Exam (elaborations) TEST BANK FOR Advanced Engineering Mathematics [Volume 1] By Herbert Kreyszig and Erwin Kreyszig (Student Solutions Manual and Study
... [Show More] Guide) P A R T A Ordinary Differential Equations (ODEs) Chap. 1 First-Order ODEs Sec. 1.1 Basic Concepts. Modeling To get a good start into this chapter and this section, quickly review your basic calculus. Take a look at the front matter of the textbook and see a review of the main differentiation and integration formulas. Also, Appendix 3, pp. A63–A66, has useful formulas for such functions as exponential function, logarithm, sine and cosine, etc. The beauty of ordinary differential equations is that the subject is quite systematic and has different methods for different types of ordinary differential equations, as you shall learn. Let us discuss some Examples of Sec. 1.1, pp. 4–7. Example 2, p. 5. Solution by Calculus. Solution Curves. To solve the first-order ordinary differential equation (ODE) y = cos x means that we are looking for a function whose derivative is cos x. Your first answer might be that the desired function is sin x, because (sin x) = cos x. But your answer would be incomplete because also (sin x+2) =cos x, since the derivative of 2 and of any constant is 0. Hence the complete answer is y= cos x+c, where c is an arbitrary constant. As you vary the constants you get an infinite family of solutions. Some of these solutions are shown in Fig. 3. The lesson here is that you should never forget your constants! Example 4, pp. 6–7. Initial Value Problem. In an initial value problem (IVP) for a first-order ODE we are given an ODE, here y =3y, and an initial value condition y(0)=5.7. For such a problem, the first step is to solve the ODE. Here we obtain y(x)=ce3x as shown in Example 3, p. 5. Since we also have an initial condition, we must substitute that condition into our solution and get y(0)=ce3·0 = ce0 =c · 1= c=5.7. Hence the complete solution is y(x)=5.7e3x. The lesson here is that for an initial value problem you get a unique solution, also known as a particular solution. 2 Ordinary Differential Equations (ODEs) Part A Modeling means that you interpret a physical problem, set up an appropriate mathematical model, and then try to solve the mathematical formula. Finally, you have to interpret your answer. Examples 3 (exponential growth, exponential decay) and 5 (radioactivity) are examples of modeling problems. Take a close look at Example 5, p. 7, because it outlines all the steps of modeling. Problem Set 1.1. Page 8 3. Calculus. From Example 3, replacing the independent variable t by x we know that y =0.2y has a solution y=0.2ce0.2x. Thus by analogy, y =y has a solution 1 · ce1·x = cex, where c is an arbitrary constant. Another approach (to be discussed in details in Sec. 1.3) is to write the ODE as dy dx = y, and then by algebra obtain dy = y dx, so that 1 y dy = dx. Integrate both sides, and then apply exponential functions on both sides to obtain the same solution as above 1 y dy = dx, ln|y| = x + c, eln |y| = ex+c, y = ex · ec = c∗ex, (where c∗ = ec is a constant). The technique used is called separation of variables because we separated the variables, so that y appeared on one side of the equation and x on the other side before we integrated. 7. Solve by integration. Integrating y =cosh 5.13x we obtain (chain rule!) y= cosh 5.13x dx = 1 5.13 (sinh 5.13x)+c. Check: Differentiate your answer: 1 5.13 (sinh 5.13x) + c = 1 5.13 (cosh 5.13x) · 5.13 = cosh 5.13x, which is correct. 