Review of Differentiation and Integration for Ordinary Differential Equations

Transcription

1 Schreyer Fall 208 Review of Differentiation an Integration for Orinary Differential Equations In this course you will be expecte to be able to ifferentiate an integrate quickly an accurately. Many stuents take this course after having taken their previous course many years ago, at another institution where certain topics may have been omitte, or just feel uncomfortable with particular techniques. Because unerstaning this material is so important to being successful in this course, we have put together this brief review packet. In this packet you will fin sample questions an a brief iscussion of each topic. If you fin the material in this pamphlet is not sufficient for you, it may be necessary for you to use aitional resources, such as a calculus textbook or online materials. Because this is consiere prerequisite material, it is ultimately your responsibility to learn it. The topics to be covere inclue Differentiation an Integration. Differentiation Exercises:. Fin the erivative of y = x 3 sin(x). 2. Fin the erivative of y = ln(x) cos(x). 3. Fin the erivative of y = ln(sin(e 2x )). Discussion: It is expecte that you know, without looking at a table, the following ifferentiation rules: [(kx)n ] = kn(kx) n () [ ] e kx = ke kx (2) [ln(kx)] = x (3) [sin(kx)] = k cos kx (4) [cos(kx)] = k sin x (5) [uv] = u v + uv (6)

2 [ ] u = u v uv (7) v v 2 [u(v(x))] = u (v)v (x). (8) We put in the constant k into () - (5) because a very common mistake to make is something like: e2x = e2x 2 (when the correct answer is 2e2x ). Equation (6) is known as the prouct rule, Equation (7) is known as the quotient rule, an Equation (8) is known as the chain rule. From these, you can erive the erivative of many other functions, such as the tangent: tan(x) = sin(x) cos(x) cos(x) cos(x) sin(x)( sin(x)) [tan(x)] = (cos(x)) 2 = cos2 (x) + sin 2 (x) = cos 2 (x) = sec 2 (x) cos 2 (x) where we have use the quotient rule an simplifie. The chain rule is applie when there is a function of a function, i.e. f(g(x)). The iea is to take the erivative of the outsie function first, leaving its argument alone. Then multiply that by the erivative of the next outermost function, leaving it s argument alone. The process is repeate until there is nothing left of which to take the erivative. So for example, to take the erivative of sin 2 (5x), we nee to first etermine the outsie function. If we re-write it as (sin(5x)) 2 it is quickly etermine that the outsie function is something square, where something in this case is sin(5x). The erivative of something square is 2 times that something times the erivative of that something. Thus we have [ sin 2 (5x) ] = [ ] (sin(5x)) 2 = 2 sin(5x) [sin(5x)] = 2 sin(5x) cos(5x) [5x] = 2 sin(5x) cos(5x)5 = 0 sin(5x) cos(5x) Solution to Exercises:. For this problem, we nee the prouct rule, (6), since two functions are being multiplie. In this case, u(x) = x 3 an v(x) = sin(x). Thus, [ x 3 sin(x) ] = 3x 2 sin(x) + x 3 [cos(x)] 2

3 = 3x 2 sin(x) + x 3 cos(x). 2. This is clearly a quotient of functions, so that the quotient rule applies, (7). We have u(x) = ln(x) an v(x) = cos(x), which implies: [ ] ln(x) cos(x) = = = cos(x) ln(x)[ sin(x)] x (cos(x)) 2 cos(x) + ln(x) sin(x) x cos 2 (x) cos(x) + x ln(x) sin(x) x cos 2 (x) 3. This is a case of a function of a function of a function of a function, f(g(h(i(x)))). We apply the chain rule, always working from the outsie function in. In this case the (very) outsie function is f() = ln of something ; the next most outsie function is, g() = sin of something, the next most outsie function is, h() = e to the something, an the insie most function is i(x) = 2x. Applying the chain rule (8) we have 2 Integration [ ln(sin(e 2x )) ] = = = = [ sin(e 2x ) ] sin(e 2x ) sin(e 2x ) cos(e2x ) [ ] e 2x sin(e 2x ) cos(e2x )e 2x [2x] sin(e 2x ) cos(e2x )e 2x 2 = 2e2x cos(e 2x ) sin(e 2x ) = 2e 2x cot(e 2x ) Solving ifferential equations requires integration - there s just no getting aroun it. What follows is a brief review. If you nee supplemental material, please see your calculus text. Exercises:. Evaluate 2. Evaluate x x 2 +. sin(x) cos(x). 3

