Q28E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 25 – 28, use the elimination method to find a general solution for the given system of three equations in the three unknown functions x(t), y(t), and z(t).

 $$\begin{gathered} \mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{z}\mathbf{,} \\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{6}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{,} \\ \mathbf{z}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{-}\mathbf{z} \end{gathered}$$

Step-by-Step Solution

Answer

The solutions for the given linear system are $\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{3}}{\mathbf{Be}}^{\mathbf{3}\mathbf{t}}\mathbf{+}\frac{\mathbf{C}}{\mathbf{6}}$, 

$\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{3}\mathbf{t}}\mathbf{+}\mathbf{C}$and $\mathbf{z}\left(\mathbf{t}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}{\mathbf{Be}}^{\mathbf{3}\mathbf{t}}\mathbf{-}\frac{\mathbf{13}}{\mathbf{6}}\mathbf{C}$.

Step 1: General form

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 $\mathbf{D}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}$

 

1. Make sure that the system is written in operator form.

 

1. 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.

 

1. (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.

 

1. Eliminate x from the system and solve for y(t). [Solving for y(t) gives more constants----in fact, twice as many as needed.]

 

1. 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) and y(t) in terms of the remaining constants.

Step 2: Evaluate the given equation

Given that,

 

$\mathbf{x}\mathbf{\text{'}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{z}$          … (1)

$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{6}\mathbf{x}\mathbf{-}\mathbf{y}$                 … (2)

$\mathbf{z}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{-}\mathbf{z}$           … (3)

 Let us rewrite the given system of equations into operator form.

 

$\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{-}\mathbf{2}\left[\mathbf{y}\right]\mathbf{-}\left[\mathbf{z}\right]\mathbf{=}\mathbf{0}$    … (4)

 $\mathbf{6}\mathbf{x}\mathbf{-}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0}$           … (5)

$\left[\mathbf{x}\right]\mathbf{+}\mathbf{2}\left[\mathbf{y}\right]\mathbf{+}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{z}\right]\mathbf{=}\mathbf{0}$  … (6)

 

Multiply D+1 on equation (4). Then, add with equation (6).

$$\begin{gathered} \left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{-}\mathbf{2}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{-}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{z}\right]\mathbf{+}\left[\mathbf{x}\right]\mathbf{+}\mathbf{2}\left[\mathbf{y}\right]\mathbf{+}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{z}\right]\mathbf{=}\mathbf{0} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{+}\left[\mathbf{x}\right]\mathbf{-}\mathbf{2}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{+}\mathbf{2}\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{+}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{+}\left(\mathbf{-}\mathbf{2}\mathbf{D}\mathbf{-}\mathbf{2}\mathbf{+}\mathbf{2}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ {\mathbf{D}}^{\mathbf{2}}\left[\mathbf{x}\right]\mathbf{-}\mathbf{2}\mathbf{D}\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ {\mathbf{D}}^{\mathbf{2}}\left[\mathbf{x}\right]\mathbf{-}\mathbf{2}\mathbf{D}\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \dots \left(7\right) \end{gathered}$$ 

 Now multiply D² on equation (5) and 6 on equation (7). Then, subtract them.

$$\begin{gathered} \mathbf{6}{\mathbf{D}}^{\mathbf{2}}\left[\mathbf{x}\right]\mathbf{-}\mathbf{12}\mathbf{D}\left[\mathbf{y}\right]\mathbf{-}\mathbf{6}{\mathbf{D}}^{\mathbf{2}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{D}}^{\mathbf{2}}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \left({\mathbf{D}}^{\mathbf{3}}\mathbf{+}{\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{12}\mathbf{D}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \mathbf{D}\left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{12}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \mathbf{D}\left(\mathbf{D}\mathbf{+}\mathbf{4}\right)\left(\mathbf{D}\mathbf{-}\mathbf{3}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \end{gathered}$$ 

Step 3: Substitution method

Since the auxiliary equation to the corresponding homogeneous equation is ${\mathbf{r}}^{\mathbf{3}}\mathbf{+}{\mathbf{r}}^{\mathbf{2}}\mathbf{-}\mathbf{12}\mathbf{r}\mathbf{=}\mathbf{0}$. The roots are $\mathbf{r}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{r}\mathbf{=}\mathbf{-}\mathbf{4}$ and $\mathbf{r}\mathbf{=}3$.

