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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

# Two large tanks, each holding 100 L of liquid, are interconnected by pipes, with the liquid flowing from tank A into tank B at a rate of 3 L/min and from B into A at a rate of 1 L/min (see Figure 5.2). The liquid inside each tank is kept well stirred. A brine solution with a concentration of 0.2 kg/L of salt flows into tank A at a rate of 6 L/min. The (diluted) solution flows out of the system from tank A at 4 L/min and from tank B at 2 L/min. If, initially, tank A contains pure water and tank B contains 20 kg of salt, determine the mass of salt in each tank at a time ${\mathbf{t}}{⩾}{\mathbf{0}}$.

The mass of salt in each tank at the time $\mathbf{t}⩾\mathbf{0}$ is;

$\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\left(\mathbf{10}\mathbf{-}\frac{\mathbf{20}}{\sqrt{\mathbf{7}}}\right){\mathbf{e}}^{\left(\frac{\mathbf{-}\mathbf{5}\mathbf{+}\sqrt{\mathbf{7}}}{\mathbf{100}}\right)\mathbf{t}}\mathbf{-}\left(\mathbf{10}\mathbf{+}\frac{\mathbf{20}}{\sqrt{\mathbf{7}}}\right){\mathbf{e}}^{\left(\frac{\mathbf{-}\mathbf{5}\mathbf{-}\sqrt{\mathbf{7}}}{\mathbf{100}}\right)\mathbf{t}}\mathbf{+}\mathbf{20}$

and $\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{30}}{\sqrt{\mathbf{7}}}{\mathbf{e}}^{\left(\frac{\mathbf{-}\mathbf{5}\mathbf{+}\sqrt{\mathbf{7}}}{\mathbf{100}}\right)\mathbf{t}}\mathbf{+}\frac{\mathbf{30}}{\sqrt{\mathbf{7}}}{\mathbf{e}}^{\left(\frac{\mathbf{-}\mathbf{5}\mathbf{-}\sqrt{\mathbf{7}}}{\mathbf{100}}\right)\mathbf{t}}\mathbf{+}\mathbf{20}$

.

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), and 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- 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.

Vieta’s formulas for finding roots:

For function y(t) to be bounded when $t\to +\infty$ we need for both roots of the auxiliary equation to be non-positive if they are reals and, if they are complex, then the real part has to be non-positive. In other words

1. If ${r}_{1},{r}_{2}\in R$, then ${r}_{1}·{r}_{2}⩾0,{r}_{1}+{r}_{2}⩽0$,
2. If ${r}_{1},{r}_{2}=\alpha ±\beta i$, $\beta \ne 0$ , then $\alpha =\frac{{r}_{1}+{r}_{2}}{2}⩽0$.

## Step 2: Evaluate the given equation

Given that, the volume of both tanks is 100 L. Then, the fluid is flowing from tank A to tank B at the rate of 3 L/min and from B into A at a rate of 1 L/min.

A brine solution with a concentration of 0.2 kg/L of salt flows into tank A at a rate of 6 L/min.

The solution flows out of the system from tank A at 4 L/min and from tank B at 2 L/min.

Let us take, the amount of salt in tank A be x(t) kg and the amount of salt in tank B be y(t) kg.

Then, $x\left(0\right)=0$ and $y\left(0\right)=20$.

Let us create the system of equations first.

For tank A:

Rate of inflow $=6×0.2+1×\frac{y\left(t\right)}{100}=1.2+0.01y$

Rate of outflow $=3×\frac{x\left(t\right)}{100}+4×\frac{x\left(t\right)}{100}=0.07x$

$\text{Rate of change}x=\text{Rate of inflow}--\text{Rate of outflow}$

$Dx=1.2+0.01y-0.07x\phantom{\rule{0ex}{0ex}}\left(D+0.07\right)\left[x\right]-0.01y=1.2\phantom{\rule{0ex}{0ex}}\left(D+0.07\right)\left[x\right]-0.01y=1.2 ......\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

For tank B:

Rate of inflow $=3×\frac{x\left(t\right)}{100}=0.03x$

Rate of outflow $=1×\frac{y\left(t\right)}{100}+2×\frac{y\left(t\right)}{100}=0.03y$

$\text{Rate of change}y=\text{Rate of inflow}--\text{Rate of outflow}$

$Dy=0.03x-0.03y\phantom{\rule{0ex}{0ex}}0.03x-\left(D+0.03\right)\left[y\right]=0\phantom{\rule{0ex}{0ex}}0.03x-\left(D+0.03\right)\left[y\right]=0 ......\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

## Step 3: Solve the equations

Multiply 0.03 on equation (3) and multiply D+0.07 on equation (4). Then, subtract them together.

