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3.7: Exact Equations - Mathematics

3.7: Exact Equations - Mathematics


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In this section it is convenient to write first order differential equations in the form

[label{eq:3.8.1} M(x,y),dx+N(x,y),dy=0.]

This equation can be interpreted as

[label{eq:3.8.2} M(x,y)+N(x,y),{dyover dx}=0,]

where (x) is the independent variable and (y) is the dependent variable, or as

[label{eq:3.8.3} M(x,y),{dxover dy}+N(x,y)=0,]

where (y) is the independent variable and (x) is the dependent variable. Since the solutions of Equation ef{eq:3.8.2} and Equation ef{eq:3.8.3} will often have to be left in implicit form we will say that (F(x,y)=c) is an implicit solution of Equation ef{eq:3.8.1} if every differentiable function (y=y(x)) that satisfies (F(x,y)=c) is a solution of Equation ef{eq:3.8.2} and every differentiable function (x=x(y)) that satisfies (F(x,y)=c) is a solution of Equation ef{eq:3.8.3}

Here are some examples:

Equation ef{eq:3.8.1}Equation ef{eq:3.8.2}Equation ef{eq:3.8.3}
(3x^2y^2,dx+2x^3y,dy =0)(3x^2y^2+2x^3y, {dyover dx} =0)(3x^2y^2, {dxover dy}+2x^3y=0)
((x^2+y^2),dx +2xy,dy=0)((x^2+y^2)+2xy, {dyover dx}=0)((x^2+y^2), {dxover dy} +2xy=0)
(3ysin x,dx-2xycos x,dy =0)(3ysin x-2xycos x, {dyover dx} =0)(3ysin x, {dxover dy}-2xycos x =0)

Table (PageIndex{1}): Examples of Exact Differential Equations in three forms

Note that a separable equation can be written as Equation ef{eq:3.8.1} as

[M(x),dx+N(y),dy=0. onumber]

we will develop a method for solving Equation ef{eq:3.8.1} under appropriate assumptions on (M) and (N). This method is an extension of the method of separation of variables. Before stating it we consider an example.

Example (PageIndex{1})

Show that

[label{eq:3.8.4} x^4y^3+x^2y^5+2xy=c ]

is an implicit solution of

[label{eq:3.8.5} (4x^3y^3+2xy^5+2y),dx+(3x^4y^2+5x^2y^4+2x),dy=0. ]

Solution

Regarding (y) as a function of (x) and differentiating Equation ef{eq:3.8.4} implicitly with respect to (x) yields

[(4x^3y^3+2xy^5+2y)+(3x^4y^2+5x^2y^4+2x),{dyover dx}=0. onumber]

Similarly, regarding (x) as a function of (y) and differentiating Equation ef{eq:3.8.4} implicitly with respect to (y) yields

[(4x^3y^3+2xy^5+2y){dxover dy}+(3x^4y^2+5x^2y^4+2x)=0. onumber]

Therefore Equation ef{eq:3.8.4} is an implicit solution of Equation ef{eq:3.8.5} in either of its two possible interpretations.

You may think Example (PageIndex{1}) is pointless, since concocting a differential equation that has a given implicit solution is not particularly interesting. However, it illustrates the next important theorem, which we will prove by using implicit differentiation, as in Example (PageIndex{1}).

Theorem (PageIndex{1})

If (F=F(x,y)) has continuous partial derivatives (F_x) and (F_y), then

[label{eq:3.8.6} F(x,y)=c ]

(with (c) as a constant) is an implicit solution of the differential equation

[label{eq:3.8.7} F_x(x,y),dx+F_y(x,y),dy=0.]

Proof

Regarding (y) as a function of (x) and differentiating Equation ef{eq:3.8.6} implicitly with respect to (x) yields

[F_x(x,y)+F_y(x,y),{dyover dx}=0. onumber]

On the other hand, regarding (x) as a function of (y) and differentiating Equation ef{eq:3.8.6} implicitly with respect to (y) yields

[F_x(x,y),{dxover dy}+F_y(x,y)=0. onumber]

Thus, Equation ef{eq:3.8.6} is an implicit solution of Equation ef{eq:3.8.7} in either of its two possible interpretations.

