Q33E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problem 31, assume that no solution flows out of the system from tank B, only 1 L/min flows from A into B, and only 4 L/min of brine flows into the system at tank A, other data being the same. Determine the mass of salt in each tank at the time $\mathbf{t}⩾\mathbf{0}$.

Step-by-Step Solution

Answer

The mass of salt in each tank at the time $t⩾0$  is

$x\left(t\right)=\left(-2+\sqrt{5}\right)\frac{10}{\sqrt{5}}{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)t}-\left(-2-\sqrt{5}\right)\frac{10}{\sqrt{5}}{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)t}+20$ and $y\left(t\right)=\frac{10}{\sqrt{5}}{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)t}-\frac{10}{\sqrt{5}}{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)t}+20$

.

Step 1: General form

Elimination Procedure for 2 x 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 L₄ 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----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.

 

Vieta’s formulas for finding roots:

 

For the function 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 $r_1,r_2=\alpha \pm \beta i$, $\mathbf{\beta }\ne 0$ , then $\alpha =\frac{r_1+r_2}{2}⩽0$.

Step 2: Evaluate the given equation

Given that, the fluid is flowing from tank A to tank B at the rate of $1L/min$ and from B into A at a rate of $4L/min$.

 

Referring to problem 31: 

 

The volume of both tanks is 100L. 

 

A brine solution with a concentration of $0.2kg/L$ of salt flows into tank A at a rate of $4L/min$.

 

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

 

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

 

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 $=4\times 0.2+1\times \frac{y\left(t\right)}{100}=0.8+0.01y$

Rate of outflow $=4\times \frac{x\left(t\right)}{100}+1\times \frac{x\left(t\right)}{100}=0.05x$

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

$$\begin{gathered} Dx=0.8+0.01y-0.05x \\ \left(D+0.05\right)\left[x\right]-0.01y=0.8 \\ \left(D+0.05\right)\left[x\right]-0.01y=0.8 ......\mathbf{\left(}\mathbf{1}\mathbf{\right)} \end{gathered}$$

 

For tank B:

 

Rate of inflow $=1\times \frac{x\left(t\right)}{100}=0.01x$

Rate of outflow $=1\times \frac{y\left(t\right)}{100}=0.01y$

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

$$\begin{gathered} Dy=0.01x-0.01y \\ 0.01x-\left(D+0.01\right)\left[y\right]=0 \\ 0.01x-\left(D+0.01\right)\left[y\right]=0 ......\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

 

Step 3: Solve the equations

Multiply 0.01 on equation (3) and multiply D+0.05 on equation (4). Then, subtract them together.

$$\begin{gathered} 0.01\left(D+0.05\right)\left[x\right]-0.0001y-\left(0.01\left(D+0.05\right)x-\left(D+0.05\right)\left(D+0.01\right)\left[y\right]\right)=0.008 \\ \left(D+0.05\right)\left(D+0.01\right)\left[y\right]-0.0001\left[y\right]=0.008 \\ \left(D^2+0.06D+0.0005-0.0001\right)\left[y\right]=0.008 \\ \left(D^2+0.06D+0.0004\right)\left[y\right]=0.008 \end{gathered}$$ 

$\left(D^2+0.06D+0.0004\right)\left[y\right]=0.008......\mathbf{\left(}\mathbf{5}\mathbf{\right)}$

Step 4: Substitution method

Since the auxiliary equation to the corresponding homogeneous equation is . 

$r^2+0.06r+0.0004=0$ 

Then,

$$\begin{gathered} r=\frac{-0.06\pm \sqrt{{\left(0.06\right)}^2-4\times 0.0004}}{2} \\ =\frac{-0.06\pm \sqrt{0.0036-0.0016}}{2} \\ =\frac{-0.06\pm \sqrt{0.0020}}{2} \\ =\frac{-3\pm \sqrt{5}}{100} \end{gathered}$$ 

 So, the roots are $r=\frac{-3+\sqrt{5}}{100}$ and $r=\frac{-3-\sqrt{5}}{100}$.

