AMA126 - 8 -
AMA126 Differential and difference equations
University of Sheffield, Dept of Applied Mathematics
2007
Revision of Integration (Indefinite Integration)
When a function f(x) is known we can differentiate it to obtain its derivative df/dx. The reverse process is to obtain the function f(x) from knowledge of its derivative. This process is called integration and it has numerous applications in all areas of sciences.
Suppose we differentiate the function y=x2. We obtain
Integration reverses this process and we say that the integral of 2x is x2. Schematically we can regard the process of integration in the following way
In reality the situation is a bit more complicated because differentiating any of
yields 2x. In fact, any function of the form x2+c, where c is a constant, will be the answer to our question since when we differentiate the constant term we obtain zero. Consequently, when we reverse the process of differentiation, we do not know what the original constant term might have been. So we include in our answer an unknown constant, c say, called the constant of integration and we state that the integral of 2x is x2+c. This constant must always be included when finding an indefinite integral. The solution containing the constant c defines a family of curves all being solutions of the integral. If one of these curves is a solution of the integral, they all are. To select between them we need more information. This problem defines so called initial conditions, a topic you will meet later when solving differential equations.
The indefinite integration is represented by the symbol ∫ known as an integral sign. Accompanying the integral sign is always a term of the form dx, which indicates the independent variable involved in the integration, in this case x. So we write
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Example
(a) State the derivative of x3
(b) Hence find the indefinite integral of 3x2
Solution
(a) From our knowledge of differentiation, the derivative of x3 is 3x2.
(b) Indefinite integration reverses the process of differentiation, and so we write
We always include the additional constant of integration when finding indefinite integrals. Note that our answer can be checked by differentiating x3+c to obtain 3x2.
More generally, we have the following relationship between derivatives and indefinite integrals:
In the expression ∫ f(x) dx, the function f(x) is referred to as the integrand. When we have calculated ∫ f(x) dx, we say f(x) has been integrated with respect to x to yield F(x)+c.
In what follows we are going to discuss some techniques which can be used to evaluate many types of integrals.
I. Integration by parts
If f and g are differentiable functions, then by the Product Rule
where the dash means the differentiation of the function with its argument. Integrating both sides of the previous equation gives us
The simplest is to perform the first integral on the right side and the integral is f(x)g(x)+C (why???). Since another constant of integration results from the second integral, it is unnecessary to include C in the formula; that is
If we let u=f(x) and v=g(x), so that du=f′(x)dx and dv=g′(x)dx, then the preceding formula may be written (integration by part)
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Example: Find
Solution
There are four possible choices for dv, namely dx, xdx, e2xdx or xe2xdx. If we let dv= e2xdx, then the remaining part of the integrand is u; that is u=x. To find v we integrate dv to obtain v=e2x/2. Note that a constant of integration is not added at this stage of the solution. Since u=x we see that du=dx. For ease of reference it is convenient to display these expressions as follows
Substituting these expressions in the definition equation, we obtain
The integral on the right hand side may be found using the integrals of exponential functions (see in the table below). This gives us
It takes considerable practice to become proficient in making a suitable choice for dv. To illustrate, if we had chosen dv=xdx, then it would have been necessary to let u=e2x, giving us
Integrating by parts we obtain
Since the exponent associated with x has increased, the integral on the right is more complicated than the given integral. This indicates an incorrect choice of dv.
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II. Trigonometric integrals
In many cases evaluating an integral we can meet trigonometric functions. For example, integrals of the type ∫sinnxdx require new method of solving. If n is an odd positive integer, we begin by writing
Since the integer n-1 is even, we may then use the fact that sin2x=1-cos2x to obtain a form which is easy to integrate.
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Example
Evaluate ∫sin5xdx.
Solution
As discussed earlier we have
We next employ the method of substitution, letting u=cosx and du=-sinxdx. Thus
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A similar technique can be employed for odd powers of cos x, specifically, we write
and use the fact that cos2x=1-sin2x in order to obtain an integrable form. If the integrand is sinnx or cosnx and n is even, then the half-angle formulas
may be used to simplify the integrand.
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Example
Evaluate ∫sin4xdx
Solution
We apply a half-angle formula again and write
Substituting in the last integral and simplifying gives us
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Integrals of the form ∫sinmxcosnxdx where m and n are positive integers may be found by using variations of the previous techniques. If m and n are both even, then half-angle formulas should be employed first. In n is odd, we can write
and express cosn-1x in terms of sinx by using the identity cos2x=1-sin2x. The substitution u=sinx then leads to an integrand which can be handled easily. A similar technique can be used if m is odd.
III. Trigonometric substitutions
If an integrand contains the expression where a>0, then the trigonometric substitution x=asinθ leads to
When making this substitution or the other trigonometric substitutions in the next examples, we shall assume that θ is in the range of the corresponding inverse trigonometric function. Thus, for the sine substitution above –π/2≤θ≤π/2. Consequently cosθ≥0 and . Of course, if occurs in a denominator we make the further restriction –π/2<θ<π/2.
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Example
Evaluate
Solution
Let x=asinθ , where –π/2<θ<π/2. It follows that
Since x=asinθ, we have dx=acosθdθ. Substituting in the given integral
It is now necessary to return to the original variable of integration x. A simple method of doing so is to use a geometrical approach. If 0<θ<π/2, then since sinθ=x/a, we may interpret θ as an acute angle of a triangle having opposite side and hypotenuse of lengths x and a, respectively. The length of the adjacent side is calculated by means of the Pythagorean Theorem.
