Definition

6.1. Definition#

Many dynamic processes can occur simultaneously, giving more than one ODE. For example, consider two tanks. Tank 1 has concentration \(c_1\) of a solute and Tank 2 has concentration \(c_2\). Tank 1 flows into Tank 2, which is then passed through a pump and recycled back to Tank 1.

Concentrations in two tanks

Eventually, the two concentrations should equalize, but how do they evolve to over time? \(c_1\) and \(c_2\) can be modeled using unsteady mass balances. This is a system of first-order ODEs that can be solved simultaneously.

The general, explicit form of a system of first-order ODEs is:

(6.1)#\[\begin{align} y_1' &= f_1(t, y_1, \cdots, y_n)\\ \vdots \\ y_n' &= f_n(t, y_1, \cdots, y_n) \end{align}\]

where t is the independent variable, \((y_1, \cdots, y_n)\) are the n dependent variables, and \((f_1, \cdots, f_n)\) are the n right-hand side functions of the n ODEs. We will notate this system more compactly using vectors

(6.2)#\[\begin{equation} \vv{y}' = \vv{f}(t, \vv{y}) \end{equation}\]

where y is the column vector of dependent variables, f is the column vector of right-hand side functions, and \(\vv{y}'\) is a shorthand for the column vector of first derivatives of all dependent variables.

6.1.1. Skill builder problems#

  1. Solve the IVP

    (6.3)#\[\begin{align} y_1' &= 1-2y_1, & y_1(0) &= 0\\ y_2' &= 2y_1 - y_1y_2, & y_2(0) &= 0 \end{align}\]
    Solution

    We begin by solving for \(y_1\) in terms of x. The ODE is separable and therefore can be solved using separation of variables.

    (6.4)#\[\begin{align} \int \frac{dy_1}{1-2y_1} &= \int dx \\ -\frac{1}{2} \ln{\left(1-2y_1\right)} &= x + c_0 \\ \ln{\left(1-2y_1\right)} &= -2x + c_1 \\ 1 - 2y_1 &= c_2 e^{-2x} \\ y_1 &= \frac{1}{2} \left(1- c_2 e^{-2x}\right) \\ \end{align}\]

    Now, use the initial condition \(y_1(0)= 0\) to solve for \(c_2\):

    (6.5)#\[\begin{equation} y_1(0) = \frac{1}{2}\left(1-c_2\right) = 0 \end{equation}\]

    so \(c_2 = 1\) and

    (6.6)#\[\begin{equation} y_1 = \frac{1}{2} \left(1-e^{-2x}\right) \end{equation}\]

    Now, we can solve for \(y_2\) knowing \(y_1\) and recognizing that \(y_2\) can also be solved via separation of variables

    (6.7)#\[\begin{align} y_2' &= y_1\left(2-y_2\right) \\ \int \frac{dy_2}{2-y_2} &= \int \frac{1}{2}\left(1-e^{-2x}\right) \d{x} \\ -\ln{\left(2-y_2\right)} &= \frac{1}{2}\left(x + \frac{1}{2}e^{-2x}\right) + c_0\\ 2 - y_2 &= c_1 \exp\left[-\frac{1}{2}\left(x+ \frac{1}{2}e^{-2x}\right)\right] \\ y_2 &= 2 - c_1 \exp\left[-\frac{1}{2}\left(x+\frac{1}{2}e^{-2x}\right)\right] \\ \end{align}\]

    Again, use the initial condition \(y_2(0)= 0\) to solve for \(c_1\):

    (6.8)#\[\begin{align} y_2(0) &= 2 - c_1 e^{-1/4} = 0 \\ c_1 &= 2e^{1/4} \end{align}\]

    Therefore,

    (6.9)#\[\begin{equation} y_2 = 2\left[1-\exp\left(\frac{1}{4}-\frac{x}{2}-\frac{1}{4}e^{-2x}\right)\right] \end{equation}\]
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