 

Then, the general solution of y is;

$\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{3}\mathbf{t}}\mathbf{+}\mathbf{C} \dots \left(8\right)$ 

 Now substitute equation (8) in equation (5).

 $$\begin{gathered} \mathbf{6}\mathbf{x}\mathbf{-}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{0} \\ \mathbf{6}\mathbf{x}\mathbf{=}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{y}\right] \\ \mathbf{=}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{3}\mathbf{t}}\mathbf{+}\mathbf{C}\right] \\ \mathbf{=}\mathbf{-}\mathbf{4}{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{+}\mathbf{3}{\mathbf{Be}}^{\mathbf{3}\mathbf{t}}\mathbf{+}{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{3}\mathbf{t}}\mathbf{+}\mathbf{C} \\ \mathbf{=}\mathbf{-}\mathbf{3}{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{+}\mathbf{4}{\mathbf{Be}}^{\mathbf{3}\mathbf{t}}\mathbf{+}\mathbf{C} \\ \mathbf{=}\frac{\mathbf{-}\mathbf{3}{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{+}\mathbf{4}{\mathbf{Be}}^{\mathbf{3}\mathbf{t}}\mathbf{+}\mathbf{C}}{\mathbf{6}} \\ \mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{3}}{\mathbf{Be}}^{\mathbf{3}\mathbf{t}}\mathbf{+}\frac{\mathbf{C}}{\mathbf{6}} \\ \mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{3}}{\mathbf{Be}}^{\mathbf{3}\mathbf{t}}\mathbf{+}\frac{\mathbf{C}}{\mathbf{6}} \dots \left(9\right) \end{gathered}$$

 Now substitute the equation (8) and (9) in equation (4)

$$\begin{gathered} \left(\mathrm{D}-1\right)\left[\mathrm{x}\right]-2\left[\mathrm{y}\right]-\left[\mathrm{z}\right]=0 \\ \mathrm{z}=\left(\mathrm{D}-1\right)\left[\mathrm{x}\right]-2\left[\mathrm{y}\right] \\ \mathrm{z}=\left(\mathrm{D}-1\right)\left[-\frac{1}{2}{\mathrm{Ae}}^{-4\mathrm{t}}+\frac{2}{3}{\mathrm{Be}}^{3\mathrm{t}}+\frac{\mathrm{C}}{6}\right]-2\left[{\mathrm{Ae}}^{-4\mathrm{t}}+{\mathrm{Be}}^{3\mathrm{t}}+\mathrm{C}\right] \\ =\frac{4}{2}{\mathrm{Ae}}^{-4\mathrm{t}}+\frac{6}{3}{\mathrm{Be}}^{3\mathrm{t}}+\frac{1}{2}{\mathrm{Ae}}^{-4\mathrm{t}}-\frac{2}{3}{\mathrm{Be}}^{3\mathrm{t}}-\frac{\mathrm{C}}{6}-2{\mathrm{Ae}}^{-4\mathrm{t}}-2{\mathrm{Be}}^{3\mathrm{t}}-2\mathrm{C} \\ =2{\mathrm{Ae}}^{-4\mathrm{t}}+2{\mathrm{Be}}^{3\mathrm{t}}+\frac{1}{2}{\mathrm{Ae}}^{-4\mathrm{t}}-\frac{2}{3}{\mathrm{Be}}^{3\mathrm{t}}-\frac{\mathrm{C}}{6}-2{\mathrm{Ae}}^{-4\mathrm{t}}-2{\mathrm{Be}}^{3\mathrm{t}}-2\mathrm{C} \\ =\frac{1}{2}{\mathrm{Ae}}^{-4\mathrm{t}}-\frac{2}{3}{\mathrm{Be}}^{3\mathrm{t}}-\frac{13}{6}\mathrm{C} \end{gathered}$$ 

So, the solution is founded.