Referring to the triangle we see that
It can be shown that this formula is also valid if –π/2<θ<0. Thus the above figure can be used whether θ is positive or negative. Substituting the new form of cotθ in the result obtained for the integral, we have
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If an integrand contains , where a>0, then the substitution x=atanθ will eliminate the radical sign. When using this substitution it will be assumed that θ is in the range of the inverse tangent function; that is –π/2<θ<π/2. After making this substitution and evaluating the resulting trigonometric integral, it is necessary to return to the original variable, x. The preceding formulas show that
For integrands containing we substitute x=asecθ, where θ is chosen in the range of the inverse secant function; that is either 0≤θ≤π/2 or π≤θ<3π/2. After evaluating the integral, we have to return to the original variable and we use
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Example
Evaluate
Solution
Let us substitute as follows
Consequently
and therefore
Since secθ=x/3 we may refer to the changing rule into the old variable and we write
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III. Partial Fractions
It is easy to verify that
The expression on the right side of the equation is called the partial fraction decomposition of the expression on the left side. This decomposition may be used to find the indefinite integral of the expression on the left side. We merely integrate each of the fractions which make up the decomposition independently, obtaining
It is theoretically possible to write any rational expression f(x)/g(x) as a sum of rational expressions whose denominators involve powers of polynomials of degree not greater than two. More specifically, if f(x) and g(x) are polynomials and the degree of f(x) is less than the degree of g(x), then it follows from a theorem in algebra that
where each of Fi has one of the forms
for some non-negative integers m and n, and when ax2+bx+c is irreductible, in the sense that this quadratic expression has no real roots, that is b2-4ac<0.
To find the fraction decomposition of a rational expression f(x)/g(x) it is essential that f(x) have lower degree than g(x). If this is not the case, then long division should be employed to arrive at such expression. For example, given
we obtain, by long division,
The partial fraction decomposition is then found for the last term in the right side.
In order to obtain the decomposition of a rational expression of the form f(x)/g(x), we begin by expressing the denominator g(x) as a product of factors px+q or irreductible quadratic factors ax2+bx+c. Repeated factors are then collected so that g(x) is a product of different factors of the form (px+q)m or (ax2+bx+c)n, where m and n are non-negative integers and the quadratic forms are irreductible. We then apply the following rules
Rule 1. For each factor of the form (px+q)m where m≥1, the partial fraction decomposition contains a sum of m partial fractions of the form
where each of Ai is a real number.
Rule 2. For each of the factor of the form (ax2+bx+c)n where n≥1 and b2-4ac<0, the partial fraction decomposition contains n partial fractions of the form
where each of Ai and Bi are real numbers.
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Example
Evaluate
Solution
By Rule 1, there is a partial fraction of the form A/(x+1) corresponding to the factor x+1 in the denominator of the integrand. For the factor (x-2)3 we apply Rule 1 (with m=3), obtaining a sum of three partial fractions B/(x-2), C/(x-2)2, and D/(x-2)3. Consequently, the partial fraction decomposition has the form
Multiplying both sides by (x+1)(x-2)3 gives us
Two of the unknown constants may be determined easily. If we let x=2 then
24-72+58-4=3D, 6=3D, and D=2
Similarly, letting x=-1 we obtain
-3-18-29-4=-27A, -54=-27A, and A=2
The remaining constants may be found by comparing the coefficients. If the right side of the decomposition is expanded and like powers of x collected, we see that the coefficient of x3 is A+B. This must equal the coefficient of x3 on the left, that is
A+B=3
Since A=2, it follows that B=1. Finally, we compare the constant terms by letting x=0. This gives us
-4=-8A+4B-2C+D
Substituting the values we have already found for A, B and D leads to
-4=-16+4-2C+2
which has the solution C=-3. The partial fraction decomposition is, therefore
To find the given integral we integrate each of the partial fractions on the right side of the last equation. This gives us
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Obviously, there are many characteristic types of indefinite integrals with their own rules. To see them all, please consult any book on calculus.
Differential equations
Definition: A differential equation is an equation which involves derivatives or differentials. Since the equations we are going to study contain only one variable, these equations are called ordinary differential equations. The primary objective of this lecture is to develop techniques for solving certain basic types of ordinary differential equations.
Introduction
An equation of the form
, 2.1
where F is a function of n+2 variables, y is a function of x, and y(k) denotes the k-th derivative of y with respect to x, is called an ordinary differential equation of order n. The following are examples of ordinary differential equations of orders 1, 2, 3, and 4, respectively:
If, in (2.1), F is a polynomial function, then by definition the degree of the differential equation is the greatest exponent associated with the highest order derivative y(n). The degrees of the preceding differential equations are 1, 1, 4, and 2, respectively.
If a function f has the property that when f(x) is substituted for y in a differential equation, the resulting expression is an identity for all x in some interval, then f(x) (or simply f) is called a solution of the differential equation. For example, if C is any real number, then a solution of y’=2x is
because substitution of f(x) for y leads to the identity 2x=2x. We call x2+C the general solution of y’=2x, since every solution has this form.
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Example
If C1 and C2 are any real numbers, prove that
is a solution of y’’-25y=0.
Solution
Since
we have f″(x)-25f(x)=0. This shows that f(x) is a solution of the equation y″-25y=0.
The solution given in the above Example is called the general solution of y″-25y=0. Observe that the differential equation is of the order 2 and the general solution contains two arbitrary parameters C1 and C2 (for connections to integrals, see why these parameters arise in the recap about integrals).