11. Initial value problem (IVP). (a) Differentiation of y=(x+c)ex by product rule and definition of y gives y = ex + (x + c)ex = ex + y. But this looks precisely like the given ODE y =ex +y. Hence we have shown that indeed y=(x + c)ex is a solution of the given ODE. (b) Substitute the initial value condition into the solution to give y(0)=(0+c)e0 =c · 1= 12 . Hence c= 12 so that the answer to the IVP is y = (x + 12 )ex. (c) The graph intersects the x-axis at x=0.5 and shoots exponentially upward. Chap. 1 First-Order ODEs 3 19. Modeling: Free Fall. y =g =const is the model of the problem, an ODE of second order. Integrate on both sides of the ODE with respect to t and obtain the velocity v=y =gt +c1 (c1 arbitrary). Integrate once more to obtain the distance fallen y= 12 gt2 +c1t +c2 (c2 arbitrary). To do these steps, we used calculus. From the last equation we obtain y= 12 gt2 by imposing the initial conditions y(0)=0 and y(0)=0, arising from the stone starting at rest at our choice of origin, that is the initial position is y=0 with initial velocity 0. From this we have y(0)=c2 =0 and v(0)= y(0) = c1 = 0. Sec. 1.2 Geometric Meaning of y =f (x, y). Direction Fields, Euler’s Method Problem Set 1.2. Page 11 1. Direction field, verification of solution. You may verify by differentiation that the general solution is y=tan(x+c) and the particular solution satisfying y(14 π)=1 is y=tan x. Indeed, for the particular solution you obtain y = 1 cos2x = sin2x + cos2x cos2x = 1 + tan2x = 1 + y2 and for the general solution the corresponding formula with x replaced by x+c. 1 –1 –2 2 –1 –0.5 0 0.5 1 y x y(x) Sec. 1.2 Prob. 1. Direction Field 15. Initial value problem. Parachutist. In this section the usual notation is (1), that is, y =f (x, y), and the direction field lies in the xy-plane. In Prob. 15 the ODE is v =f (t, v)=g −bv2/m, where v suggests velocity. Hence the direction field lies in the tv-plane.With m=1 and b=1 the ODE becomes v = g −v2. To find the limiting velocity we find the velocity for which the acceleration equals zero. This occurs when g −v2 = 9.80−v2 =0 or v=3.13 (approximately). For v<3.13 you have v >0 (increasing curves) and for v>3.13 you have v <0 (decreasing curves). Note that the isoclines are the horizontal parallel straight lines g −v2 =const, thus v=const. 4 Ordinary Differential Equations (ODEs) Part A Sec. 1.3 Separable ODEs. Modeling Problem Set 1.3. Page 18 1. CAUTION! Constant of integration. It is important to introduce the constant of integration immediately, in order to avoid getting the wrong answer. For instance, let y = y. Then ln |y| = x + c, y = c∗ex (c∗ = ec), which is the correct way to do it (the same as in Prob. 3 of Sec. 1.1 above) whereas introducing the constant of integration later yields y = y, ln|y| = x, y = ex + C which is not a solution of y =y when C =0. 5. General solution. Separating variables, we have y dy=−36x dx. By integration, 12 y2 = −18x2 + ˜c, y2 = 2˜c − 36x2, y = ± c − 36x2 (c = 2˜c). With the plus sign of the square root we get the upper half and with the minus sign the lower half of the ellipses in the answer on p. A4 in Appendix 2 of the textbook. For y=0 (the x-axis) these ellipses have a vertical tangent, so that at points of the x-axis the derivative y does not exist (is infinite). 17. Initial value problem. Using the extended method (8)–(10), let u=y/x. Then by product rule y =u+xu. Now y = y + 3x4cos2(y/x) x = y x + 3x3 cos y x = u + 3x3 cos2 u = u + x(3x2 cos2 u) so that u = 3x2 cos2 u. Separating variables, the last equation becomes du cos2 u = 3x2dx. Integrate both sides, on the left with respect to u and on the right with respect to x, as justified in the text then solve for u and express the intermediate result in terms of x and y tan u = x3 + c, u = y x = arctan (x3 + c), y = xu = x arctan (x3 + c). Substituting the initial condition into the last equation, we have y(1) = 1 arctan (13 + c) = 0, hence c = −1. Together we obtain the answer y = x arctan (x3 − 1). 23. Modeling. Boyle–Mariotte’s law for ideal gases. From the given information on the rate of change of the volume dV dP = −V P . Chap. 1 First-Order ODEs 5 Separating variables and integrating gives dV V = −dP P , 1 V dV = − 1 P dP, ln|V| = −ln |P| + c. Applying exponents to both sides and simplifying eln |V| = e−ln |P|+c = e−ln |P| · ec = 1 eln |P| · ec = 1 |P|ec. Hence we obtain for nonnegative V and P the desired law (with c∗ =ec, a constant) V · P = c∗ . Sec. 1.4 Exact ODEs. Integrating Factors Use (6) or (6∗), on p. 22, only if inspection fails. Use only one of the two formulas, namely, that in which the integration is simpler. For integrating factors try both Theorems 1 and 2, on p. 25. Usually only one of them (or sometimes neither) will work. There is no completely systematic method for integrating factors, but these two theorems will help in many cases. Thus this section is slightly more difficult. Problem Set 1.4. Page 26 1. Exact ODE.We proceed as in Example 1 of Sec. 1.4.We can write the given ODE as M dx + N dy = 0 where M = 2xy and N = x2. Next we compute ∂M ∂y =2x (where, when taking this partial derivative, we treat x as if it were a constant) and ∂N ∂x =2x (we treat y as if it were a constant). (See Appendix A3.2 for a review of partial derivatives.) This shows that the ODE is exact by (5) of Sec. 1.4. From (6) we obtain by integration u = M dx + k(y) = 2xy dx + k(y) = x2y + k(y). To find k(y) we differentiate this formula with respect to y and use (4b) to obtain ∂u ∂y = x2 + dk dy = N = x2. From this we see that dk dy = 0, k = const. The last equation was obtained by integration. Insert this into the equation for u, compare with (3) of Sec. 1.4, and obtain u=x2y+c∗. Because u is a constant, we have x2y = c, hence y = c/x2. 6 Ordinary Differential Equations (ODEs) Part A 5. Nonexact ODE. From the ODE, we see that P =x2 +y2 and Q=2xy. Taking the partials we have ∂P ∂y =2y and ∂Q ∂x =−2y and, since they are not equal to each other, the ODE is nonexact. Trying Theorem 1, p. 25, we have R = (∂P/∂y − ∂Q/∂x) Q = 2y + 2y −2xy = 4y −2xy = −2 x which is a function of x only so, by (17), we have F(x)= exp R(x) dx. Now R(x) dx = −2 1 x dx = −2 ln x = ln (x−2) so that F(x) = x−2. Then M = FP = 1 + x−2y2 and N = FQ = −2x−1y. Thus ∂M ∂y = 2x−2y = ∂N ∂x . This shows that multiplying by our integrating factor produced an exact ODE.We solve this equation using 4(b), p. 21.We have u = −2x−1y dy = −2x−1 y dy = −x−1y2 + k(x). From this we obtain ∂u ∂x = x−2y2 + dk dx = M = 1 + x−2y2, so that dk dx = 1 and k = dx = x + c∗ . Putting k into the equation for u, we obtain u(x, y) = −x−1y2 + x + c∗ and putting it in the form of (3) u = −x−1y2 + x = c. Solving explicitly for y requires that we multiply both sides of the last equation by x, thereby obtaining (with our constant=−constant on p. A5) −y2 + x2 = cx, y = (x2 − cx)1/2. 9. Initial value problem. In this section we usually obtain an implicit rather than an explicit general solution. The point of this problem is to illustrate that in solving initial value problems, one can proceed directly with the implicit solution rather than first converting it to explicit form. The given ODE is exact because (5) gives My = ∂ ∂y (2e2x cos y) = −2e2x sin y = Nx. From this and (6) we obtain, by integration, u = M dx = 2 e2x cos y dx = e2x cos y + k(y). uy =N now gives uy = −e2x sin y + k(y) = N, k(y) = 0, k(y) = c∗ = const. Chap. 1 First-Order ODEs 7 Hence an implicit general solution is u = e2x cos y = c. To obtain the desired particular solution (the solution of the initial value problem), simply insert x=0 and y=0 into the general solution obtained: e0 cos 0 = 1 · 1 = c. Hence c=1 and the answer is e2x cos y = 1. This implies cos y = e−2x, thus the explicit form y = arccos (e−2x). 15. Exactness.We have M =ax+by, N =kx+ly. The answer follows from the exactness condition (5), p. 21. The calculation is My = b = Nx = k, M = ax + ky, u = M dx = 1 2 ax2 + kxy + κ(y) with κ(y) to be determined from the condition uy = kx + κ (y) = N = kx + ly, hence κ = ly. Integration gives κ = 12 ly2.With this κ, the function u becomes u = 12 ax2 + kxy + 12 ly2 = const. (If we multiply the equation by a factor 2, for beauty, we obtain the answer on p. A5). Sec. 1.5 Linear ODEs. Bernoulli Equation. Population Dynamics Example 3, pp. 30–31. Hormone level. The integral I = eKt cos πt 12 dt can be evaluated by integration by parts, as is shown in calculus, or, more simply, by undetermined coefficients, as follows.We start from eKt cos πt 12 dt = eKt a cos πt 12 + b sin πt 12 with a and b to be determined. Differentiation on both sides and division by eKt gives cos πt 12 = K a cos πt 12 + b sin πt 12 − aπ 12 sin πt 12 + bπ 12 cos πt 12 . We now equate the coefficients of sine and cosine on both sides. The sine terms give 0 = Kb − aπ 12 , hence a = 12K π b. 8 