# 2.3: Modeling with First Order Differential Equations

Whenever there is a process to be investigated, a mathematical model becomes a possibility. Since most processes involve something changing, derivatives come into play resulting in a differential equation. We will investigate examples of how differential equations can model such processes.

Example (PageIndex{1}): Pollution

A pond initially contains 500,000 gallons of unpolluted water has an outlet that releases 10,000 gallons of water per day. A stream flows into the pond at 12,000 gallons per day containing water with a concentration of 2 grams per gallon of a pollutant. Find a differential equation that models this process and determine what the concentration of pollutant will be after 10 days.

Solution

We let (x(t)) be amount of pollutant in grams in the pond after (t) days.

We use a fundament property of rates:

[Total Rate = Rate In - Rate Out.]

To find the rate in we use

[egin{align} dfrac{ ext{grams}}{ ext{day}} &= dfrac{ ext{gallons}}{ ext{day}} dfrac{ ext{grams}}{ ext{gallon}} &= dfrac{12,000}{1} dfrac{2}{1} &= 24,000 ext{ grams per day} end{align}]

To find the rate out we first notice that since there was initially 500,000 gallons of water in the lake and the water level is increasing at a rate of 2,000 gallons per day, the total number of gallons of water in the lake after (t) days is

[gallons = 500,000 + 2,000 t.]

The units for the rate out is grams per day. We write

[egin{align} dfrac{grams}{day} = dfrac{gallons}{day} dfrac{grams}{gallon} &= dfrac{10,000}{1} dfrac{x}{500,000 + 2,000 , t} &= dfrac{10x}{500 + 2t} ext{grams per day}. end{align} ]

Putting this all together, we get

[ dfrac{dx}{dt} = 24,000 - dfrac{10x}{500 +2t}.]

This is a first order linear differential equation with

[ p(t) = dfrac{10}{500 + 2t} ;;; ext{and} ;;; g(t) = 24,000. ]

We have

[egin{align} large mu &= e^{int frac{10}{500 + 2t}dt} &= e^{5, ln (500 + 2t)} & = (500 + 2t)^5. end{align} ]

Multiplying by the integrating factor and using the reverse product rule gives

[ ((500 + 2t)^5x)' = 24,000(500 + 2t)^5. ]

Now integrate both sides to get

[ (500 + 2t)^5x = 2,000(500 + 2t)^6 + C ]

[ implies x = 2000(500+2t) + dfrac{C}{(500+2t)^5}. ]

Now use the initial condition to get

[ x = 2000(500)+dfrac{C}{(500)^5} ]

[implies C = -3.125 imes 10^{19}. ]

Now plug in 10 for (t) and calculate (x)

[ egin{align} x &= 2000(500+2(10)) + dfrac{-3.125, ext{x},10^{19}}{(500+2(10))^5} &=,218,072, ext{grams}. end{align}]

A graph is given below

Example (PageIndex{2}): Getting Lucky

You just won the lottery. You put your $5,000,000 in winnings into a fund that has a rate of return of 4%. Each year you use$300,000. How much money will you have twenty years from now?

Solution

This is also a

[Total Rate = Rate In - Rate Out ]

problem. Let

[x = ext{the balance after $t$ years.}]

The rate out is

[300,000]

and the rate in is

[0.04x.]

We have the differential equation

[ dfrac{dx}{dt} = 0.04x - 300,000.]

This is both linear first order and separable. We separate and integrate to obtain

[egin{align} int dfrac{dx}{0.04x - 300,000} &= int dt implies 25, ln, (0.04, x - 300,000) &= t + C_1 implies 0.04x - 300,000 &= C_2 e^{frac{t}{25}} implies x &= Ce^{frac{t}{25}} + 7,500,000. end{align}]

Now use the initial condition that when (t = 0), (x = 5,000,000)

[ 5,000,000 = C + 7,500,000. ]

So that

[ x = -2,500,000, e^{t/25} + 7,500,000. ]

Plugging in 20 for (t) gives

[ x = 1,936,148. ]