4 3. Evaluate xe 3x 4. Given u n ln(u) u = un+ ln(u) n + un+ (n + ) + C, evaluate 2 x 2 ln(2x). The iscussion section is rather long, an the solution to these exercises are given at the en of this section. Discussion: 2. Basic Integration Formulas Not only is it important to be familiar with various integration techniques, but it is also important that we be quick an efficient when evaluating integrals so that we can concentrate on the concepts as oppose to the mechanics of integration. Examples. We can evaluate the inefinite integral e 2x by oing a u-substitution. However, we can become more efficient at evaluating integrals of this type by obtaining a general formula. Let f(x) = e ax, where a is equal to a constant. We woul like to obtain a general formula for e ax. We can accomplish this by oing a u-substitution. Let u = ax, then u = a u =. Substitution yiels a e ax = a e u u = a eu + C = a eax + C Now we have a general formula that we can use again an again without going to the trouble of oing the u-substitution each time. For example, (a) y(x) = e 2x e 2x = 2 e2x + C (b) y(x) = e πx e πx = π eπx + C 2. Let us follow the proceure in Example to fin the general formula for integrals of the form cos (ax), where a is equal to a constant. Let u = ax, then u = a =. Substitution yiels u a 4

5 cos(ax) = a cos(u) u = sin(u) + C = sin(ax) + C. a a Here are some examples of using this general formula. (a) y(x) = cos(4x) cos(4x) = 4 sin(4x) + C (b) y(x) = cos 2π x ( ) cos 2π x = 2π sin 2πx + C Exercises Follow the examples above to obtain a general formula for the integral given, then use it to evaluate parts (a) an (b). As above, a is equal to a constant.. sin(ax) (a) (b) sin(6x) ( ) sin 2 x 2. ln(ax) (a) (b) ln(πx) ln( x) π 3. tan(ax) (a) (b) tan(3x) ( ) tan 3 x 4. sec(ax) (a) sec(2.78x) (b) sec(68x) 5. arctan(ax) 5

6 (a) arctan(πx) (b) arctan ( π x) 2.2 u-substitution In general, u-substitutions are not as straight forwar as the ones in the previous section. When oing a u-substitution you want to look for the part of the integral whose erivative is elsewhere in the integral (up to a constant). Formally, if we have an integral of the form f(g(x))g (x), we let u = g(x), then u = g (x), substitution yiels f(g(x))g (x) = f(u)u Essentially, we have transforme the space in which we are evaluating the integral. We evaluate the integral in this new space an then substitute u back in to obtain a solution in the original space. Similar techniques are often employe to solve ifferential equations. Examples. x 5 e x6 First, let u = x 6, then u = 6x 5 u 6 = x5. Substitution yiels x 5 e x6 = 6 e u u = 6 eu + C = 6 ex6 + C. 2. (x 2 + ) 2 (2x) First, let u = x 2 +, then u = 2x. Substitution yiels (x 2 + ) 2 (2x) = u 2 u = 3 u3 + C = 3 (x2 + ) 3 + C. 6

7 Exercises Use u-substitution to evaluate the following inefinite integrals.. sin 2 (3x) cos(3x) 2. cos ( ) θ 2 θ θ 3. sin x cos 2 x 4. e x (e x + ) 2 5. tan 4 x sec 2 x 2.3 Integration by Parts Integration by parts is applicable to a plethora of functions which we may nee to integrate. Formally, if u an v are functions of x an have continuous erivatives, then uv = uv vu Choosing which part is equal to u may be facilitate by remembering the acronym: LIATE, which stans for: Logarithm, Inverse trig, Algebraic, Trigonometric, Exponential. This means that whichever of these expressions appears first in the acronym, that is the expression you shoul let u be. So if you want to evaluate (x 2 + 5x 2)e 5x, we see we have an algebraic expression, x 2 + 5x 2, times an exponential function, e 5x. By this acronym, since A appears before E, we set u = x 2 + 5x 2. Examples. Evaluate xe x First, let u = x u = an let v = e x v = e x. Using the integration by parts formula we obtain xe x = xe x e x = xe x e x + C. 2. Evaluate arcsin x First, let u = arcsin x u = x 2 an let v = v = x. arcsin x = x arcsin x x x 2 7

8 = x arcsin x + 2 w 2 w = x arcsin x + w 2 + C = x arcsin x + ( x 2 ) 2 + C = x arcsin x + x 2 + C. Where the secon equality comes from oing a u substitution (w in this case) where w = x 2 3. Sometimes it is necessary to o integration by parts more than once. For example, x 2 e x. First, let u = x 2 u = 2x an let v = e x v = e x. Substitution yiels x 2 e x = x 2 e x 2 xe x = x 2 e x 2(xe x e x ) + C = x 2 e x 2xe x 2e x + C. where the secon equality comes from our previous calculation in Example. 4. Here is another example where integration by parts will be use repeately to evaluate an integral. Evaluate y(x) = e x cos x. First, let u = e x u = e x an let v = cos x v = sin x. Substitution yiels e x sin x = e x sin x e x sin x. Using integration by parts again, let u = e x u = e x an let v = sin x v = cos x. Substitution yiels 2 [ ] e x cos x = e x sin x e x cos x + e x cos x = e x sin x + e x cos x e x cos x e x cos x = e x sin x + e x cos x e x cos x = 2 (ex sin x + e x cos x) + C. Exercises Use integration by parts to evaluate the following inefinite integrals.. t ln (t + ) t 2. (ln x) 2 x 3. arccos x 8

