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Section4.4Solving Systems Analytically

Objectives

  • To understand that a system of the form

    \begin{align*} \frac{dx}{dt} & =f(x)\\ \frac{dy}{dt} & = g(y) \end{align*}

    is decoupled and can be solved by solving each equation independently.

  • To understand that a system of the form

    \begin{align*} \frac{dx}{dt} & =f(x),\\ \frac{dy}{dt} & = g(x, y) \end{align*}

    is partially coupled system and can be solved by first solving the first equation and then substituting the solution into the second equation, which can then be solved.

Mixing problems model how substances flow back and forth between two or more compartments. These problems often arise in applicationsfor example, we might want to model how greenhouse gases flow back and forth between different layers of the earth's atomosphere[17], how chemicals move between tanks in a refinery or a brewery, or how pollutants move between a series of lakes or ponds. Systems of differential equations can prove very useful when it comes to modeling such situations.

Subsection4.4.1Partially Coupled Systems

We will use linear systems of differential equations such as

\begin{align*} \frac{dx}{dt} & = ax + by\\ \frac{dy}{dt} & = cx + dy \end{align*}

to illustrate how we can use systems of differential equations to model how substances flow back and forth between two or more compartments. Suppose that we have two tanks (\(A\) and \(B\)) between which a mixture of brine flows (Figure4.4.1). Tank \(A\) contains 300 liters of water in which 100 kilograms of salt has been dissolved and Tank \(B\) contains 300 liters of pure water. Fresh water is pumped into Tank \(A\) at the rate of 500 liters per hour, and brine is pumped into Tank \(B\) from Tank \(A\) at the rate of 500 liters per hour. Brine is also drained at a rate 500 liters of brine per hour from Tank \(B\text{.}\) All brine mixtures are well-stirred. If we let \(x = x(t)\) be the amount of salt in Tank \(A\) at time \(t\) and \(y = y(t)\) be the amount of salt in Tank \(B\) at time \(t\text{,}\) then we know that

\begin{align*} x(0) & = 100\\ y(0) & = 0 \end{align*}

We know that the salt concentrations in the two tanks are \(x/300\) kilograms per liter and \(y/300\) kilograms per liter. Thus, we can describe the rate of change in each tank with a differential equation,

\begin{align*} \frac{dx}{dt} & = - 500 \cdot \frac{x}{300} = - \frac{5}{3} x,\\ \frac{dy}{dt} & = 500 \cdot \frac{x}{300} - 500 \cdot \frac{y}{300} = \frac{5}{3} x - \frac{5}{3} y. \end{align*}
two three hundred liter tanks connected with a pipe flowing from the first to the second tank with the first tank having an input pipe and the second having a drain pipe
Figure4.4.1Mixing example with two tanks

We can now ask how we might solve the system of equations

\begin{align*} \frac{dx}{dt} & = - \frac{5}{3} x,\\ \frac{dy}{dt} & = \frac{5}{3} x - \frac{5}{3} y. \end{align*}

The task of solving the system

\begin{align*} \frac{dx}{dt} & =f(x,y),\\ \frac{dy}{dt} & = g(x,y), \end{align*}

may be quite difficult or even impossible. However, we can find solutions in certain cases. For example, if we have a system of the form

\begin{align*} \frac{dx}{dt} & =f(x),\\ \frac{dy}{dt} & = g(y), \end{align*}

then each equation is an autonomous first-order equation. To solve our system, we only need to solve two first-order equations. Such a system is said to be decoupled. Generalizing slightly, we say that a partially coupled system is a system of the form

\begin{align*} \frac{dx}{dt} & =f(x),\\ \frac{dy}{dt} & = g(x, y). \end{align*}

Since the first equation is an autonomous first-order equation in \(x\text{,}\) we can solve this equation separately, and substitute our solution into the second equation.