You will have about $2 million left. ## Euler's Method The simplest numerical method for approximating solutions of differential equations is Euler's method. Consider a first order differential equation with an initial condition: The procedure for Euler's method is as follows: Contruct the equation of the tangent line to the unknown function$y(t)$at$t=t_0$: where$y'(t_0) = f(y_0,t_0)$is the slope of$y(t)$at$t=t_0$. Use the tangent line to approximate$y(t)$at a small time step$t_1 = t_0 + h$:$ y_1 = y_0 + f(y_0,t_0)(t_1 - t_0) $where$y_1 approx y(t_1)$. Construct the tangent line at the point$(t_1,y_1)$and repeat. The formula for Euler's method defines a recursive sequence:$ y_ = y_n + f(y_n,t_n)(t_ - t_n) , y_0 = y(t_0) $where$y_n approx y(t_n)$for each$n$. If we choose equally spaced$t$values then the formula becomes$ y_ = y_n + f(y_n,t_n)h , y_0 = y(t_0) , t_n = t_0 + nh $with time step$h = t_ - t_n$. Note two very important things about Euler's method and numerical methods in general: • A smaller time step$h$reduces the error in the approximation. • A smaller time step$h$requires more computations! ## 2.3: Oscillatory Solutions to Differential Equations The boundary conditions for the string held to zero at both ends argue that (u(x,t)) collapses to zero at the extremes of the string (Figure (PageIndex<1>)). Figure (PageIndex<1>): Standing waves in a string (both spatially and temporally). from Wikipedia. Unfortunately, when (K>0), the general solution (Equation 2.2.7) results in a sum of exponential decays and growths that cannot achieve the boundary conditions (except for the trivial solution) hence (K<0). This means we must introduce complex numbers due to the (sqrt) terms in Equation 2.2.5. So we can rewrite (K): The general solution to differential equations of the form of Equation ef <2.3.2>is Verify that Equation ( ef<2.3.3>) is the general form for differential equations of the form of Equation ( ef<2.3.2>), which when substituted with Equation ( ef<2.3.1>) give Expand the complex exponentials into trigonometric functions via Euler formula ((e^ = cos heta + isin heta)) [X(x) = A left[cos (px) + i sin (px) ight] + B left[ cos (px) - i sin (px) ight] onumber] [X(x) = (A + B ) cos (px) + i (A - B) sin (px) label<2.3.6>] Introduce new complex constants (c_1=A+B) and (c_2=i(A-B)) so that the general solution in Equation ( ef<2.3.6>) can be expressed as oscillatory functions [X(x) = c_1 cos (px) + c_2 sin (px) label<2.3.7>] Now let's apply the boundary conditions from Equation 2.2.7 to determine the constants (c_1) and (c_2). Substituting the first boundary condition ((X(x=0)=0)) into the general solutions of Equation ( ef<2.3.7>) results in [ egin X(x=0) = c_1 cos (0) + c_2 sin (0) &=0 onumber [4pt] c_1 + 0 &= 0 onumber [4pt] c_1 &=0 label <2.3.8c>end] and substituting the second boundary condition ((X(x=L)=0)) into the general solutions of Equation ( ef<2.3.7>) results in [ X(x=L) = c_1 cos (pL) + c_2 sin (pL) = 0 label<2.3.9>] we already know that (c_1=0) from the first boundary condition so Equation ( ef<2.3.9>) simplifies to Given the properties of sines, Equation ( ef<2.3.9>) simplifies to with (n=0) is the trivial solution that we ignore so (n = 1, 2, 3. ). Substituting Equations ( ef<2.3.12>) and ( ef<2.3.8c>) into Equation ( ef<2.3.7>) results in [X(x) = c_2 sin left(dfrac ight) onumber] [X(x) = c_2 sin left( omega x ight) onumber] A similar argument applies to the other half of the ansatz ((T(t))). Given two traveling waves: [ psi_1 = sin <(c_1 x+c_2 t)> extrm < and > psi_2 = sin <(c_1 x-c_2 t)> onumber] 1. Find the wavelength and the wave velocity of ( psi_1 ) and ( psi_2 ) 2. Find the following and identify nodes: [ psi_+ = psi_1 + psi_2 extrm < and > psi_- = psi_1 - psi_2 onumber] (psi_1 ) is a sin function. At every integer ( n pi ) where (n=0,pm 1, pm 2, . ), a sin function will be zero. Thus, ( psi_1 = 0 ) when (c_1 x + c_2 t = pi n ). Solving for the x, while ignoring trivial solutions: The velocity of this wave is: Similarly for ( psi_2 ). At every integer ( n pi ) where (n=0,pm 1, pm 2, . ), a sin function will be zero. Thus, ( psi_2 = 0 ) when (c_1 x - c_2 t = pi n ). Solving for x, for ( psi_2 ): The velocity of this wave is: The wavelength for each wave is twice