Ordinary Differential Equations (ODEs) Part A The cosine terms give 1 = Ka + π 12 b = 12K2 π + π 12 b = 144K2 + π2 12π b. Hence, b = 12π 144K2 + π2 , a = 144K 144K2 + π2 . From this we see that the integral has the value eKt a cos πt 12 + b sin πt 12 = 12π 144K2 + π2 eKt 12K π cos πt 12 + sin πt 12 . This value times B (a factor we did not carry along) times e−Kt (the factor in front of the integral on p. 31) is the value of the second term of the general solution and of the particular solution in the example. Example 4, pp. 32–33. Logistic equation, Verhulst equation. This ODE y = Ay − By2 = Ay 1 − B A y is a basic population model. In contrast to the Malthus equation y = ky, which for a positive initial population models a population that grows to infinity (if k > 0) or to zero (if k < 0), the logistic equation models growth of small initial populations and decreasing populations of large initial populations. You can see directly from the ODE that the dividing line between the two cases is y = A/B because for this value the derivative y is zero. Problem Set 1.5. Page 34 5. Linear ODE. Multiplying the given ODE (with k = 0) by ekx, you obtain (y + ky)ekx = e−kxeks = e0 = 1. The left-hand side of our equation is equal to (yekx), so that we have (yekx) = 1. Integration on both sides gives the final answer. yekx = x + c, y = (x + c)e−kx. The use of (4), p. 28, is simple, too, namely, p(x) = k, h = p(x) dx = k dx = kx. Furthermore, r = e−kx. This gives y = e−kx ekxe−kxdx + c = e−kx 1 dx + c = e−kx(x + c). Chap. 1 First-Order ODEs 9 9. Initial value problem. For the given ODE y + y sin x = ecos x we have in (4) p(x) = sin x so that by integration h = sin x dx = −cos x Furthermore the right-hand side of the ODE r = ecos x. Evaluating (4) gives us the general solution of the ODE. Thus y = ecos x e−cos x · ecos x dx + c = ecos x(x + c). We turn to the initial condition and substitute it into our general solution and obtain the value for c y(0) = ecos 0(0 + c) = −2.5, c = −2.5 e Together the final solution to the IVP is y = ecos x x − 2.5 e . 23. Bernoulli equation. In this ODE y + xy = xy−1 we have p(x) = x, g(x) = x and a = −1. The new dependent variable is u(x) = [y(x)]1−a = y2. The resulting linear ODE (10) is u + 2xu = 2x. To this ODE we apply (4) with p(x) = 2x, r(x) = 2x hence h = 2x dx = x2, −h = −x2 so that (4) takes the form u = e−x2 ex2(2x) dx + c . In the integrand, we notice that (ex2 ) = (ex2 ) · 2x, so that the equation simplifies to u = e−x2 (ex2 + c) = 1 + ce−x2 . Finally, u(x) = y2 so that y2 = 1 + ce−x2 . From the initial condition [y(0)]2 = 1 + c = 32. It follows that c = 8. The final answer is y = 1 + 8e−x2 . 31. Newton’s law of cooling. Take a look at Example 6 in Sec. 1.3, pp. 15–16. Newton’s law of cooling is given by dT dt = K(T − TA). 10 Ordinary Differential Equations (ODEs) Part A In terms of the given problem, Newton’s law of cooling means that the rate of change of the temperature T of the cake at any time t is proportional to the difference of temperature of the cake and the temperature TA of the room. Example 6 also solves the equation by separation of variables and arrives at T(t) = TA + cekt . At time t = 0, we have T(0) = 300 = 60 + c · e0·k = 60 + c, which gives that c = 240. Insert this into the previous equation with TA = 60 and obtain T(t) = 60 + 240ekt . Ten minutes later is t = 10 and we know that the cake has temperature T(10) = 200 [◦F]. Putting this into the previous equation we have T(10) = 60 + 240e10k = 200, ek = 7 12 1/10 , k = 1 10 ln 7 12 = −0.0539. Now we can find out the time t when the cake has temperature of T(t) = 61◦F.We set up, using the computed value of k from the previous step, 60 + 240e−0.0539t = 61, e−0.0539t = 1 240 , t = −ln (240) −0.0539 = −5.48 −0.0539 = 102 min. Sec. 1.6 Orthogonal Trajectories The method is rather general because one-parameter families of curves can often be represented as general solutions of an ODE of first order. Then replacing y = f (x, y) by ˜y = −1/f (x, ˜y) gives the ODE of the trajectories to be solved because two curves intersect at a right angle if the product of their