$0.03\left(D+0.07\right)\left[x\right]-0.0003y-\left(0.03\left(D+0.07\right)x-\left(D+0.07\right)\left(D+0.03\right)\left[y\right]\right)=0.036\phantom{\rule{0ex}{0ex}}\left(D+0.07\right)\left(D+0.03\right)\left[y\right]-0.0003\left[y\right]=0.036\phantom{\rule{0ex}{0ex}}\left({D}^{2}+0.1D+0.0021-0.0003\right)\left[y\right]=0.036\phantom{\rule{0ex}{0ex}}\left({D}^{2}+0.1D+0.0018\right)\left[y\right]=0.036\phantom{\rule{0ex}{0ex}}$

$\left({D}^{2}+0.1D+0.0018\right)\left[y\right]=0.036 ......\mathbf{\left(}\mathbf{5}\mathbf{\right)}$

## Step 4: Substitution method

Since the auxiliary equation to the corresponding homogeneous equation is .

${r}^{2}+0.1r+0.0018=0$

Then,

$r=\frac{-0.1±\sqrt{{\left(0.1\right)}^{2}-4×0.0018}}{2}\phantom{\rule{0ex}{0ex}}=\frac{-0.1±\sqrt{0.01-0.0072}}{2}\phantom{\rule{0ex}{0ex}}=\frac{-0.1±\sqrt{0.0028}}{2}\phantom{\rule{0ex}{0ex}}=\frac{-\frac{1}{10}±\sqrt{\frac{28}{10000}}}{2}\phantom{\rule{0ex}{0ex}}=\frac{-5±\sqrt{7}}{100}$

So, the roots are $r=\frac{-5+\sqrt{7}}{100}$ and $r=\frac{-5-\sqrt{7}}{100}$.

Then, the general solution of y is ${y}_{h}\left(t\right)=A{e}^{\left(\frac{-5+\sqrt{7}}{100}\right)t}+B{e}^{\left(\frac{-5-\sqrt{7}}{100}\right)t} ......\mathbf{\left(}\mathbf{6}\mathbf{\right)}$

Let us assume that, ${y}_{p}\left(t\right)=C ......\mathbf{\left(}\mathbf{7}\mathbf{\right)}$

Substitute equation (7) in equation (5).

$\left({D}^{2}+0.1D+0.0018\right)\left[y\right]=0.036\phantom{\rule{0ex}{0ex}}\left({D}^{2}+0.1D+0.0018\right)\left[C\right]=0.036\phantom{\rule{0ex}{0ex}}0.0018C=0.036\phantom{\rule{0ex}{0ex}}C=\frac{0.036}{0.0018}\phantom{\rule{0ex}{0ex}}=20$

Substitute the value of C in equations (7) and y(t).

$y\left(t\right)={y}_{h}\left(t\right)+{y}_{p}\left(t\right)\phantom{\rule{0ex}{0ex}}=A{e}^{\left(\frac{-5+\sqrt{7}}{100}\right)t}+B{e}^{\left(\frac{-5-\sqrt{7}}{100}\right)t}+20$

So, $y\left(t\right)=A{e}^{\left(\frac{-5+\sqrt{7}}{100}\right)t}+B{e}^{\left(\frac{-5-\sqrt{7}}{100}\right)t}+20 ......\mathbf{\left(}\mathbf{8}\mathbf{\right)}$

## Step 5: Substitution method

Now substitute equation (8) in equation (4).

$0.03x-\left(D+0.03\right)\left[y\right]=0\phantom{\rule{0ex}{0ex}}0.03x=\left(D+0.03\right)\left[y\right]\phantom{\rule{0ex}{0ex}}0.03x=\left(D+0.03\right)\left[A{e}^{\left(\frac{-5+\sqrt{7}}{100}\right)t}+B{e}^{\left(\frac{-5-\sqrt{7}}{100}\right)t}+20\right]\phantom{\rule{0ex}{0ex}}=\frac{-5+\sqrt{7}}{100}A{e}^{\left(\frac{-5+\sqrt{7}}{100}\right)t}+\frac{-5-\sqrt{7}}{100}B{e}^{\left(\frac{-5-\sqrt{7}}{100}\right)t}+0.03A{e}^{\left(\frac{-5+\sqrt{7}}{100}\right)t}+0.03B{e}^{\left(\frac{-5-\sqrt{7}}{100}\right)t}+0.6$

$=\frac{-5+3+\sqrt{7}}{100}A{e}^{\left(\frac{-5+\sqrt{7}}{100}\right)t}+\frac{-5+3-\sqrt{7}}{100}B{e}^{\left(\frac{-5-\sqrt{7}}{100}\right)t}+0.6\phantom{\rule{0ex}{0ex}}=\frac{-2+\sqrt{7}}{100}A{e}^{\left(\frac{-5+\sqrt{7}}{100}\right)t}+\frac{-2-\sqrt{7}}{100}B{e}^{\left(\frac{-5-\sqrt{7}}{100}\right)t}+0.6\phantom{\rule{0ex}{0ex}}x=\frac{-2+\sqrt{7}}{100}\left(\frac{100}{3}\right)A{e}^{\left(\frac{-5+\sqrt{7}}{100}\right)t}+\frac{-2-\sqrt{7}}{100}\left(\frac{100}{3}\right)B{e}^{\left(\frac{-5-\sqrt{7}}{100}\right)t}+\left(\frac{0.6×100}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{-2+\sqrt{7}}{3}A{e}^{\left(\frac{-5+\sqrt{7}}{100}\right)t}+\frac{-2-\sqrt{7}}{3}B{e}^{\left(\frac{-5-\sqrt{7}}{100}\right)t}+20$