We will say that the equation

[label{eq:3.8.8} M(x,y),dx+N(x,y),dy=0]

is exact on an an open rectangle (R) if there’s a function (F=F(x,y)) such (F_x) and (F_y) are continuous, and

[label{eq:3.8.9} F_x(x,y)=M(x,y) quad ext{and} quad F_y(x,y)=N(x,y)]

for all ((x,y)) in (R). This usage of “exact” is related to its usage in calculus, where the expression

[F_x(x,y),dx+F_y(x,y),dy onumber ]

(obtained by substituting Equation ef{eq:3.8.9} into the left side of Equation ef{eq:3.8.8}) is the exact differential of (F).

Example (PageIndex{1}) shows that it is easy to solve Equation ef{eq:3.8.8} if it is exact and we know a function (F) that satisfies Equation ef{eq:3.8.9}. The important questions are:

  • Question 1. Given an equation Equation ef{eq:3.8.8}, how can we determine whether it is exact?
  • Question 2. If Equation ef{eq:3.8.8} is exact, how do we find a function (F) satisfying Equation ef{eq:3.8.9}?

To discover the answer to Question 1, assume that there’s a function (F) that satisfies Equation ef{eq:3.8.9} on some open rectangle (R), and in addition that (F) has continuous mixed partial derivatives (F_{xy}) and (F_{yx}). Then a theorem from calculus implies that [label{eq:3.8.10} F_{xy}=F_{yx}.] If (F_x=M) and (F_y=N), differentiating the first of these equations with respect to (y) and the second with respect to (x) yields

[label{eq:3.8.11} F_{xy}=M_y quad ext{and} quad F_{yx}=N_x.]

From Equation ef{eq:3.8.10} and Equation ef{eq:3.8.11}, we conclude that a necessary condition for exactness is that (M_y=N_x). This motivates the next theorem, which we state without proof.

Theorem (PageIndex{2}): The Exactness Condition

Suppose (M) and (N) are continuous and have continuous partial derivatives (M_y) and (N_x) on an open rectangle (R.) Then

[M(x,y),dx+N(x,y),dy=0 onumber ]

is exact on (R) if and only if

[label{eq:3.8.12} M_y(x,y)=N_x(x,y)]

for all ((x,y)) in (R.).

To help you remember the exactness condition, observe that the coefficients of (dx) and (dy) are differentiated in Equation ef{eq:3.8.12} with respect to the “opposite” variables; that is, the coefficient of (dx) is differentiated with respect to (y), while the coefficient of (dy) is differentiated with respect to (x).

Example (PageIndex{2})

Show that the equation

[3x^2y,dx+4x^3,dy=0 onumber ]

is not exact on any open rectangle.

Solution

Here

[M(x,y)=3x^2y quad ext{and} quad N(x,y)=4x^3 onumber]

so

[M_y(x,y)=3x^2 quad ext{and} N_x(x,y)=12 x^2. onumber]

Therefore (M_y=N_x) on the line (x=0), but not on any open rectangle, so there’s no function (F) such that (F_x(x,y)=M(x,y)) and (F_y(x,y)=N(x,y)) for all ((x,y)) on any open rectangle.

The next example illustrates two possible methods for finding a function (F) that satisfies the condition (F_x=M) and (F_y=N) if (M,dx+N,dy=0) is exact.

Example (PageIndex{3})

Solve

[label{eq:3.8.13} (4x^3y^3+3x^2),dx+(3x^4y^2+6y^2),dy=0.]