slopes at the point of intersection equals −1; in the present case, y ˜y = −1. Problem Set 1.6. Page 38 9. Orthogonal trajectories. Bell-shaped curves. Note that the given curves y = ce−x2 are bell-shaped curves centered around the y-axis with the maximum value (0, c) and tangentially approaching the x-axis for increasing |x|. For negative c you get the bell-shaped curves reflected about the x-axis. Sketch some of them. The first step in determining orthogonal trajectories usually is to solve the given representation G(x, y, c) = 0 of a family of curves for the parameter c. In the present case, yex2 = c. Differentiation with respect to x then gives (chain rule!) yex2 + 2xyex2 = 0, y + 2xy = 0. where the second equation results from dividing the first by ex2 . Hence the ODE of the given curves is y = −2xy. Consequently, the trajectories have the ODE ˜y = 1/(2x˜y). Separating variables gives 2˜y d ˜y = dx/x. By integration, 2˜y2/2 = −ln |x| + c1, ˜y2 = −ln |x| + c1. Taking exponents gives e˜y2 = x · c2. Thus, x = ˜ce˜y2 where the last equation was obtained by letting ˜c = 1/c2. These are curves that pass through (˜c, 0) and grow extremely rapidly in the positive x direction for positive ˜c with the x-axis serving as an axis of symmetry. For negative ˜c the curves open sideways in the negative x direction. Sketch some of them for positive and negative ˜c and see for yourself. Chap. 1 First-Order ODEs 11 12. Electric field. To obtain an ODE for the given curves (circles), you must get rid of c. For this, multiply (y − c)2 out. Then a term c2 drops out on both sides and you can solve the resulting equation algebraically for c. The next step then is differentiation of the equation just obtained. 13. Temperature field. The given temperature field consists of upper halfs of ellipses (i.e., they do not drop below the x-axis).We write the given equation as G(x, y, c) = 4x2 + 9y2 − c = 0 y > 0. Implicit differentiation with respect to x, using the chain rule, yields 8x + 18yy = 0 and y = −4x 9y . Using (3) of Sec. 1.6, we get ˜y = − 1 4x/9˜y = 9˜y 4x so that d ˜y dx = 9˜y 4x and d ˜y 1 9˜y = dx 1 4x . Integrating both sides gives 1 9 1 ˜y d ˜y = 1 4 1 x dx and 1 9 ln |˜y| = 1 4 ln |x| + c1. Applying exponentiation on both sides and using (1) of Appendix 3, p. A63, gives the desired result y = x9/4 · ˜c, as on p. A5. The curves all go through the origin, stay above the x-axis, and are symmetric to the y-axis. Sec. 1.7 Existence and Uniqueness of Solutions for Initial Value Problems Since absolute values are always nonnegative, the only solution of |y| + |y| = 0 is y = 0 (y(x) ≡ 0 for all x) and this function cannot satisfy the initial condition y(0) = 1 or any initial condition y(0) = y0 with y0 =0. The next ODE in the text y = 2x has the general solution y = x2 + c (calculus!), so that y(0) = c = 1 for the given initial condition. The third ODE xy = y − 1 is separable, dy y − 1 = dx x . By integration, ln |y − 1| = ln |x| + c1, y − 1 = cx, y = 1 + cx, a general solution which satisfies y(0) = 1 with any c because c drops out when x = 0. This happens only at x = 0.Writing the ODE in standard form, with y as the first term, you see that y − 1 x y = −1 x , showing that the coefficient 1/x of y is infinite at x = 0. 12 Ordinary Differential Equations (ODEs) Part A Theorems 1 and 2, pp. 39–40, concern initial value problems y = f (x, y), y(x) = y0. It is good to remember the two main facts: 1. Continuity of f (x, y) is enough to guarantee the existence of a solution of (1), but is not enough for uniqueness (as is shown in Example 2 on p. 42). 