the distance between two successive nodes. In other words, Find ( psi_+ = psi_1 + psi_2 extrm < and > psi_- = psi_1 - psi_2 ). [ egin psi_+ &= sin (c_1 x + c_2 t) + sin (c_1 x - c_2 t) [4pt] &= sin (c_1 x ) cos (c_2 t) + cancel + sin (c_1 x ) cos (c_2 t) - cancel [4pt] &= 2sin (c_1 x ) cos (c_2 t) end] This should have a node at every ( x= n pi / c_1 ) and [ egin psi_- &= sin (c_1 x + c_2 t) - sin (c_1 x - c_2 t) [4pt] &= cancel + cos(c_1 x) sin(c_1 t) - cancel + cos(c_1 x) sin(c_1 t) [4pt] &= 2cos (c_1 x ) sin (c_2 t) end] Q = amount of salt in tank. dQ/dt has units of mass/time right? So the "rate in" and "rate out" must also have these same units. Now, pure water enters the tank at the rate of 12L/min. That rate in contains no salt, therefore rate in = 0. The rate out is tricky. The rate of salt leaving is the concentration of salt times the the rate solution leaves. Notice you have (kg/L)*(L/min), which gives kg/min--exactly the units. The concentration is the amount of salt( which is Q at all times) over the volume of the solution (which is initially 2000L, but then changes by 6L/min - 12L/min). You should see it from here. I don't get it, how do I account for it? The incoming rate is pure water, the outgoing rate is solution, I can't mix those two rates can I? It wouldn't make sense unit-wise. Is this problem meant to be reduced to a rate-in - rate-out business or should I let go of that idea? I don't get it, how do I account for it? The incoming rate is pure water, the outgoing rate is solution, I can't mix those two rates can I? It wouldn't make sense unit-wise. Is this problem meant to be reduced to a rate-in - rate-out business or should I let go of that idea? It's clear the rate of salt entering the tank is 0, correct? The rate leaving depends on the concentration and the rate of solution leaving, correct? The concentration depends on the amount of salt and the amount of solution. What is the amount of solution (aka the volume of the tank) at all times? You need this in order to know the concentration at all times, t. Here's a start. dQ/dt = rate salt enters - rate salt leaves That's for the salt. Now for the volume aka amount of solution: dV/dt = rate solution enters - rate solution leaves. We know solution enters at 6L/min (this is pure water, but it's still affecting the volume of tank), and solution leaves at 12L/min (this is a concentration, but it's still affecting the volume of the tank. Therefore, dV/dt = 6 - 12 = -6 Use that (and a given initial condition) to find an expression for V at all times. Where does V go into your dQ/dt diff eq? ## A First Course in Differential Equations with Modeling Applications 11th Edition eTextbook • FREE return shipping at the end of the semester. • Access codes and supplements are not guaranteed with rentals. Fulfillment by Amazon (FBA) is a service we offer sellers that lets them store their products in Amazon's fulfillment centers, and we directly pack, ship, and provide customer service for these products. Something we hope you'll especially enjoy: FBA items qualify for FREE Shipping and . If you're a seller, Fulfillment by Amazon can help you grow your business. Learn more about the program. Enter your mobile number or email address below and we'll send you a link to download the free Kindle App. Then you can start reading Kindle books on your smartphone, tablet, or computer - no Kindle device required. ## Contents Because most chemical reactions that are important for a scientific understanding of our world involve complex mechanisms, the development of mathematical models of such reactions must be preceded by a substantial amount of theoretical and experimental work in Chemistry aimed at gathering an understanding of the mechanisms. In this module, we will restrict our attention to the study of simple chemical reactions. Simple reactions are reactions that do not involve complex mechanisms. The study of simple reactions is a good starting point for learning some of the mathematics that also pertains to the study of more complex reactions. ### 1.1 Units of Measurement and Notation Since molecules are very small, quantities of molecules are measured in units of moles. One mole of molecules is an Avogadro's number of molecules. Avogadro's