9 4. e 2x sin x 5. e x sin x 2.4 Integration using Partial Fraction Decomposition Partial fraction ecomposition is applicable to rational functions which we may nee to integrate. A rational function is one which is a ratio of two polynomials: f(x) = P (x) Q(x). We consier two cases: The egree of P (x) < egree of Q(x) an the egree of P (x) egree of Q(x). If the egree of P (x) Q(x), then we use long ivision to write the rational polynomial as a polynomial plus a rational polynomial where the egree of the numerator is less than the egree of the enominator. Long Division Examples. x 3 x 3 2 x 2 2 We note that the egree of the numerator, 3, is larger than the egree of the enominator, 2. Using long ivision: (x 2 2) x 3 x 3 2 Think: since x 2 goes into x 3 x times, an we have: x 2 2 x x 3 x 3 2 (x 3 2x) x 3 2 Since the egree of x 3 2 is less than the egree of x 2 2 we stop an we have: x 3 x 3 2 x 2 2 = x + x 3 2 x x 5 3 x 3 x 2 We note that the egree of the numerator, 5, is larger than the egree of the enominator, 3. Using long ivision: (x 3 x 2 ) x 5 x 3 9

10 Think: x 3 goes into x 5 x 2 times, an we have x 2 x 3 x 2 x5 3 (x 5 x 4 ) x 4 3 Now x 3 goes into x 4 x times an we continue: x 2 + x + x 3 x 2 x5 3 (x 5 x 4 ) x 4 3 (x 4 x 3 ) x 3 3 (x 3 x 2 ) x 2 3 an so we have x 5 3 x 3 x = 2 x2 + x + + x2 3 x 3 x. 2 We can check our answer by oing a quick polynomial aition where we put everything uner the same enominator: (x 2 + x + )(x 3 x 2 ) x 3 x 2 + x2 3 x 3 x 2 = x5 3 x 3 x 2. Factoring Polynomials Since integrating polynomials is (hopefully) easy, we now concentrate on integrating rational functions where the egree of the numerator is strictly less than the egree of the enominator. We first note that All polynomials with real coefficients can be factore into linear an irreucible quaratic factors: Q(x) = (α x β ) n (α 2 x β 2 ) n 2...(a x 2 + b x + c ) m (a 2 x 2 + b 2 x + c 2 ) m 2... An irreucible quaratic factor is one that has imaginary roots, i.e. from the quaratic formula, b 2 4ac < 0. So for example x 4 5x 2 5 = (x 2 )(x 2 + 4) = (x )(x + )(x 2 + 4) 0

11 Here the linear factors are (x ) an (x + ) an the irreucible quaratic factor is (x 2 + 4) since the roots of (x 2 + 4) are ±2i (complex). Note that (x 2 ) is not irreucible, because it has real roots (±). Another example: 2x 4 + 2x 3 + 3x 2 = x 2 (5x 2 + 2x + 3). Here we have a repeate linear root: x = 0 appears twice (think (x 0) 2 ), an an irreucible quaratic: for 5x 2 + 2x + 3, b 2 4ac = 5 2 4(5)(3) = 35 < 0, so this quaratic has only complex roots. Partial Fraction Decomposition We use the following examples to guie our selection of how to ecompose a rational function where the egree of the numerator is strictly less than the egree of the enominator. Note that the enominators have alreay been factore into linear an irreucible quaratic terms. Examples (x )(x 2)(x 3) = A x + B x 2 + C x 3 (x ) 3 (x 2) = A x + B (x ) + C 2 (x ) + D 3 x 2 x 2 (x 2 + ) = A x + B x + Cx + D 2 x 2 + (x ) 2 (x 2 + ) = A 2 x + B (x ) + Cx + D 2 x Ex + F (x 2 + ) 2 The goal is to etermine the values of the parameters A, B,... F an then we can integrate each term using techniques we alreay know. Note that in each case, the number of unknown parameters is equal to the egree of the original polynomial in the enominator. In example, on the left sie, the enominator is a polynomial of egree 3, an there are 3 parameters, A, B, an C. In example 4, the, the egree of the polynomial in the enominator is 6, an there are 6 parameters, etc. This is a sanity check, to be sure we have not mae a blatant error in setting up the partial fraction ecomposition. There are two methos that are use to etermine the unknown parameters, an we will o two examples illustrating each metho. In some problems, the first metho is easiest to use, an in others the secon. Only experience will help you etermine which metho to use. Examples. Fin the partial fraction ecomposition of (x )(x 2)(x 3).

