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### 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

# Verify that the solution to the initial value problem${\mathbf{x}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{5}}{\mathbf{x}}{\mathbf{-}}{\mathbf{3}}{\mathbf{y}}{\mathbf{-}}{\mathbf{2}}{\mathbf{;}}{ }{ }{ }{ }{\mathbf{x}}\left(\mathbf{0}\right){\mathbf{=}}{\mathbf{2}}{\mathbf{,}}\phantom{\rule{0ex}{0ex}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{4}}{\mathbf{x}}{\mathbf{-}}{\mathbf{3}}{\mathbf{y}}{\mathbf{-}}{\mathbf{1}}{\mathbf{;}}{ }{ }{ }{\mathbf{y}}\left(\mathbf{0}\right){\mathbf{=}}{\mathbf{0}}$Satisfies $|x\left(t\right)|\mathbf{+}|y\left(t\right)|\mathbf{\to }\mathbf{+}\mathbf{\infty }$ as $\mathbf{t}\mathbf{\to }\mathbf{+}\mathbf{\infty }$

The solutions for the given initial value problem are $x\left(t\right)=-\frac{5}{4}{e}^{-t}+\frac{9}{4}{c}_{2}{e}^{3t}+1$

and $y\left(t\right)=-\frac{5}{2}{e}^{-t}+\frac{3}{2}{e}^{3t}+1$. Then, it satisfies the $\underset{t\to \infty }{lim}\left(\left|x\left(t\right)\right|+\left|y\left(t\right)\right|\right)=\infty$ also.

See the step by step solution

## Step 1: General form

Elimination Procedure for 2 × 2 Systems:

To find a general solution for the system;

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

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,

$x\text{'}=5x-3y-2 ......\mathbf{\left(}\mathbf{1}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}y\text{'}=4x-3y-1 ......\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

To verify: $\left|x\left(t\right)\right|+\left|y\left(t\right)\right|\to +\infty$ as $t\to +\infty$.

Let us rewrite this system of operators in operator form:

$\left(D-5\right)\left[x\right]+3y=-2 ......\mathbf{\left(}\mathbf{3}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}4x-\left(D+3\right)\left[y\right]=1 ......\mathbf{\left(}\mathbf{4}\mathbf{\right)}$

Multiply 4 on equation (3) and D-5 on equation (4). Then, add them together to get,

$\left(4\left(D-5\right)\left[x\right]+12y\right)-\left(4\left(D-5\right)x-\left(D-5\right)\left(D+3\right)\left[y\right]\right)=-8-\left(D-5\right)1\phantom{\rule{0ex}{0ex}}\left(D-5\right)\left(D+3\right)\left[y\right]-12y=-8+5\phantom{\rule{0ex}{0ex}}\left({D}^{2}-2D-15+12\right)\left[y\right]=-3\phantom{\rule{0ex}{0ex}}\left({D}^{2}-2D-3\right)\left[y\right]=-3$

$\left({D}^{2}-2D-3\right)\left[y\right]=-3 ......\mathbf{\left(}\mathbf{5}\mathbf{\right)}$

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

Then, the homogeneous solution is ${\mathbf{y}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}} ......\mathbf{\left(}\mathbf{6}\mathbf{\right)}$

Let us take the undetermined coefficients and assume that ${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{C} ......\mathbf{\left(}\mathbf{7}\mathbf{\right)}$

Now derivate the equation (7)

$D\left[{y}_{p}\left(t\right)\right]=0$

Substitute the derivation in equation (5).

$\left({D}^{2}-2D-3\right)\left[C\right]=-3\phantom{\rule{0ex}{0ex}}0-0-3C=-3\phantom{\rule{0ex}{0ex}}C=1$

So, ${y}_{p}\left(t\right)=1$.

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

## Step 3: Substitution method

Substitute equation (8) in equation (4).