Solution (Method 1)

Here [M(x,y)=4x^3y^3+3x^2,quad N(x,y)=3x^4y^2+6y^2, onumber ] and [M_y(x,y)=N_x(x,y)=12 x^3y^2 onumber ] for all ((x,y)). Therefore Theorem (PageIndex{2}) implies that there’s a function (F) such that

[label{eq:3.8.14} F_x(x,y)=M(x,y)=4x^3y^3+3x^2]

and

[label{eq:3.8.15} F_y(x,y)=N(x,y)=3x^4y^2+6y^2]

for all ((x,y)). To find (F), we integrate Equation ef{eq:3.8.14} with respect to (x) to obtain

[label{eq:3.8.16} F(x,y)=x^4y^3+x^3+phi(y),]

where (phi (y)) is the “constant” of integration. (Here (phi) is “constant” in that it is independent of (x), the variable of integration.) If (phi) is any differentiable function of (y) then (F) satisfies Equation ef{eq:3.8.14}. To determine (phi) so that (F) also satisfies Equation ef{eq:3.8.15}, assume that (phi) is differentiable and differentiate (F) with respect to (y). This yields

[F_y(x,y)=3x^4y^2+phi'(y). onumber]

Comparing this with Equation ef{eq:3.8.15} shows that

[phi'(y)=6y^2. onumber]

We integrate this with respect to (y) and take the constant of integration to be zero because we are interested only in finding some (F) that satisfies Equation ef{eq:3.8.14} and Equation ef{eq:3.8.15}. This yields

[phi (y)=2y^3. onumber]

Substituting this into Equation ef{eq:3.8.16} yields

[label{eq:3.8.17} F(x,y)=x^4y^3+x^3+2y^3.]

Now Theorem (PageIndex{1}) implies that [x^4y^3+x^3+2y^3=c onumber ] is an implicit solution of Equation ef{eq:3.8.13}. Solving this for (y) yields the explicit solution

[y=left(c-x^3over2+x^4 ight)^{1/3}. onumber]

Solution (Method 2)

Instead of first integrating Equation ef{eq:3.8.14} with respect to (x), we could begin by integrating Equation ef{eq:3.8.15} with respect to (y) to obtain

[label{eq:3.8.18} F(x,y)=x^4y^3+2y^3+psi (x),]

where (psi) is an arbitrary function of (x). To determine (psi), we assume that (psi) is differentiable and differentiate (F) with respect to (x), which yields

[F_x(x,y)=4x^3y^3+psi'(x). onumber]

Comparing this with Equation ef{eq:3.8.14} shows that

[psi'(x)=3x^2. onumber]

Integrating this and again taking the constant of integration to be zero yields

[psi(x)=x^3. onumber]

Substituting this into Equation ef{eq:3.8.18} yields Equation ef{eq:3.8.17}.

Figure (PageIndex{1}) shows a direction field and some integral curves of Equation ef{eq:3.8.13}.

Here’s a summary of the procedure used in Method 1 of this example. You should summarize procedure used in Method 2.

HOWTO: Procedure For Solving An Exact Equation (Method 1)

  • Step 1. Check that the equation [label{eq:3.8.19} M(x,y),dx+N(x,y),dy=0] satisfies the exactness condition (M_y=N_x). If not, don’t go further with this procedure.
  • Step 2. Integrate [{partial F(x,y)overpartial x}=M(x,y) onumber ] with respect to (x) to obtain [label{eq:3.8.20} F(x,y)=G(x,y)+phi(y),] where (G) is an antiderivative of (M) with respect to (x), and (phi) is an unknown function of (y).
  • Step 3. Differentiate Equation ef{eq:3.8.20} with respect to (y) to obtain [{partial F(x,y)overpartial y}={partial G(x,y)overpartial y}+phi'(y). onumber]
  • Step 4. Equate the right side of this equation to (N) and solve for (phi'); thus, [{partial G(x,y)overpartial y}+phi'(y)=N(x,y), quad ext{so} quad phi'(y)=N(x,y)-{partial G(x,y)overpartial y}. onumber]
  • Step 5. Integrate (phi') with respect to (y), taking the constant of integration to be zero, and substitute the result in Equation ef{eq:3.8.20} to obtain (F(x,y)).
  • Step 6. Set (F(x,y)=c) to obtain an implicit solution of Equation ef{eq:3.8.19}. If possible, solve for (y) explicitly as a function of (x).

It’s a common mistake to omit Step 6 in the procedure above. However, it is important to include this step, since F isn’t itself a solution of Equation ef{eq:3.8.19}. Many equations can be conveniently solved by either of the two methods used in Example (PageIndex{3}). However, sometimes the integration required in one approach is more difficult than in the other. In such cases we choose the approach that requires the easier integration.