number is approximately 6.022×10 23 . Hence, for example, two moles is the same as 1.2044×10 24 molecules. Concentrations of molecules in a solution are measured in units of molarities (M). One molarity is one mole of solute per liter of solution. For example, a 2 M aqueous solution of sodium chloride (NaCl) is a solution consisting of two moles of NaCl per each liter of solution. The notation [A] denotes the concentration (in molarities) of a molecule A in solution. Thus, if we write [NaCl] = 2 M, we mean that we have a solution with a 2 M concentration of sodium chloride. ## Alternative Method 1: Finite Interval Method By years of experience, I’ve come to learn that not everyone can setup and solve a differential equation. Instead of actually solving the differential equation, we can just make an educated estimate by using the finite interval method which we have discussed in detail in the previous article. Here is the summary of the procedure. 1. Decide on the time interval. 2. Calculate the new concentration after the addition of the new solution. This concentration should be equal to the concentration of the outgoing solution. Make sure that the volume added is consistent with the time interval used, i.e. if the volumetric rate is 5 L per minute and the time interval is 0.5 minutes, the volume added per interval is (5)(0.5) = 2.5 L. 3. Calculate for the amount of salt remaining after this interval. 4. Repeat steps 2 and 3 until the desired time or amount is achieved. Do note that the method requires a lot of time and effort especially if you are only using a scientific calculator. I have created a sample spreadsheet template for this simulation just in case you’d like to practice (see link below). For the sample problem above, we have the following inputs: For one-minute intervals, we get (Q(10)=10.97) and (Q(57.00) = 15). We also see that (Q(t)) approaches 30 as (t) approaches infinity. Personally, I believe this approach is the best when you are having a hard time figuring out how to setup or how to solve the DE model. As long as you understand the principle, you may be able to solve (well, technically make decent approximations) to mixing problems. ## Solutions for Chapter 2.3: Modeling with First-Order Differential Equations Solutions for Chapter 2.3: Modeling with First-Order Differential Equations • 2.3.1: Consider a tank used in certain hydrodynamic experiments. After one. • 2.3.2: A tank initially contains 120 L of pure water. A mixture containing. • 2.3.3: A tank contains 100 gal of water and 50 oz of salt. Water containin. • 2.3.4: Suppose that a tank containing a certain liquid has an outlet near . • 2.3.5: Suppose that a sum S0 is invested at an annual rate of return r com. • 2.3.6: A young person with no initial capital invests k dollars per year a. • 2.3.7: A certain college graduate borrows$8000 to buy a car. The lender c.
• 2.3.8: A recent college graduate borrows \$150,000 at an interest rate of 6.
• 2.3.9: An important tool in archeological research is radiocarbon dating, .
• 2.3.10: Suppose that a certain population has a growth rate that varies wit.
• 2.3.11: Suppose that a certain population satisfies the initial value probl.
• 2.3.12: Newtons law of cooling states that the temperature of an object cha.
• 2.3.13: Heat transfer from a body to its surroundings by radiation, based o.
• 2.3.14: Consider an insulated box (a building, perhaps) with internal tempe.
• 2.3.15: Consider a lake of constant volume V containing at time t an amount.
• 2.3.16: A ball with mass 0.15 kg is thrown upward with initial velocity 20 .
• 2.3.17: Assume that the conditions are as in except that there is a force d.
• 2.3.18: Assume that the conditions are as in except that there is a force d.
• 2.3.19: A body of constant mass m is projected vertically upward with an in.
• 2.3.20: A body of mass m is projected vertically upward with an initial vel.
• 2.3.21: A body falling in a relatively dense fluid, oil for example, is act.
• 2.3.22: Let v(t) and w(t) be the horizontal and vertical components, respec.
• 2.3.23: A more realistic model (than that in 22) of a baseball in flight in.
• 2.3.24: Brachistochrone Problem. One of the famous problems in the history .
##### ISBN: 9781119256007