Consider the system

\begin{align*} \frac{dx}{dt} & = x \\ \frac{dy}{dt} & = x + y. \end{align*}

We can easily solve the first equation, \(dx/dt = x\text{,}\) to obtain \(x = a e^t\text{.}\) Using this information in the second equation, we have

\begin{equation*} \frac{dy}{dt} - y = a e^t \end{equation*}

which is a first-order linear equation. This equation has an integrating factor \(\mu(t) = e^{-t}\text{,}\) and

\begin{equation*} \frac{d}{dt} (e^{-t} y) = \mu(t) \left(\frac{dy}{dt} - y \right)= a e^t \mu(t) = a. \end{equation*}

Therefore, the solution to our second equation is

\begin{equation*} y(t) = at e^t + be^t. \end{equation*}

Revisiting the mixing problem that we posed at the beginning of this section, we have the following initial value problem,

\begin{align*} \frac{dx}{dt} & = - \frac{5}{3} x,\\ \frac{dy}{dt} & = \frac{5}{3} x - \frac{5}{3} y.\\ x(0) & = 100,\\ y(0) & = 0. \end{align*}

Solving \(dx/dt = - (5/3) x\) is easy. We can quickly determine that \(x(t) = c_1 e^{-5t/3}\text{.}\) Applying the initial condition \(x(0) = 100\text{,}\) we can determine that \(c_1 = 100\) and \(x(t) = 100 e^{-5t/3}\text{.}\) Our second equation now becomes

\begin{equation*} \frac{dy}{dt} = \frac{5}{3} x - \frac{5}{3} y = \frac{500}{3} e^{-5t/3} - \frac{5}{3} y. \end{equation*}

This last equation is a first-order linear equation

\begin{equation*} \frac{dy}{dt} + \frac{5}{3} y = \frac{500}{3} e^{-5t/3}. \end{equation*}

Multiplying both sides of this last equation by the integrating factor \(\mu(t) = e^{5t/3}\) yields

\begin{equation*} \frac{d}{dt} (e^{5t/3}y) = e^{5t/3}\frac{dy}{dt} + e^{5t/3}\frac{5}{3} y = \frac{500}{3}. \end{equation*}

Integrating both sides of this last equation gives us

\begin{equation*} e^{5t/3}y = \frac{500}{3} t + c_2. \end{equation*}

Using our initial condition \(y(0) = 0\text{,}\) we can determine that \(c_2 = 0\text{.}\) Thus,

\begin{equation*} y = \frac{500}{3} t e^{-5t/3} \end{equation*}

Consider the partially coupled system

\begin{align*} x' & = 2x\\ y' & = x + 3y \end{align*}

Notice that we already have the tools to solve \(x' = 2x\text{.}\) In fact, the solution is \(x(t) = c_1 e^{2t}\text{.}\) We can use this information to solve the second equation, \(y' = x + 3y\text{.}\) That is, if we use the fact that \(x(t) = c_1 e^{2t}\text{,}\) the second equation becomes

\begin{equation*} y' = c_1 e^{2t} + 3y. \end{equation*}

We can rewrite this equation as

\begin{equation*} y' - 3y = c_1 e^{2t}, \end{equation*}

which is a first-order linear equation. If we multiply both sides of the equation by \(\mu(t) = e^{-3t}\text{,}\) we have

\begin{equation*} \frac{d}{dt}\left( e^{-3t}y \right) = e^{-3t}y' - 3e^{-3t}y = c_1 e^{-t}. \end{equation*}

Integrating, we have

\begin{equation*} e^{-3t}y = - c_1 e^{-t} + c_2. \end{equation*}

Solving for \(y\text{,}\) yields \(y = - c_1 e^{2t} + c_2e^{3t}\text{.}\) Thus the solution to our system is

\begin{align*} x & = c_1 e^{2t}\\ y & = - c_1 e^{2t} + c_2e^{3t}. \end{align*}
Activity4.4.1Solving Partially Coupled Systems

Solve each of the following systems of differential equations.

(a)
\begin{align*} x' & = -x\\ y' & = x - 3y \end{align*}
(b)
\begin{align*} x' & = -x - y\\ y' & = - 3y \end{align*}
(c)
\begin{align*} x' & = 2x\\ y' & = 3x + 2y \end{align*}
(d)
\begin{align*} x' & = 3x + 4y\\ y' & = -2y \end{align*}