2. Continuity of f and of its partial derivative with respect to y is enough to have uniqueness of the solution of (1), p. 39. Problem Set 1.7. Page 42 1. Linear ODE. In this case the solution is given by the integral formula (4) in Sec. 1.5, which replaces the problem of solving an ODE by the simpler task of evaluating integrals – this is the point of (4). Accordingly, we need only conditions under which the integrals in (4) exist. The continuity of f and r are sufficient in this case. 3. Vertical strip as “rectangle.” In this case, since a is the smaller of the numbers a and b/K and K is constant and b is no longer restricted, the answer |x − x0| < a given on p. A6 follows. Chap. 2 Second-Order Linear ODEs Chapter 2 presents different types of second-order ODEs and the specific techniques on how to solve them. The methods are systematic, but it requires practice to be able to identify with what kind of ODE you are dealing (e.g., a homogeneous ODE with constant coefficient in Sec. 2.2 or an Euler–Cauchy equation in Sec. 2.5, or others) and to recall the solution technique. However, you should know that there are only a few ideas and techniques and they are used repeatedly throughout the chapter. More theoretical sections are interspersed with sections dedicated to modeling applications (e.g., forced oscillations, resonance in Sec. 2.8, electric circuits in Sec. 2.9). The bonus is that, if you understand the methods of solving secondorder linear ODEs, then you will have no problem in solving such ODEs of higher order in Chap. 3. Sec. 2.1 Homogeneous Linear ODEs of Second Order Take a look at pp. 46–47. Here we extend concepts defined in Chap. 1 for first-order ODEs, notably solution and homogeneous and nonhomogeneous, to second-order ODEs. To see this, look into Secs. 1.1 and 1.5 before we continue. We will see in this section that a homogeneous linear ODE is of the form (2) y + p(x)y + q(x)y = 0. An initial value problem for it will consist of two conditions, prescribing an initial value and an initial slope of the solution, both at the same point x0. But, on the other hand, a general solution will now involve two arbitrary constants for which some values can be determined from the two initial conditions. Indeed, a general solution is of the form (5) y = c1y1 + c2y2 where y1 and y2 are such that they cannot be pooled together with just one arbitrary constant remaining. The technical term for this is linear independence.We call y1 and y2 “linearly independent,” meaning that they are not proportional on the interval on which a solution of the initial value problem is sought. Problem Set 2.1. Page 53 As noted in Probs. 1 and 2, there are two cases of reduction of order that we wish to consider: Case A: x does not appear explicitly and Case B: y does not appear explicitly. The most general second-order ODE is of the form F(x, y, y, y) = 0. The method of solution starts the same way in both cases. They can be reduced to first order by setting z = y = dy/dx.With this change of variable and the chain rule, y = dy dx = dy dy · dy dx = dz dy z. The third equality is obtained by noting that, when we substitute y = z into the term dy /dy, we get dy /dy = dz/dy . Furthermore, y = dy/dx = z, so that together y = dy /dy · dy/dx = (dz/dy)z. 3. Reduction to first order. Case B: y does not appear explicitly. The ODE y + y = 0 is Case B, so, from the above, set z = y = dy/dx, to give dz dy z = −z. 14 Ordinary Differential Equations (ODEs) Part A Separation of variables (divide by z) and integrating gives z dz = − dy, thus z = −y + c1. But z = y, so our separation of variables gives us the following linear ODE: y + y = c1. We can solve this ODE by (4) of Sec. 1.5 with p = 1 and r = c1. Then h = p dx = dx = x and y(x) = e−x exc1dx + c2 = e−x(c1ex + c2), which simplifies to the answer given on p. A6 (except for what we called the constants). It is important to remember the trick of reducing the second derivative by setting z = y. 7. Reduction to first order. Case A: x does not appear explicitly. The ODE y + y 3 sin y = 0 is Case A, as explained above. After writing the ODE as y = −y 3 sin y, we reduce it, by setting z = y = dy/dx, to give y = dz dy z = −z3 sin y. Division by −z3 (in the second equation) and separation of variables yields −dz dy 1 z2 = sin y, −dz z2 = sin y dy. Integration gives 1 z = −cos y + c1. Next use z = dy/dx, hence 1/z = dx/dy, and separate again, obtaining dx = (−cos y + c1) dy. By integration, x = −sin y + c1y + c2. This is given on p. A6. The derivation shows that the two arbitrary constants result from the two integrations, which finally gave us the answer. Again, the trick of reducing the second derivative should be remembered. 