 

Then, the general solution of y is $y_h\left(t\right)=A{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)t}+B{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)t} ......\mathbf{\left(}\mathbf{6}\mathbf{\right)}$

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

 

Substitute equation (7) in equation (5).

 $$\begin{gathered} \left(D^2+0.06D+0.0004\right)\left[y\right]=0.008 \\ \left(D^2+0.06D+0.0004\right)\left[C\right]=0.008 \\ 0.0004C=0.008 \\ C=\frac{0.008}{0.0004} \\ =20 \end{gathered}$$

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

$$\begin{gathered} y\left(t\right)=y_h\left(t\right)+y_p\left(t\right) \\ =A{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)t}+B{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)t}+20 \end{gathered}$$ 

 Hence,  $y\left(t\right)=A{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)t}+B{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)t}+20 ......\mathbf{\left(}\mathbf{8}\mathbf{\right)}$

Step 5: Substitution method

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

 $$\begin{gathered} 0.01x-\left(D+0.01\right)\left[y\right]=0 \\ 0.01x=\left(D+0.01\right)\left[y\right] \\ 0.01x=\left(D+0.01\right)\left[A{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)t}+B{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)t}+20\right] \\ =\frac{-3+\sqrt{5}}{100}A{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)t}+\frac{-3-\sqrt{5}}{100}B{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)t}+0.01A{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)t}+0.01B{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)t}+0.2 \end{gathered}$$

$$\begin{gathered} =\frac{-3+1+\sqrt{5}}{100}A{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)t}+\frac{-3+1-\sqrt{5}}{100}B{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)t}+0.2 \\ =\frac{-2+\sqrt{5}}{100}A{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)t}+\frac{-2-\sqrt{5}}{100}B{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)t}+0.2 \\ x=\frac{-2+\sqrt{5}}{100}\left(\frac{100}{1}\right)A{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)t}+\frac{-2-\sqrt{5}}{100}\left(\frac{100}{1}\right)B{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)t}+\left(\frac{0.2\times 100}{1}\right) \\ =\left(-2+\sqrt{5}\right)A{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)t}+\left(-2-\sqrt{5}\right)B{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)t}+20 \end{gathered}$$

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

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):

$$\begin{gathered} x\left(t\right)=\left(-2+\sqrt{5}\right)A{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)t}+\left(-2-\sqrt{5}\right)B{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)t}+20 \\ x\left(0\right)=\left(-2+\sqrt{5}\right)A{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)0}+\left(-2-\sqrt{5}\right)B{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)0}+20 \\ 0=\left(-2+\sqrt{5}\right)A+\left(-2-\sqrt{5}\right)B+20 \end{gathered}$$ 

 So, $\left(-2+\sqrt{5}\right)A+\left(-2-\sqrt{5}\right)B=-20 ......\mathbf{\left(}\mathbf{a}\mathbf{\right)}$

 

Case (2):

 $$\begin{gathered} y\left(t\right)=A{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)t}+B{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)t}+20 \\ y\left(0\right)=A{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)0}+B{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)0}+20 \\ 20=A+B+20 \\ 0=A+B \end{gathered}$$

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

 

Solve the equation (a) and (b).

$$\begin{gathered} \left(-2+\sqrt{5}\right)A+\left(-2-\sqrt{5}\right)B-\left(-2-\sqrt{5}\right)A-\left(-2-\sqrt{5}\right)B=-20 \\ \left(-2+\sqrt{5}+2+\sqrt{5}\right)A=-20 \\ 2\sqrt{5}A=-20 \\ A=\frac{20}{2\sqrt{5}} \\ =\frac{10}{\sqrt{5}} \end{gathered}$$ 

 Substitute the value of A in equation (b).

$$\begin{gathered} A+B=0 \\ \frac{10}{\sqrt{5}}+B=0 \\ B=-\frac{10}{\sqrt{5}} \end{gathered}$$

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

 $x\left(t\right)=\left(-2+\sqrt{5}\right)\frac{10}{\sqrt{5}}{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)t}-\left(-2-\sqrt{5}\right)\frac{10}{\sqrt{5}}{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)t}+20$

 $y\left(t\right)=\frac{10}{\sqrt{5}}{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)t}-\frac{10}{\sqrt{5}}{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)t}+20$

$y\left(t\right)=\frac{10}{\sqrt{5}}{e}^{\left(\frac{-3+\sqrt{5}}{100}\right)t}-\frac{10}{\sqrt{5}}{e}^{\left(\frac{-3-\sqrt{5}}{100}\right)t}+20$ 

Therefore, the solution is founded