Chapter 2.3: Modeling with First-Order Differential Equations includes 24 full step-by-step solutions. This textbook survival guide was created for the textbook: Elementary Differential Equations and Boundary Value Problems, edition: 11. Elementary Differential Equations and Boundary Value Problems was written by and is associated to the ISBN: 9781119256007. Since 24 problems in chapter 2.3: Modeling with First-Order Differential Equations have been answered, more than 23745 students have viewed full step-by-step solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions.

The n roots are the eigenvalues of A.

Remove row i and column j multiply the determinant by (-I)i + j •

Must have n independent eigenvectors (in the columns of S automatic with n different eigenvalues). Then S-I AS = A = eigenvalue matrix.

A = S-1 AS. A = eigenvalue matrix and S = eigenvector matrix of A. A must have n independent eigenvectors to make S invertible. All Ak = SA k S-I.

A(B + C) = AB + AC. Add then multiply, or mUltiply then add.

A sequence of row operations that reduces A to an upper triangular U or to the reduced form R = rref(A). Then A = LU with multipliers eO in L, or P A = L U with row exchanges in P, or E A = R with an invertible E.

0,1,1,2,3,5, . satisfy Fn = Fn-l + Fn- 2 = (A7 -A

)I()q -A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ > A].

Columns without pivots these are combinations of earlier columns.

Set of n nodes connected pairwise by m edges. A complete graph has all n(n - 1)/2 edges between nodes. A tree has only n - 1 edges and no closed loops.

The subspace spanned by b, Ab, . , Aj-Ib. Numerical methods approximate A -I b by x j with residual b - Ax j in this subspace. A good basis for K j requires only multiplication by A at each step.

The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b - Ax is orthogonal to all columns of A.

The lowest degree polynomial with meA) = zero matrix. This is peA) = det(A - AI) if no eigenvalues are repeated always meA) divides peA).

The n by m matrix that "inverts" A from column space back to row space, with N(A+) = N(AT). A+ A and AA+ are the projection matrices onto the row space and column space. Rank(A +) = rank(A).

= number of pivots = dimension of column space = dimension of row space.

Unit vector u is reflected to Qu = -u. All x intheplanemirroruTx = o have Qx = x. Notice QT = Q-1 = Q.

Space of all (v in V) + (w in W). Direct sum: V n W = to>.

Entries AL = Ajj. AT is n by In, AT A is square, symmetric, positive semidefinite. The transposes of AB and A-I are BT AT and (AT)-I.

1 Introduction 1

1.1 Some Basic Mathematical Models Direction Fields 1

1.2 Solutions of Some Differential Equations 9

1.3 Classification of Differential Equations 16

2 First-Order Differential Equations 24

2.1 Linear Differential Equations Method of Integrating Factors 24

2.2 Separable Differential Equations 33

2.3 Modeling with First-Order Differential Equations 39

2.4 Differences Between Linear and Nonlinear Differential Equations 51

2.5 Autonomous Differential Equations and Population Dynamics 58

2.6 Exact Differential Equations and Integrating Factors 70

2.7 Numerical Approximations: Euler&rsquos Method 76

2.8 The Existence and Uniqueness Theorem 83

2.9 First-Order Difference Equations 91

3 Second-Order Linear Differential Equations 103

3.1 Homogeneous Differential Equations with Constant Coefficients 103

3.2 Solutions of Linear Homogeneous Equations the Wronskian 110

3.3 Complex Roots of the Characteristic Equation 120

3.4 Repeated Roots Reduction of Order 127

3.5 Nonhomogeneous Equations Method of Undetermined Coefficients 133

3.6 Variation of Parameters 142

3.7 Mechanical and Electrical Vibrations 147

3.8 Forced Periodic Vibrations 159

4 Higher-Order Linear Differential Equations 169

4.1 General Theory of n th Order Linear Differential Equations 169

4.2 Homogeneous Differential Equations with Constant Coefficients 174

4.3 The Method of Undetermined Coefficients 181

4.4 The Method of Variation of Parameters 185

5 Series Solutions of Second-Order Linear Equations 189

5.1 Review of Power Series 189

5.2 Series Solutions Near an Ordinary Point, Part I 195

5.3 Series Solutions Near an Ordinary Point, Part II 205

5.4 Euler Equations Regular Singular Points 211

5.5 Series Solutions Near a Regular Singular Point, Part I 219

5.6 Series Solutions Near a Regular Singular Point, Part II 224

6 The Laplace Transform 241

6.1 Definition of the Laplace Transform 241

6.2 Solution of Initial Value Problems 248

6.4 Differential Equations with Discontinuous Forcing Functions 264

6.6 The Convolution Integral 275

7 Systems of First-Order Linear Equations 281

7.3 Systems of Linear Algebraic Equations Linear Independence, Eigenvalues, Eigenvectors 295