12 We first note that the egree of the numerator (0) is less than the egree of the enominator (3). The enominator is alreay factore, an from the iscussion above we know that the form of the partial fraction ecomposition is (x )(x 2)(x 3) = A x + B x 2 + C x 3 (9) First get the enominators of each term the same by multiplying each term on the right-han sie by special forms of : (x )(x 2)(x 3) = A x (x 2)(x 3) (x 2)(x 3) + B (x )(x 3) x 2 (x )(x 3) + C (x )(x 2) x 3 (x )(x 2).(0) Because the enominators are the same, in orer to satisfy the above equation, we now only nee to be sure the numerators are equal. Examining the numerators only we have = A(x 2)(x 3) + B(x )(x 3) + C(x )(x 2). () The above equation must hol for all x. Since it hols for all x, it must hol for particular values of x, an we choose these values of x to make it easy to solve for A, B, an C. Consier x =. Substituting in for x into equation () we get: = A( )( 2) + B(0)( 2) + C(0)( ) = 2A A = 2 Likewise, let us consier x = 2 an x = 3. Substituting these values into equation () gives us respectively, = A(0)( ) + B()( ) + C()(0) = B B = = A()(0) + B(2)(0) + C(2)() = 2C C = 2. Thus we have (going back to equation (9)): (x )(x 2)(x 3) = /2 x + x 2 + /2 x 3 where we see that the right-han sie is much easier to integrate than the left sie. We can check our answer by aing the fractions on the right-han sie (requiring us to get a common enominator) an checking to be sure we get the left-han sie. Note that in general, if we want to solve for 2 unknowns, we nee 2 x values; if we want to solve for 3 unknowns, we nee 3 x values; etc. This is because if we want to solve for n unknowns we nee n equations that are not reunant. 2

13 2. Fin the partial fraction ecomposition of 3x 3 + x 2 x 2 (x 2 + ) We first note that the egree of the numerator (3) is less than the egree of the enominator (4). The enominator is alreay factore, an from the iscussion above we know that the form of the partial fraction ecomposition is 3x 3 + x 2 x 2 (x 2 + ) = A x + B x + Cx + D 2 x 2 + First get the enominators of each term the same by multiplying each term on the right-han sie by special forms of : 3x 3 + x 2 x 2 (x 2 + ) = A x x(x 2 + ) x(x 2 + ) + B (x 2 + ) x 2 (x 2 + ) + Cx + D x 2 + (2) x 2 x 2. (3) Because the enominators are the same, in orer to satisfy the above equation, we now only nee to be sure the numerators are equal. Examining the numerators only we have 3x 3 + x 2 = Ax(x 2 + ) + B(x 2 + ) + (Cx + D)x 2 (4) = Ax 3 + Ax + Bx 2 + B + Cx 3 + Dx 2 (5) The above equation says that we must fin the values of A, B, C, an D so that the two polynomials (the one on the left sie an the one on the right) are equal. Two polynomials are equal if an only if their coefficients are equal. So we look at the coefficients of x 3, x 2, x, an x 0 (constants), to give us 4 equations for the 4 unknowns: x 3 : x 2 : x : x 0 : 3 = A + C 0 = B + D = A 2 = B It turns out that these 4 equations are easy to solve for our unknowns, but in general one can use the technique of elimination to solve for the 4 unknowns. In this case, the thir an fourth equations tell us that A = an B = 2. Using these values in the first two equations gives C = 3 = 2 an D = B = 2. Using these values, we can now write equation (2) as: 3x 3 + x 2 x 2 (x 2 + ) = x + 2 x + 2x x 2 + To integrate the last term we rewrite the fraction as: 2x + 2 x 2 + = 2x x x 2 +. The first of these terms can be integrate using u-substitution, an the secon is of the form u 2 + a which has an integral of u 2 a tan a + C. 3

14 Problems. Fin the partial fraction ecomposition of x 3 + (x 2 + ) 2 Solve for the unknown coefficients, but o not integrate. 2. Evaluate 3. Integrate x 3 x 2 3x + 3 x (x 2 + x + )(x ) (Hint: to integrate one of the terms, consier completing the square) 4. Integrate x 3 x 3 2 x What happens when you try to solve for A, B, an C for the partial fraction ecomposition: (x 2 )(x ) = A x + Bx + C x 2? Try to solve for A, B, an C, an show all work. What is wrong with this formulation? 4