Example (PageIndex{4})

Solve the equation

[label{eq:3.8.21} left( y e ^ { x y } an x + e ^ { x y } sec ^ { 2 } x ight) d x + x e ^ { x y } an x , dy = 0]

Solution

We leave it to you to check that (M_y = N_x) on any open rectangle where ( an x) and (sec x) are defined. Here we must find a function F such that

[label{eq:3.8.22} F_x(x, y) = ye^{xy} an x + e^{xy} sec^2 x]

and

[label{eq:3.8.23} F_y(x, y) = xe^{xy} an x. ]

It’s difficult to integrate Equation ef{eq:3.8.22} with respect to (x), but easy to integrate Equation ef{eq:3.8.23} with respect to (y). This yields

[label{eq:3.8.24} F(x, y) = e^{xy} an x + psi(x). ]

Differentiating this with respect to (x) yields

[F_x(x, y) = y e^{xy} an x + e^{xy} sec^2 x + psi'(x). onumber]

Comparing this with Equation ef{eq:3.8.22} shows that (psi'(x) = 0). Hence, (psi) is a constant, which we can take to be zero in Equation ef{eq:3.8.24}, and

[e^{xy} an x = c, onumber]

is an implicit solution of Equation ef{eq:3.8.21}.

Attempting to apply our procedure to an differential equation that is not exact will lead to failure in Step 4, since the function

[N - frac { partial G } { partial y } onumber]

will not be independent of (x) if (M_y eq N_x), and therefore cannot be the derivative of a function of (y) alone. Example (PageIndex{5}) illustrates this.

Example (PageIndex{5})

Verify that the equation

[label{eq:3.8.25} 3x^2y^2,dx+6x^3y,dy=0]

is not exact, and show that the procedure for solving exact equations fails when applied to Equation ef{eq:3.8.25}.

Solution

Here [M_y(x,y)=6x^2y quad ext{and} quad N_x(x,y)=18x^2y, onumber ]

so Equation ef{eq:3.8.25} is not exact. Nevertheless, let us try to find a function (F) such that

[label{eq:3.8.26} F_x(x,y)=3x^2y^2]

and

[label{eq:3.8.27} F_y(x,y)=6x^3y.]

Integrating Equation ef{eq:3.8.26} with respect to (x) yields

[F(x,y)=x^3y^2+phi(y), onumber]

and differentiating this with respect to (y) yields

[F_y(x,y)=2x^3y+phi'(y). onumber]

For this equation to be consistent with Equation ef{eq:3.8.27},

[6x^3y=2x^3y+phi'(y), onumber]

or

[phi'(y)=4x^3y. onumber]

This is a contradiction, since (phi') must be independent of (x). Therefore the proce


Exact and Nonexact Equations

(3) Integrate either the first equation with respect of the variable x or the second with respect of the variable y . The choice of the equation to be integrated will depend on how easy the calculations are. Let us assume that the first equation was chosen, then we get

The function should be there, since in our integration, we assumed that the variable y is constant. (4) Use the second equation of the system to find the derivative of . Indeed, we have

Note that is a function of y only. Therefore, in the expression giving the variable, x , should disappear. Otherwise something went wrong! (5) Integrate to find (6) Write down the function F ( x , y ) (7) All the solutions are given by the implicit equation

(8) If you are given an IVP, plug in the initial condition to find the constant C .

You may ask, what do we do if the equation is not exact? In this case, one can try to find an integrating factor which makes the given differential equation exact.


Given a simply connected and open subset D of R 2 and two functions I and J which are continuous on D, an implicit first-order ordinary differential equation of the form

I ( x , y ) d x + J ( x , y ) d y = 0 ,

is called an exact differential equation if there exists a continuously differentiable function F, called the potential function, [1] [2] so that

An exact equation may also be presented in the following form:

I ( x , y ) + J ( x , y ) y ′ ( x ) = 0

where the same constraints on I and J apply for the differential equation to be exact.