$x\left(t\right)=\frac{-2+\sqrt{7}}{3}A{e}^{\left(\frac{-5+\sqrt{7}}{100}\right)t}+\frac{-2-\sqrt{7}}{3}B{e}^{\left(\frac{-5-\sqrt{7}}{100}\right)t}+20 ......\mathbf{\left(}\mathbf{9}\mathbf{\right)}$

## Step 6: Initial value problem

Given that, $x\left(0\right)=0$ and $y\left(0\right)=20$.

Substitute the values in equations (8) and (9).

Case (1):

$x\left(t\right)=\frac{-2+\sqrt{7}}{3}A{e}^{\left(\frac{-5+\sqrt{7}}{100}\right)t}+\frac{-2-\sqrt{7}}{3}B{e}^{\left(\frac{-5-\sqrt{7}}{100}\right)t}+20\phantom{\rule{0ex}{0ex}}x\left(0\right)=\frac{-2+\sqrt{7}}{3}A{e}^{\left(\frac{-5+\sqrt{7}}{100}\right)0}+\frac{-2-\sqrt{7}}{3}B{e}^{\left(\frac{-5-\sqrt{7}}{100}\right)0}+20\phantom{\rule{0ex}{0ex}}0=\frac{-2+\sqrt{7}}{3}A+\frac{-2-\sqrt{7}}{3}B+20$

Hence, $\frac{-2+\sqrt{7}}{3}A+\frac{-2-\sqrt{7}}{3}B=-20 ......\mathbf{\left(}\mathbf{a}\mathbf{\right)}$

Case (2):

$y\left(t\right)=A{e}^{\left(\frac{-5+\sqrt{7}}{100}\right)t}+B{e}^{\left(\frac{-5-\sqrt{7}}{100}\right)t}+20\phantom{\rule{0ex}{0ex}}y\left(0\right)=A{e}^{\left(\frac{-5+\sqrt{7}}{100}\right)0}+B{e}^{\left(\frac{-5-\sqrt{7}}{100}\right)0}+20\phantom{\rule{0ex}{0ex}}20=A+B+20\phantom{\rule{0ex}{0ex}}0=A+B$

Thereafter, $A+B=0 ......\mathbf{\left(}\mathbf{b}\mathbf{\right)}$

Solve the equation (a) and (b).

$\frac{-2+\sqrt{7}}{3}A+\frac{-2-\sqrt{7}}{3}B-\frac{-2-\sqrt{7}}{3}A-\frac{-2-\sqrt{7}}{3}B=-20\phantom{\rule{0ex}{0ex}}\frac{-2+\sqrt{7}+2+\sqrt{7}}{3}A=-20\phantom{\rule{0ex}{0ex}}\frac{2\sqrt{7}}{3}A=-20\phantom{\rule{0ex}{0ex}}A=\frac{-20×3}{2\sqrt{7}}\phantom{\rule{0ex}{0ex}}=\frac{-30}{\sqrt{7}}$

Substitute the value of A in equation (b).

$A+B=0\phantom{\rule{0ex}{0ex}}-\frac{30}{\sqrt{7}}+B=0\phantom{\rule{0ex}{0ex}}B=\frac{30}{\sqrt{7}}$

Finally, substitute the values of A and B in equations (8) and (9).

$x\left(t\right)=\frac{-2+\sqrt{7}}{3}\left(\frac{-30}{\sqrt{7}}\right){e}^{\left(\frac{-5+\sqrt{7}}{100}\right)t}+\frac{-2-\sqrt{7}}{3}\left(\frac{30}{\sqrt{7}}\right){e}^{\left(\frac{-5-\sqrt{7}}{100}\right)t}+20\phantom{\rule{0ex}{0ex}}=-\left(10-\frac{20}{\sqrt{7}}\right){e}^{\left(\frac{-5+\sqrt{7}}{100}\right)t}-\left(10+\frac{20}{\sqrt{7}}\right){e}^{\left(\frac{-5-\sqrt{7}}{100}\right)t}+20\phantom{\rule{0ex}{0ex}}y\left(t\right)=-\frac{30}{\sqrt{7}}{e}^{\left(\frac{-5+\sqrt{7}}{100}\right)t}+\frac{30}{\sqrt{7}}{e}^{\left(\frac{-5-\sqrt{7}}{100}\right)t}+20$

Therefore, the solution is founded.