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Q10E

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Found in: Page 249

### 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 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.${\mathbf{2}}{\mathbf{x}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{x}}{\mathbf{-}}{\mathbf{y}}{\mathbf{=}}{{\mathbf{e}}}^{\mathbf{-}\mathbf{t}}{\mathbf{,}}\phantom{\rule{0ex}{0ex}}{\mathbf{x}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{2}}{\mathbf{x}}{\mathbf{+}}{\mathbf{y}}{\mathbf{=}}{{\mathbf{e}}}^{{\mathbf{t}}}$

The solutions for the given linear system are $x\left(t\right)=A\mathrm{sin}t+B\mathrm{cos}t$ and $y\left(t\right)={c}_{1}{e}^{t}-\frac{A+B}{2}\mathrm{sin}t+\frac{3\left(A-B\right)}{2}\mathrm{cos}t-\frac{1}{2}{e}^{-t}$

See the step by step solution

## Step 1: General form

Elimination Procedure for 2 × 2 Systems.

To find a general solution for the system

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

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.

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,

$2x\text{'}+y\text{'}-x-y={e}^{-t} ......\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

$x\text{'}+y\text{'}+2x+y={e}^{t} ......\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Let us rewrite this system of operators in operator form:

$\left(2D-1\right)\left[x\right]+\left(D-1\right)\left[y\right]={e}^{-t} ......\mathbf{\left(}\mathbf{3}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\left(D+2\right)\left[x\right]+\left(D+1\right)\left[y\right]={e}^{t} ......\mathbf{\left(}\mathbf{4}\mathbf{\right)}$

Multiply (D+1) on equation (3).

$\left(D+1\right)\left(2D-1\right)\left[x\right]+\left(D+1\right)\left(D-1\right)\left[y\right]=\left(D+1\right){e}^{-t}\phantom{\rule{0ex}{0ex}}\left(2{D}^{2}-D+2D-1\right)\left[x\right]+\left({D}^{2}-1\right)\left[y\right]=-{e}^{-t}+{e}^{-t}\phantom{\rule{0ex}{0ex}}\left(2{D}^{2}+D-1\right)\left[x\right]+\left({D}^{2}-1\right)\left[y\right]=0 ......\mathbf{\left(}\mathbf{5}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}$

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

$\left(D-1\right)\left(D+2\right)\left[x\right]+\left(D-1\right)\left(D+1\right)\left[y\right]=\left(D-1\right){e}^{t}\phantom{\rule{0ex}{0ex}}\left({D}^{2}+2D-D-2\right)\left[x\right]+\left({D}^{2}-1\right)\left[y\right]={e}^{t}-{e}^{t}\phantom{\rule{0ex}{0ex}}\left({D}^{2}+D-2\right)\left[x\right]+\left({D}^{2}-1\right)\left[y\right]=0 ......\mathbf{\left(}\mathbf{6}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}$

Then subtract equation (5) and (6) together one gets,

$\left(2{D}^{2}+D-1\right)\left[x\right]+\left({D}^{2}-1\right)\left[y\right]-\left(\left({D}^{2}+D-2\right)\left[x\right]+\left({D}^{2}-1\right)\left[y\right]\right)=0\phantom{\rule{0ex}{0ex}}\left(2{D}^{2}+D-1-{D}^{2}-D+2\right)\left[x\right]=0\phantom{\rule{0ex}{0ex}}\left({D}^{2}+1\right)\left[x\right]=0\phantom{\rule{0ex}{0ex}}\left({D}^{2}+1\right)\left[x\right]=0 ......\mathbf{\left(}\mathbf{7}\mathbf{\right)}$

Now solve equation (7).

$\left({D}^{2}+1\right)\left[x\right]=0\phantom{\rule{0ex}{0ex}}x\text{'}\text{'}+x=0\phantom{\rule{0ex}{0ex}}x=A\mathrm{sin}t+B\mathrm{cos}t$

## Step 3: Substitution method

Substitute the value of x in equation (3)

$\left(2D-1\right)\left[x\right]+\left(D-1\right)\left[y\right]={e}^{-t}\phantom{\rule{0ex}{0ex}}\left(2D-1\right)\left[A\mathrm{sin}t+B\mathrm{cos}t\right]+\left(D-1\right)\left[y\right]={e}^{-t}\phantom{\rule{0ex}{0ex}}2A\mathrm{cos}t-2B\mathrm{sin}t-A\mathrm{sin}t-B\mathrm{cos}t+\left(D-1\right)\left[y\right]={e}^{-t}\phantom{\rule{0ex}{0ex}}\left(2A-B\right)\mathrm{cos}t-\left(A+2B\right)\mathrm{sin}t+\left(D-1\right)\left[y\right]={e}^{-t}$

$\left(D-1\right)\left[y\right]={e}^{-t}+\left(A+2B\right)\mathrm{sin}t-\left(2A-B\right)\mathrm{cos}t ......\mathbf{\left(}\mathbf{8}\mathbf{\right)}$

Since the auxiliary equation to the corresponding homogeneous equation is $r-1=0$.

The root is r=1.

Then, the homogeneous solution of u is ${y}_{h}\left(t\right)={c}_{1}{e}^{t} ......\mathbf{\left(}\mathbf{9}\mathbf{\right)}$

Let us take the undetermined coefficients and assume that

${y}_{p}\left(t\right)=a\mathrm{sin}t+b\mathrm{cos}t+{c}_{2}{e}^{-t} ......\mathbf{\left(}\mathbf{10}\mathbf{\right)}$

Now derivate the equation (7)

$D\left[{y}_{p}\left(t\right)\right]=a\mathrm{cos}t-b\mathrm{sin}t-{c}_{2}{e}^{-t} ......\mathbf{\left(}\mathbf{11}\mathbf{\right)}$

## Step 4: Substitution method

Subtract the equations (11) and (10).

$D\left[{y}_{p}\left(t\right)\right]-{y}_{p}\left(t\right)=a\mathrm{cos}t-b\mathrm{sin}t-{c}_{2}{e}^{-t}-a\mathrm{sin}t-b\mathrm{cos}t-{c}_{2}{e}^{-t}\phantom{\rule{0ex}{0ex}}=\left(a-b\right)\mathrm{cos}t-\left(a+b\right)\mathrm{sin}t-2{c}_{2}{e}^{-t}$

Now, equalize the like terms of equation (8).

$A+2B=-\left(a+b\right)\phantom{\rule{0ex}{0ex}}-2A+B=\left(a-b\right)\phantom{\rule{0ex}{0ex}}1=-2{c}_{2}\phantom{\rule{0ex}{0ex}}{c}_{2}=-\frac{1}{2}$

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

$3B-3A=-2b\phantom{\rule{0ex}{0ex}}b=\frac{3\left(A-B\right)}{2}$

Then,

$A+B=-2a\phantom{\rule{0ex}{0ex}}a=-\frac{A+B}{2}$

So, ${y}_{p}\left(t\right)=-\frac{A+B}{2}\mathrm{sin}t+\frac{3\left(A-B\right)}{2}\mathrm{cos}t-\frac{1}{2}{e}^{-t} ......\mathbf{\left(}\mathbf{12}\mathbf{\right)}$

Use equations (9) and (12) to get,

$y\left(t\right)={y}_{h}\left(t\right)+{y}_{p}\left(t\right)\phantom{\rule{0ex}{0ex}}y\left(t\right)={c}_{1}{e}^{t}-\frac{A+B}{2}\mathrm{sin}t+\frac{3\left(A-B\right)}{2}\mathrm{cos}t-\frac{1}{2}{e}^{-t}$

So, the solution is founded.