The nomenclature of "exact differential equation" refers to the exact differential of a function. For a function F ( x 0 , x 1 , . . . , x n − 1 , x n ) ,x_<1>. x_,x_)> , the exact or total derivative with respect to x 0 > is given by

Example Edit

is a potential function for the differential equation

In physical applications the functions I and J are usually not only continuous but even continuously differentiable. Schwarz's Theorem then provides us with a necessary criterion for the existence of a potential function. For differential equations defined on simply connected sets the criterion is even sufficient and we get the following theorem:

Given a differential equation of the form (for example, when F has zero slope in the x and y direction at F(x,y)):

I ( x , y ) d x + J ( x , y ) d y = 0 ,

with I and J continuously differentiable on a simply connected and open subset D of R 2 then a potential function F exists if and only if

Given an exact differential equation defined on some simply connected and open subset D of R 2 with potential function F, a differentiable function f with (x, f(x)) in D is a solution if and only if there exists real number c so that

we can locally find a potential function by

for y, where c is a real number, we can then construct all solutions.

The concept of exact differential equations can be extended to second order equations. [3] Consider starting with the first-order exact equation:


Introduction to Equations

By an equation we mean a mathematical sentence that states that two algebraic expressions are equal. For example, a (b + c) =ab + ac, ab = ba, and x 2 -1 = (x-1)(x+1) are all equations that we have been using. We recall that we defined a variable as a letter that may be replaced by numbers out of a given set, during a given discussion. This specified set of numbers is sometimes called the replacement set. In this chapter we will deal with equations involving variables where the replacement set, unless otherwise specified, is the set of all real numbers for which all the expressions in the equation are defined.

If an equation is true after the variable has been replaced by a specific number, then the number is called a solution of the equation and is said to satisfy it. Obviously, every solution is a member of the replacement set. The real number 3 is a solution of the equation 2x-1 = x+2, since 2*3-1=3+2. while 1 is a solution of the equation (x-1)(x+2) = 0. The set of all solutions of an equation is called the solution set of the equation.

In the first equation above <3>is the solution set, while in the second example <-2,1>is the solution set. We can verify by substitution that each of these numbers is a solution of its respective equation, and we will see later that these are the only solutions.

A conditional equation is an equation that is satisfied by some numbers from its replacement set and not satisfied by others. An identity is an equation that is satisfied by all numbers from its replacement set.

Example 1 Consider the equation 2x-1 = x+2

The replacement set here is the set of all real numbers. The equation is conditional since, for example, 1 is a member of the replacement set but not of the solution set.

Example 2 Consider the equation (x-1)(x+1) =x 2 -1

The replacement set is the set of all real numbers. From our laws of real numbers if a is any real number, then (a-1)(a+1) = a 2 -1

Therefore, every member of the replacement set is also a member of the solution set. Consequently this equation is an identity.

Example 3 Consider

The replacement set for this equation is the set of real numbers except 0, since 1/x and (1- x)/x are not defined for x = 0. If a is any real number in the replacement set, then

so that the original equation is an identity.

Example 4 Consider

The replacement set is the set of all non-negative real numbers, since is not a real number if x is negative. The equation is conditional since, for example, 4 is a member of the replacement set but not of the solution set.


3.7: Exact Equations - Mathematics

Chapter 4 – Simple Equations

Q.1. After 20 years, Manoj will be 5 times as old as he is now. Find his present age.

Q.2. If 45 is added to half a number, the result is triple the number. Find the number.

Q.3. In a family, the consumption of wheat is 4 times that of rice. The total consumption of the two cereals is 80 kg. Find the quantities of rice and wheat consumed in the family.

Q.4. Anamika thought of a number. She multiplied it by 2, added 5 to the product and obtained 17 as the result. What is the number she had thought of ?

Q.5. One of the two numbers is twice the other. The sum of the numbers is 12. Find the numbers.

Q.6. A number when divided by 6 gives the quotient 6. What is the number?

Q.7. The perimeter of a rectangle is 40m. The length of the rectangle is 4 m less than 5 times its breadth. Find the length of the rectangle.

Q.8. The sum of two consecutive multiples of 2 is 18. Find the numbers.