Subsection4.4.2Harmonic Oscillators

The equation

\begin{equation*} \frac{d^2 x}{dt^2} + 3 \frac{dx}{dt} + 2 x = 0 \end{equation*}

is a specific case of a damped harmonic oscillator, where \(m = 1\text{,}\) the spring constant is 2, and the damping constant is 3. We can rewrite this equation as a first-order linear system,

\begin{align*} \frac{dx}{dt} & = v,\\ \frac{dv}{dt} & = - 2x - 3v. \end{align*}

Suppose that \(x(0) = 0\) is the initial position of the mass and \(v(0) = 1\) is the initial velocity. To solve the system, one can simply use the characteristic equation to solve the original second order homogeneous equation \(\frac{d^2 x}{dt^2} + 3 \frac{dx}{dt} + 2 x = 0\text{.}\) The solution to the system is

\begin{align*} x(t) & = e^{-t} - e^{-2t},\\ v(t) & = - e^{-t} + 2 e^{-2t}. \end{align*}

This is an example of an over-damped harmonic oscillator (Figure4.4.3). In other words, a spring-mass system that is modeled by this system of equations has so much damping that the mass will not oscillate.

position and velocity curves that approach zero with no oscillations
Figure4.4.3An over-damped harmonic oscillator

Now let us relax the damping and increase the spring constant on our harmonic oscillator,

\begin{equation*} \frac{d^2 x}{dt^2} + 2 \frac{dx}{dt} + 10 x = 0. \end{equation*}

The corresponding linear system is

\begin{align*} \frac{dx}{dt} & = v,\\ \frac{dv}{dt} & = - 10x - 2v\\ x(0) \amp = 0\\ v(0) \amp = 1. \end{align*}

Notice that our initial conditions have not changed.

The solution is now

\begin{align*} x(t) & = \frac{1}{3} e^{-t} \sin 3t,\\ v(t) & = e^{-t} \cos 3t - \frac{1}{3} e^{-t} \sin 3t. \end{align*}

Notice that the system oscillates due to the sine and cosine terms in the solution. This system is underdamped (Figure4.4.4).

position and velocity curves that approach zero with smaller and smaller oscillations
Figure4.4.4An under-damped harmonic oscillator

Subsection4.4.3Important Lessons

  • A system of the form

    \begin{align*} \frac{dx}{dt} & =f(x)\\ \frac{dy}{dt} & = g(y) \end{align*}

    is said to be decoupled. Such a system can be solved by solving each equation independently.

  • A system of the form

    \begin{align*} \frac{dx}{dt} & =f(x),\\ \frac{dy}{dt} & = g(x, y) \end{align*}

    is a partially coupled system. Since the first equation is an autonomous first-order equation in \(x\text{,}\) we can solve this equation separately, and substitute our solution into the second equation.

Exercise4.4.5

Explain what a partially coupled system is.

Exercise4.4.6

Why would one expect that it is impossible to find explicit solutions for most systems of differential equations?

Subsection4.4.4Exercises

Solve each of the following partially coupled systems.

  1. \begin{align*} x' & = 2x\\ y' & = x + 2y \end{align*}
  2. \begin{align*} x' & = -x + 3y\\ y' & = -2y \end{align*}
  3. \begin{align*} x' & = -3x\\ y' & = 2x + 3y \end{align*}
  4. \begin{align*} x' & = 2x - 3y\\ y' & = 4y \end{align*}

Suppose that we have two tanks (\(A\) and \(B\)) between which a mixture of brine flows. Tank \(A\) contains 200 liters of water in which 20 kilograms of salt has been dissolved and Tank \(B\) contains 200 liters of water in which 10 kilograms of salt has been dissolved. Fresh water is pumped into Tank \(A\) at the rate of 200 liters per hour, and brine is pumped into Tank \(B\) from Tank \(A\) at the rate of 200 liters per hour. Brine is also drained at a rate 200 liters of brine per hour from Tank \(B\text{.}\) All brine mixtures are well-stirred. Find the amount of salt in each tank at time \(t\text{.}\)

For each of the following harmonic oscillators (1) rewrite the second-order initial value problem as a system of two first-order linear equations, (2) solve the system by solving the original second order homogeneous differential equation, and (3) classify the harmonic oscillator as underdamped, critically damped, or overdamped.

  1. \begin{align*} y'' + 7y' + 6y & = 0\\ y(0) & = 1\\ y'(0) & = 0 \end{align*}
  2. \begin{align*} y'' + 4y' + 5y & = 0\\ y(0) & = 1\\ y'(0) & = 0 \end{align*}
  3. \begin{align*} y'' + 6y' + 9y & = 0\\ y(0) & = 1\\ y'(0) & = 0 \end{align*}