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.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{,} \\ \mathbf{-}\mathbf{x}\mathbf{+}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{1} \end{gathered}$$

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

Given that,

$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}$ … (1)

$\mathbf{-}\mathbf{x}\mathbf{+}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{1}$ … (2)

Let us rewrite the system in operator form,

$\mathbf{D}\left[\mathbf{x}\right]\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}$… (3)

$\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{D}\left[\mathbf{y}\right]\mathbf{=}\mathbf{1}$… (4)

Multiply D on both sides of equation (4) then add equation (3) and (4) together one gets,

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

Since the auxiliary equation to the corresponding homogeneous equation is $\left({\mathbf{r}}^{\mathbf{2}}\mathbf{+}1\right)\mathbf{=}\mathbf{0}$. The roots are $\mathbf{r}\mathbf{=}\mathbf{\pm }\mathbf{i}$.

Then, the homogeneous solution of u is;

${\mathbf{y}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{Asint}\mathbf{+}\mathbf{Bcost}$ … (6)

Let us take the undetermined coefficients and assume that,

${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{at}}^{\mathbf{2}}\mathbf{+}\mathbf{bt}\mathbf{+}\mathbf{c}$ … (7)

Now derivate the equation (7)

$$\begin{gathered} \mathbf{D}\left[{\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\right]\mathbf{=}\mathbf{2}\mathbf{at}\mathbf{+}\mathbf{b} \\ {\mathbf{D}}^{\mathbf{2}}\left[{\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\right]\mathbf{=}\mathbf{2}\mathbf{a} \end{gathered}$$

Substitute the derivation in equation (5).

$$\begin{gathered} \left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\right)\left[{\mathbf{at}}^{\mathbf{2}}\mathbf{+}\mathbf{bt}\mathbf{+}\mathbf{c}\right]\mathbf{=}{\mathbf{t}}^{\mathbf{2}} \\ \mathbf{2}\mathbf{a}\mathbf{+}{\mathbf{at}}^{\mathbf{2}}\mathbf{+}\mathbf{bt}\mathbf{+}\mathbf{c}\mathbf{=}{\mathbf{t}}^{\mathbf{2}} \\ {\mathbf{at}}^{\mathbf{2}}\mathbf{+}\mathbf{bt}\mathbf{+}\mathbf{2}\mathbf{a}\mathbf{+}\mathbf{c}\mathbf{=}{\mathbf{t}}^{\mathbf{2}} \end{gathered}$$

Now, equalize the like terms.

$$\begin{gathered} \mathbf{a}\mathbf{=}\mathbf{1} \\ \mathbf{b}\mathbf{=}\mathbf{0} \\ \mathbf{2}\mathbf{a}\mathbf{+}\mathbf{c}\mathbf{=}\mathbf{0} \\ \mathbf{c}\mathbf{=}\mathbf{-}\mathbf{2} \end{gathered}$$

So, ${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{2}$ … (8)

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

$$\begin{gathered} \mathbf{y}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{y}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right) \\ \mathbf{y}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{+}\mathbf{Asint}\mathbf{+}\mathbf{Bcost} \dots \left(9\right) \end{gathered}$$

Now, take equation (4).

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

Therefore, the solutions for the given linear system are$\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{Acost}\mathbf{-}\mathbf{Bsint}\mathbf{-}\mathbf{1}$ and $\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{+}\mathbf{Asint}\mathbf{+}\mathbf{Bcost}$.