17. General solution. Initial value problem. Just substitute x3/2 and x−1/2 (x = 0) and see that the two given solutions satisfy the given ODE. (The simple algebraic derivation of such solutions will be shown in Sec. 2.5 starting on p. 71.) They are linearly independent (not proportional) on any interval not containing 0 (where x−1/2 is not defined). Hence y = c1x3/2 + c2 x−1/2 is a general solution of the given ODE. Set x = 1 and use the initial conditions for y and y = 32 c1x1/2 − 12 c2x−3/2, where the equation for y was obtained by differentiation of the general solution. This gives (a) y(1) = c1 + c2 = −3 (b) y(1) = 32 c1 − 12 c2 = 0. We now use elimination to solve the system of two linear equations (a) and (b). Multiplying (b) by 2 and solving gives (c) c2 = 3c1. Substituting (c) in (a) gives c1 + 3c1 = −3 so that (d) c1 = −34 . Substituting (d) in (a) gives c2 = −94 . Hence the solution of the given initial value problem (IVP) is y = −0.75x3/2 − 2.25x−1/2. Chap. 2 Second-Order Linear ODEs 15 Sec. 2.2 Homogeneous Linear ODEs with Constant Coefficients To solve such an ODE (1) y + a y + b y = 0 (a, b constant) amounts to first solving the quadratic equation (3) λ2 + aλ + b = 0 and identifying its roots. From algebra, we know that (3) may have two real roots λ1, λ2, a real double root λ, or complex conjugate roots −1 2a +iω, −12 a −iω with i = √ −1 and ω = b − 14 a2. Then the type of root (Case I, II, or III) determines the general solution to (1). Case I gives (6), Case II gives (7), and Case III gives (9). You may want to use the convenient table “Summary of Cases I–III” on p. 58. In (9) we have oscillations, harmonic if a = 0 and damped (going to zero as x increases) if a > 0. See Fig. 32 on p. 57 of the text. The key in the derivation of (9), p. 57, is the Euler formula (11), p. 58, with t = ωx, that is, eiωx = cos ωx + i sin ωx which we will also need later. Problem Set 2.2. Page 59 13. General solution. Real double root. Problems 1–15 amount to solving a quadratic equation. Observe that (3) and (4) refer to the “standard form,” namely, the case that y has the coefficient 1. Hence we have to divide the given ODE 9y − 30y + 25y = 0, by 9, so that the given ODE in standard form is y − 30 9 y + 25 9 y = 0. The corresponding characteristic equation is λ2 − 30 9 λ + 25 9 = 0. From elementary algebra we know that the roots of any quadratic equation ax2 + bx + c = 0 are x1,2 = −b± √ b2 −4 ac 2a , so that here we have, if we use this formula [instead of (4) where a and b are used in a different way!], λ1,2 = −b + √ b2 − 4ac 2a = 30 9 ± 30 9 2 − 4 · 1 · 25 9 2 = 30 9 ± 900 81 − 900 81 2 = 5 3 . (The reason we used this formula here instead of (4) of Sec. 2.2 is that it is most likely familiar to you, that is, we assume you had to memorize it at some point.) Thus we see that the characteristic equation factors λ2 − 30 9 λ + 25 9 = λ − 53 2 = 0. This shows that it has a real double root (Case II) λ = 53 . Hence, as in Example 3 on p. 56 (or use “Summary of Cases I–III” on p. 58), the ODE has the general solution y = (c1 + c2x)e5x/3. 16 Ordinary Differential Equations (ODEs) Part A 15. Complex roots. The ODE y + 0.54y + (0.0729 + π)y = 0 has the characteristic equation λ2 + 0.54λ + (0.0729 + π) = 0, whose solutions are (noting that √ −4π = √ −1 √ 4π = i · 2 √ π) λ1,2 = −0.54 ± (0.54)2 − 4 · (0.0729 + π) 2 = −0.54 ± √ 0.2916 − 0.2916 − 4π 2 = −0.27 ± i √ π. This gives the real general solution (see Example 5 on p. 57 or Case III in the table “Summary of Cases I–III” on p. 58). y = e−0.27x(A cos( √ πx) + B sin( √ πx)). This represents oscillations with a decreasing amplitude. See Graph in Prob. 29. 