Suggested languages for you:

Americas

Europe

Q27E

Expert-verified
Found in: Page 250

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# 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), z(t) ${\mathbf{x}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{4}}{\mathbf{x}}{\mathbf{-}}{\mathbf{4}}{\mathbf{z}}{\mathbf{,}}\phantom{\rule{0ex}{0ex}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{4}}{\mathbf{y}}{\mathbf{-}}{\mathbf{2}}{\mathbf{z}}{\mathbf{,}}\phantom{\rule{0ex}{0ex}}{\mathbf{z}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{2}}{\mathbf{x}}{\mathbf{-}}{\mathbf{4}}{\mathbf{y}}{\mathbf{+}}{\mathbf{4}}{\mathbf{z}}$

The solutions for the given linear system are $x\left(t\right)={c}_{1}-{c}_{2}{e}^{8t}-2{c}_{3}{e}^{4t}$, $y\left(t\right)=\frac{{c}_{1}}{2}-\frac{{c}_{2}}{2}{e}^{8t}+{c}_{3}{e}^{4t}$ and $z\left(t\right)={c}_{1}+{c}_{2}{e}^{8t}$.

See the step by step solution

## Step 1: General form

Elimination Procedure for 2 × 2 Systems:

To find a general solution for the system

${\mathbf{L}}_{\mathbf{1}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{1}}\mathbf{,}\phantom{\rule{0ex}{0ex}}{\mathbf{L}}_{\mathbf{3}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{4}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{2}}\mathbf{,}$

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 x(t). 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 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), y(t) give the desired general solution.
4. Eliminate x from the system and solve for y(t). [Solving for y(t) gives more constants----in fact, twice as many as needed.]
5. 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,

$x\text{'}=4x-4z ......\mathbf{\left(}\mathbf{1}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}y\text{'}=4y-2z ......\mathbf{\left(}\mathbf{2}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}z\text{'}=-2x-4y+4z ......\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

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

$\left(D-4\right)\left[x\right]+4\left[z\right]=0 ......\mathbf{\left(}\mathbf{4}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\left(D-4\right)\left[y\right]+2\left[z\right]=0 ......\mathbf{\left(}\mathbf{5}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}2\left[x\right]+4\left[y\right]+\left(D-4\right)\left[z\right]=0 ......\mathbf{\left(}\mathbf{6}\mathbf{\right)}$

Multiply D-4 on equation (6). Then, substitute equation (4) and (5).

$2\left(D-4\right)\left[x\right]+4\left(D-4\right)\left[y\right]+{\left(D-4\right)}^{2}\left[z\right]=0\phantom{\rule{0ex}{0ex}}-8\left[z\right]-8\left[z\right]+{\left(D-4\right)}^{2}\left[z\right]=0\phantom{\rule{0ex}{0ex}}{\left(D-4\right)}^{2}\left[z\right]-16\left[z\right]=0\phantom{\rule{0ex}{0ex}}\left({D}^{2}-8D+16-16\right)\left[z\right]=0$

$\left({D}^{2}-8D\right)\left[z\right]=0\phantom{\rule{0ex}{0ex}}\left({D}^{2}-8D\right)\left[z\right]=0 ......\mathbf{\left(}\mathbf{7}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}$

## Step 3: Substitution method

Since, the auxiliary equation to the corresponding homogeneous equation is:

${r}^{2}-8r=0$

. The roots are r =0 and r = 8.

Then, the general solution of z is

$z\left(t\right)={c}_{1}+{c}_{2}{e}^{8t} ......\mathbf{\left(}\mathbf{8}\mathbf{\right)}$

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

$\left(D-4\right)\left[y\right]+2\left[z\right]=0\phantom{\rule{0ex}{0ex}}\left(D-4\right)\left[y\right]=-2\left[z\right]\phantom{\rule{0ex}{0ex}}\left(D-4\right)\left[y\right]=-2\left[{c}_{1}+{c}_{2}{e}^{8t}\right]\phantom{\rule{0ex}{0ex}}\left(D-4\right)\left[y\right]=-2{c}_{1}-2{c}_{2}{e}^{8t}$

$\left(D-4\right)\left[y\right]=-2{c}_{1}-2{c}_{2}{e}^{8t} ......\mathbf{\left(}\mathbf{9}\mathbf{\right)}$

So, the complementary solution of the differential equation is ${y}_{h}\left(t\right)={c}_{3}{e}^{4t} ......\mathbf{\left(}\mathbf{10}\mathbf{\right)}$

Let us assume that function: ${y}_{p}\left(t\right)=A+B{e}^{8t} ......\mathbf{\left(}\mathbf{11}\mathbf{\right)}$

Find the derivation of equation (11).

$D\left[{y}_{p}\left(t\right)\right]=8B{e}^{8t}$

Use the derivation in equations (9) to get,

$\left(D-4\right)\left[y\right]=-2{c}_{1}-2{c}_{2}{e}^{8t}\phantom{\rule{0ex}{0ex}}\left(D-4\right)\left[A+B{e}^{8t}\right]=-2{c}_{1}-2{c}_{2}{e}^{8t}\phantom{\rule{0ex}{0ex}}8B{e}^{8t}-4A-4B{e}^{8t}=-2{c}_{1}-2{c}_{2}{e}^{8t}\phantom{\rule{0ex}{0ex}}-4A+4B{e}^{8t}=-2{c}_{1}-2{c}_{2}{e}^{8t}$

Now, equalise the like terms.

$4A=2{c}_{1}\phantom{\rule{0ex}{0ex}}A=\frac{{c}_{1}}{2}$

Then,

$4B=-2{c}_{2}\phantom{\rule{0ex}{0ex}}B=-\frac{{c}_{2}}{2}$

So, ${y}_{p}\left(t\right)=\frac{{c}_{1}}{2}-\frac{{c}_{2}}{2}{e}^{8t}$.

Then, $y\left(t\right)=\frac{{c}_{1}}{2}-\frac{{c}_{2}}{2}{e}^{8t}+{c}_{3}{e}^{4t} ......\mathbf{\left(}\mathbf{12}\mathbf{\right)}$

Now substitute the equation (8) and (12) in equation (6)

$2\left[x\right]+4\left[y\right]+\left(D-4\right)\left[z\right]=0\phantom{\rule{0ex}{0ex}}2\left[x\right]=-4\left[y\right]-\left(D-4\right)\left[z\right]\phantom{\rule{0ex}{0ex}}x=-2\left[y\right]-\frac{1}{2}D\left[z\right]+2\left[z\right]\phantom{\rule{0ex}{0ex}}=-2\left[\frac{{c}_{1}}{2}-\frac{{c}_{2}}{2}{e}^{8t}+{c}_{3}{e}^{4t}\right]-\frac{1}{2}D\left[{c}_{1}+{c}_{2}{e}^{8t}\right]+2\left[{c}_{1}+{c}_{2}{e}^{8t}\right]$

$=-2\frac{{c}_{1}}{2}+2\frac{{c}_{2}}{2}{e}^{8t}-2{c}_{3}{e}^{4t}-\frac{8}{2}{c}_{2}{e}^{8t}+2{c}_{1}+2{c}_{2}{e}^{8t}\phantom{\rule{0ex}{0ex}}={c}_{1}-{c}_{2}{e}^{8t}-2{c}_{3}{e}^{4t}$

So, the solution is founded.

## Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.