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

$4x=1-{c}_{1}{e}^{-t}+3{c}_{2}{e}^{3t}+3{c}_{1}{e}^{-t}+3{c}_{2}{e}^{3t}+3\phantom{\rule{0ex}{0ex}}=2{c}_{1}{e}^{-t}+6{c}_{2}{e}^{3t}+4\phantom{\rule{0ex}{0ex}}x=\frac{2{c}_{1}{e}^{-t}+6{c}_{2}{e}^{3t}+4}{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{c}_{1}{e}^{-t}+\frac{3}{2}{c}_{2}{e}^{3t}+1$

$x\left(t\right)=\frac{1}{2}{c}_{1}{e}^{-t}+\frac{3}{2}{c}_{2}{e}^{3t}+1 ......\mathbf{\left(}\mathbf{9}\mathbf{\right)}$

## Step 4: Find the initial value problem

Given, $x\left(0\right)=2, y\left(0\right)=0$.

Now substitute the values in equations (8) and (9).

$x\left(t\right)=\frac{1}{2}{c}_{1}{e}^{-t}+\frac{3}{2}{c}_{2}{e}^{3t}+1\phantom{\rule{0ex}{0ex}}x\left(0\right)=\frac{1}{2}{c}_{1}{e}^{0}+\frac{3}{2}{c}_{2}{e}^{3\left(0\right)}+1\phantom{\rule{0ex}{0ex}}1=\frac{1}{2}{c}_{1}+\frac{3}{2}{c}_{2}\phantom{\rule{0ex}{0ex}}{c}_{1}+3{c}_{2}=2 ......\mathbf{\left(}\mathbf{10}\mathbf{\right)}$

$y\left(t\right)={c}_{1}{e}^{-t}+{c}_{2}{e}^{3t}+1\phantom{\rule{0ex}{0ex}}y\left(0\right)={c}_{1}{e}^{-0}+{c}_{2}{e}^{3\left(0\right)}+1\phantom{\rule{0ex}{0ex}}-1={c}_{1}+{c}_{2}\phantom{\rule{0ex}{0ex}}{c}_{1}+{c}_{2}=-1 ......\mathbf{\left(}\mathbf{11}\mathbf{\right)}$

First, solve the equations (10) and (11).

${c}_{1}+3{c}_{2}-{c}_{1}-{c}_{2}=2+1\phantom{\rule{0ex}{0ex}}2{c}_{2}=3\phantom{\rule{0ex}{0ex}}{c}_{2}=\frac{3}{2}$

Then,

$3{c}_{1}+3{c}_{2}-{c}_{1}-3{c}_{2}=-3-2\phantom{\rule{0ex}{0ex}}2{c}_{1}=-5\phantom{\rule{0ex}{0ex}}{c}_{1}=-\frac{5}{2}$

Now substitute the values of c in equations (8) and (9).

$x\left(t\right)=-\frac{5}{4}{e}^{-t}+\frac{9}{4}{c}_{2}{e}^{3t}+1\phantom{\rule{0ex}{0ex}}y\left(t\right)=-\frac{5}{2}{e}^{-t}+\frac{3}{2}{e}^{3t}+1$

Now calculate the limits:

$\underset{t\to \infty }{lim}\left(\left|x\left(t\right)\right|+\left|y\left(t\right)\right|\right)=\underset{t\to \infty }{lim}\left(\left|-\frac{5}{4}{e}^{-t}+\frac{9}{4}{c}_{2}{e}^{3t}+1\right|+\left|-\frac{5}{2}{e}^{-t}+\frac{3}{2}{e}^{3t}+1\right|\right)\phantom{\rule{0ex}{0ex}}=\underset{t\to \infty }{lim}\left|-\frac{5}{4}{e}^{-t}+\frac{9}{4}{c}_{2}{e}^{3t}+1\right|+\underset{t\to \infty }{lim}\left|-\frac{5}{2}{e}^{-t}+\frac{3}{2}{e}^{3t}+1\right|\phantom{\rule{0ex}{0ex}}=\left|-0+\infty +1\right|+\left|-0+\infty +1\right|\phantom{\rule{0ex}{0ex}}\underset{t\to \infty }{lim}\left(\left|x\left(t\right)\right|+\left|y\left(t\right)\right|\right)=\infty$

So, the solution is founded.