Q.9. Two complementary angles differ by 20°. Find the angles.

Q.10. 150 has been divided into two parts such that twice the first part is equal to the second part. Find the parts.

Q.11. In a class of 60 students, the number of girls is one third the number of boys. Find the number of girls and boys in the class.

Q.12. A number is as much greater than 27 as it is less than 73. Find the number.

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Solution

When you solve this equation with pictures, you end up with 3 bags balancing with 1 tile. In order to do the division, you have to cut the tile, leading to the fraction 1/3, which is the solution you get symbolically.

In order to solve this equation with pictures, you have to have some way of representing the subtraction in $2x – 4$. If students have experience with integer chips, they can transfer that knowledge to this situation to show $2x + -4$, but otherwise they may struggle with the idea. The pictures give us a nice model for understanding the operations we do to solve equations, but it is only smooth for problems with “nice” numbers. This is one reason why we want to move to the symbolic approach.

A linear equation will have no solution if there are the same number of $x$’s and different constants on each side. For example: $2x + 4 = 2x + 1$. If you solve this with pictures, when you take away the $2x$ from both sides you will end up with $4 = 1$, which clearly cannot be balanced. If the equation had infinitely many solutions, you would find that you had exactly the same picture on the two sides of the balance.

The mistake is in the first step - the student divided only part of the left-hand-side of the equation by 2. You can see in the picture that splitting the equation this way will not keep the balance level (assuming the two bags are equal):


3.7: Exact Equations - Mathematics

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Negation of a Statement

Definition: A closed sentence is an objective statement which is either true or false.

Thus, each closed sentence in Example 1 has a truth value of either true or false as shown below.

1. Every triangle has three sides. true
2. Albany is the capital of New York State. true
3. No prime number is even. false

Note that the third sentence is false since 2 is a prime number. It is possible that a closed sentence will have different truth values at different times. This is demonstrated in Example 2 below.

1. Today is Tuesday.
2. Bill Clinton was the 42nd President of the United States.

Example 3: Examine the sentences below.

1. x + 3 = 7
2. She passed math.
3. y - 4 = 11
4. He is my brother.

The sentences in Example 3 are open sentences.

Definition: An open sentence is a statement which contains a variable and becomes either true or false depending on the value that replaces the variable.

Let's take another look at Example 3. This time we will Identify the variable for each open sentence.

1. x + 3 = 7 The variable is x.
2. She passed math. The variable is she.
3. y - 4 = 11 The variable is y.
4. He is my brother. The variable is he.

Now that we have identified the variables, we can analyze the meaning of these open sentences. Sentence 1 is true if x is replaced by 4, but false if x is replaced by a number other than 4. Sentence 3 is true if y is replaced by 15, but false otherwise. Sentence 2 is either true or false depending on the value of the variable "she." Similarly, sentence 4 is either true or false depending on the value of the variable "he." In summary, the truth value of each open sentence depends on what value is used to replace the variable in that sentence.

Given: Let p represent, "Baseball is a sport."
Let q represent, "There are 100 cents in a dollar."
Let r represent, "She does her homework."
Let s represent, "A dime is not a coin."
Problem: Write each sentence below using symbols and indicate if it is true, false or open.

In Example 5 we are asked to find the negation of p.

Definition: The negation of statement p is "not p." The negation of p is symbolized by "

p is the opposite of the truth value of p.

Let's look at some more examples of negation.

We can construct a truth table to determine all possible truth values of a statement and its negation.

Definition: A truth table helps us find all possible truth values of a statement. Each statement is either True (T) or False (F), but not both.

Connection: To help us remember this definition, think of a computer, which is either on or off, but not both.

Example 8: Construct a truth table for the negation of x.

In Example 8, when x is true,

x is false and when x is false,

x is true. From this truth table, we can see that a statement and its negation have opposite truth values.

Example 9: Construct a truth table for the negation of p.

We can also negate a negation. For example, the negation of

p) or p. This is illustrated in the example below.

Example 10: Construct a truth table for the negation of p, and for the negation of not p.

