{\mathbf{3}}$ and L₄are polynomials in$\mathbf{D}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}$

Given that,

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

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

Multiply (2D+1) on equation (1).

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

$\left(\mathbf{2}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{u}\right]\mathbf{-}\left(\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{1}\right)\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{v}\right]\mathbf{=}\mathbf{3}{\mathbf{e}}^{\mathbf{t}} \dots \left(3\right)$

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

$$\begin{gathered} \left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{u}\right]\mathbf{+}\left(\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{1}\right)\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{v}\right]\mathbf{=}5\left(\mathbf{D}\mathbf{+}\mathbf{1}\right) \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\right)\left[\mathbf{u}\right]\mathbf{+}\left(\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{1}\right)\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{v}\right]\mathbf{=}\mathbf{5} \dots \left(4\right) \end{gathered}$$

Then adding equations (3) and (4) together we get,

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

Since the auxiliary equation to the corresponding homogeneous equation is $\mathbf{3}{\mathbf{r}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{r}\mathbf{=}\mathbf{0}$. The roots are r=0 and r=-1.

Then, the homogeneous solution of u is;

$$\begin{gathered} {\mathbf{u}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{0\times \mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{-\mathbf{t}} \\ {\mathbf{u}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{-\mathbf{t}} \dots \left(6\right) \end{gathered}$$

Let us take the undetermined coefficients and assume that,

${\mathbf{u}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}\mathbf{Bt} \dots \left(7\right)$

Now find derivate the equation (7)

$$\begin{gathered} \mathbf{D}\left[{\mathbf{u}}_{\mathbf{p}}\left(\mathbf{t}\right)\right]\mathbf{=}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}\mathbf{B} \\ {\mathbf{D}}^{\mathbf{2}}\left[{\mathbf{u}}_{\mathbf{p}}\left(\mathbf{t}\right)\right]\mathbf{=}{\mathbf{Ae}}^{\mathbf{t}} \end{gathered}$$

Substitute the derivative in equation (5).

$$\begin{gathered} \left(\mathbf{3}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{D}\right)\left[\mathbf{u}\right]\mathbf{=}\mathbf{3}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{5} \\ \mathbf{3}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}\mathbf{3}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}\mathbf{3}\mathbf{B}\mathbf{=}\mathbf{3}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{5} \\ \mathbf{6}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}\mathbf{3}\mathbf{B}\mathbf{=}\mathbf{3}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{5} \end{gathered}$$

Now, equalize the like terms.

$$\begin{gathered} 6\mathbf{A}\mathbf{=}\mathbf{3} \\ 3\mathbf{B}\mathbf{=}5 \end{gathered}$$

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

$$\begin{gathered} \mathbf{6}\mathbf{A}\mathbf{=}\mathbf{3} \\ \mathbf{A}\mathbf{=}\frac{\mathbf{3}}{\mathbf{6}} \\ \mathbf{=}\frac{\mathbf{1}}{\mathbf{2}} \end{gathered}$$

Then,

$$\begin{gathered} \mathbf{3}\mathbf{B}\mathbf{=}\mathbf{5} \\ \mathbf{B}\mathbf{=}\frac{\mathbf{5}}{\mathbf{3}} \end{gathered}$$

So,${\mathbf{u}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\frac{1}{2}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\frac{5}{3}\mathbf{t} \dots \left(8\right)$

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

$$\begin{gathered} \mathbf{u}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{u}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{u}}_{\mathbf{p}}\left(\mathbf{t}\right) \\ \mathbf{u}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{-\mathbf{t}}\mathbf{+}\frac{1}{2}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\frac{5}{3}\mathbf{t} \dots \left(9\right) \end{gathered}$$

Multiply by 2 on equation (1) and then add with equation (2).

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

Substitute equation (9) here.

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

Thus, the solutions for the given linear system are$\mathbf{u}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{-\mathbf{t}}\mathbf{+}\frac{1}{2}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\frac{5}{3}\mathbf{t}$ and $\mathbf{v}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{t}$.