Summary: A statement is a sentence that is either true or false. A closed sentence is an objective statement which is either true or false. An open sentence is a statement which contains a variable and becomes either true or false depending on the value that replaces the variable. The negation of statement p is "not p", symbolized by "

p". A statement and its negation have opposite truth values.

Exercises

Directions: Read each question below. Select your answer by clicking on its button. Feedback to your answer is provided in the RESULTS BOX. If you make a mistake, choose a different button.


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  • Features practical simulation algorithms written in pseudocode with implemented code available online

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  • Summarising and presenting the state-of-the-art in modeling epidemics on networks with results and readily usable models signposted throughout the book
  • Presenting different mathematical approaches to formulate exact and solvable models
  • Identifying the concrete links between approximate models and their rigorous mathematical representation
  • Presenting a model hierarchy and clearly highlighting the links between model assumptions and model complexity
  • Providing a reference source for advanced undergraduate students, as well as doctoral students, postdoctoral researchers and academic experts who are engaged in modeling stochastic processes on networks
  • Providing software that can solve differential equation models or directly simulate epidemics on networks.

Dr. I.Z. Kiss: Dr. Kiss is a Reader in the Department of Mathematics at the University of Sussex with his research at the interface of network science, stochastic processes and dynamical systems. His work focuses on the modeling and analysis of stochastic epidemic processes on static and dynamic networks. His current interests include the identification of rigorous links between approximate models and their rigorous mathematical counterparts and formulating new models for more complex spreading processes or structured networks.

Dr. J.C. Miller: Dr. Miller is a Senior Research Scientist at the Institute for Disease Modeling in Seattle. He is also a Senior Lecturer at Monash University in Melbourne with a joint appointment in Mathematics and Biology. His research interests include dynamics of infectious diseases, stochastic processes on networks, and fluid flow in porous media. The majority of his work is at the intersection of infectious disease dynamics and stochastic processes on networks.

Prof. P.L. Simon: Prof. Simon is a Professor at the Institute of Mathematics, Eötvös Loránd University, Budapest. He is a member of the Numerical Analysis and Large Networks research group and the Head of Department of Applied Analysis and Computational Mathematics. His research interests include dynamical systems, partial differential equations and their applications in chemistry and biology. In particular, his work focuses on the modeling and analysis of network processes using differential equations.

“The book adds to the knowledge of epidemic modeling on networks by providing a number of rigorous mathematical arguments and confirming the validity and optimal range of applicability of the epidemic models. It serves as a good reference guide for researchers and a comprehensive textbook for graduate students.” (Yilun Shang, Mathematical Reviews, November, 2017)

“This is one of the first books to appear on modeling epidemics on networks. … This is a comprehensive and well-written text aimed at students with a serious interest in mathematical epidemiology. It is most appropriate for strong advanced undergraduates or graduate students with some background in differential equations, dynamical systems, probability and stochastic processes.” (William J. Satzer, MAA Reviews, September, 2017)


Equations with Multiple Steps

Most of the time, solving equations requires more than a single step. For example, think about the equation I introduced you to at the beginning of the section: 3x - 2 = 19. Not only is there a -2 on the same side of the equal sign as the x, but there's also a 3 clinging to that x, like a dryer sheet stuck to a pant leg. In order to isolate x (and therefore solve the equation), you'll have to get rid of both numbers, using each of the techniques you've learned so far. (It wouldn't hurt to throw in some fabric softener with static cling controller as well.)

If a solution requires more than a single step, here's the order you should follow:

Talk the Talk

Consider the equation 3y - 7y = 12. Since 3y and -7y both have the exact same variable part (y), they are called like terms and you can simplify by combining the coefficients and leaving the variable alone: 3y - 7y = -4y, since 3 - 7 = -4, so the equation is now -4y = 12.

I'll carefully define like terms and discuss them further in Introducing Polynomials.

  1. Simplify the sides of the equation separately. Each of the items added to or subtracted from one another in the equation are called terms. If two terms have the exact same variable portion, then they are called like terms, and you can combine them as though they were numbers.
  2. Separate the variable. Using addition and subtraction, move all terms containing the variable you're isolating to one side of the equation (usually the left) and move everything else to the other side (usually the right). You're finished when you have something that looks like this: ax = b (a number times the variable is equal to a number).
  3. Eliminate the coefficient. If the variable's coefficient is something other than 1, you need to either divide by it or multiply by its reciprocal (like you did earlier in this section).

Equation solving requires practice, and it's going to take some trial and error before you get good at it. Don't forget to check your answers! Even though I will only show answer checking rarely from this point forward (to save space), rest assured that I never let the chance to make sure I got the answer right go by! Eventually, you'll feel comfortable checking answers by substituting in your head and working things out mentally.

Example 2: Solve each equation.

  • (a) 3x - 2 = 19
  • Solution: You can't simplify the left side, since 3x and -2 are not like terms, so the first thing to do is to separate the variable term. Accomplish this by adding 2 to both sides.

Divide both sides by 3 to eliminate the coefficient.

  • 3x3 = 21 3
  • x = 7
  • (b) -14 = 2x + 4(x + 1)
  • Solution: You can do a bit of simplifying on the right side of the equation. Start by distributing that positive 4 into the quantity within parentheses.
  • -14 = 2x + 4 x + 4 1
  • -14 = 2x + 4x + 4
  • Simplify like terms 2x and 4x.
  • -14 = 6x + 4
  • At this point, the problem looks a lot like the equation from part (a), except the variable term appears on the right side of the equation. There's no problem with thatit's perfectly fine. In fact, if you leave the 6x on the right side, it's less work to separate the variable term. Just subtract 4 from both sides.

Divide both sides by 6 to eliminate the coefficient.

How'd You Do That?

In Example 2, part (b), I solved the equation by isolating the x on the right side, rather than the left side. To tell you the truth, I prefer x on the left side as a matter of personal taste, even though it doesn't affect the answer at all.

According to the symmetric property of algebra, you can swap sides of an equation without affecting its solution or outcome. In other words, I could have flip-flopped the sides of the equation in 2(b) to get 2x + 4(x + 1) = -14. If you solve that equation, you'll get x = -3, the exact same answer. So, if you ever wish the sides of an equation were reversed, go ahead and flip them without fear.

  • (c) -3(x + 7) = -2(x - 1) + 5
  • Solution: You can apply the distributive property on both sides of the equal sign to begin.
  • -3x - 21 = -2x + 2 + 5
  • Simplify the right side by combining the 2 and 5 (which are technically like terms, since they have the exact same variable partno variables at all).
  • -3x - 21 = -2x + 7
  • Now, it's time to separate the variable term. Do this by adding 2x to both sides (to remove all x terms from the right side of the equation) and adding 21 to both sides as well (to remove plain old numbers from the left side of the equation).
Critical Point

As demonstrated in Example 2(c), a negative variable like -w technically has an implied coefficient of -1, so you can rewrite it as -1w if you wish. (This is similar to implied exponents, where a plain old variable like w has an implied coefficient of 1, so w = 1w 1 .)

  • At this point, you have -x = 28, which means "the opposite of the answer equals 28." Therefore, the correct answer is x = -28 (since -28 is the opposite of 28).
  • Here's another way to get the final answer: Since -x = -1 x, you can rewrite the final line of the equation so it looks like it has a coefficient and divide by that -1 coefficient:
  • -1x1 = 28 1
  • x = -28
  • (d) y + 3 = 1 4y + 5
  • Solution: Since there are no like terms together on one side of the equation, skip right to separating the variable terms. Accomplish this by subtracting 1 4y and 3 on both sides of the equation. (By the way, even though the variable is y, not x, in this equation, that doesn't change the way you solve it.)
You've Got Problems

Problem 3: Solve the equations.

(b) 2x - 7 = 4x + 13

  • Since the coefficient is fractional, multiply both sides by its reciprocal to finish.
  • 4 3 ( 3 4y = ( 4 3) 2 1
  • 12 12y = 8 3
  • y 8 3
  • Since 8 and 3 have no common factors (other than 1), the improper fraction cannot be simplified, so leave it as is.

Excerpted from The Complete Idiot's Guide to Algebra 2004 by W. Michael Kelley. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc.


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1. A dime is a coin.