29. Initial value problem.We continue by looking at the solution of Prob. 15.We have additional information, that is, two initial conditions y(0) = 0 and y(0) = 1. The first initial condition we can substitute immediately into the general solution and obtain y(0) = e−0.27·0(A cos( √ π0) + B sin( √ π0)) = 1 · A = 0, thus A = 0. The second initial condition concerns y.We thus have to compute y first, from our general solution (with A = 0). Using product and chain rules we have y = −0.27e−0.27x(B sin( √ πx)) + e−0.27x( √ πB(cos( √ πx)). Subsituting x = 0 into the equation for y we get y(0) = 0 + 1 · ( √ π · B) = 1 by the second initial condition. Hence B = 1/ √ π. Together, substituting A = 0 and B = 1/ √ π into the formula for the general solution, gives y = e−0.27x(0 · cos( √ πx) + 1 √ π sin( √ πx)) = 1 √ π e−0.27x sin( √ πx) –5 –4 –3 –2 –1 0 1 2 3 4 5 –2 –1 1 y x Sec. 2.2 Prob. 29. Graph of IVP. Oscillations with a Decreasing Amplitude Chap. 2 Second-Order Linear ODEs 17 31. Linear independence. This follows by noting that ekx and xekx correspond to a basis of a homogenous ODE (1) as given in Case II of table “Summary of Cases I–III.” Being a basis means, by definition of basis, that they are independent. The corresponding ODE is y − 2ky + k2y = 0. [We start with e−ax/2 = ekx so that a = −2k. The double root is λ = −12 a = −1 2 (−2k) = k. This determines our characteristic equation and finally the ODE.] 35. Linear dependence. This follows by noting that sin 2x = 2 sin x cos x. Thus the two functions are linearly dependent. The problem is typical of cases in which some functional relation is used to show linear dependence, as discussed on p. 50 in Sec. 2.1. Sec. 2.3 Differential Operators Problem Set 2.3. Page 61 3. Differential operators. For e2x we obtain (D − 2I )e2x = 2e2x − 2e2x = 0. Since (D − 2I )e2x = 0, applying (D − 2I ) twice to e2x will also be 0, i.e., (D − 2I )2e2x = 0. For xe2x we first have (D − 2I )xe2x = Dxe2x − 2Ixe2x = e2x + 2xe2x − 2xe2x = e2x. Hence (D − 2I )xe2x = e2x. Applying D − 2I again, the right side gives (D − 2I )2xe2x = (D − 2I )e2x = 2e2x − 2e2x = 0. Hence xe2x is a solution only in the case of a real double root (see the table on p. 58, Case II), the ODE being (D − 2I )2y = (D2 − 4D + 4I )y = y − 4y + 4y = 0. For e−2x we obtain (D − 2I )2e−2x = (D2 − 4D + 4I )e−2x = 4e−2x + 8e−2x + 4e−2x = 16e−2x. Alternatively, (D − 2I )e−2x = −2e−2x − 2e−2x = −4e−2x, so that (D − 2I )2e−2x = (D − 2I )(−4e−2x) = 8e−2x + 8e−2x = 16e−2x. 9. Differential operators, general solution. The optional Sec. 2.3 introduces the operator notation and shows how it can be applied to linear ODEs with constant coefficients. The given ODE is (D2 − 4.20D + 4.41I )y = (D − 2.10I )2y = y − 4.20y + 4.41y = 0. From this we conclude that a general solution is y = (c1 + c2x)e2.10x because (D − 2.10I )2((c1 + c2x)e2.10x) = 0. We verify this directly as follows: (D − 2.10I )c1e2.10x = 2.10c1e2.10x − 2.10c1e2.10x = 0 18 Ordinary Differential Equations (ODEs) Part A and (D − 2.10I )c2xe2.10x = c2(D − 2.10I )xe2.10x = c2(e2.10x + 2.10xe2.10x − 2.10xe2.10x) = c2e2.10x, so that (D − 2.10I )2c2xe2.10x = (D − 2.10I )c2e2.10x = c2(2.10e2.10x − 2.10e2.10x) = 0. Sec. 2.4 Modeling of Free Oscillations of a Mass–Spring System Newton’s law and Hooke’s law give the model, namely, the ODE (3), on p. 63, if the damping is negligibly small over the time considered, and (5), on p. 64, if there is damping that cannot be neglected so the model must contain the damping term cy. It is remarkable that the three cases in Sec. 2.2 here correspond to three cases in terms of mechanics; see p. 65. The curves in Cases I and II look similar, but their formulas (7) and (8) are different. Case III includes, as a limiting case, harmonic oscillations (p. 63) in which no damping is present and no energy is taken from the system, so that the motion becomes periodic with the same maximum amplitude C in (4*) at all times. Equation (4*) also shows the phase shift δ. Hence it gives a better impression than the sum (4) of sines and cosines. The justification of (4*), suggested in the text, begins with y(t) = C cos(ω0t − δ) = C(cos ω0t cos δ + sin ω0t sin δ) = C cos δ cos ω0t + C sin δ sin ω0t = A cos ω0t + B sin ω0t. By comparing, we see that A = C cos δ, B = C sin δ, hence A2 + B2 = C2 cos2 δ + C2 sin2 δ = C2 and tan δ = sin δ cos δ = C sin δ C cos δ = B A . Problem Set 2.4. Page 69 5. Springs in parallel. This problem deals with undamped motion and follows the method of Example 1.We have m = 5 [kg]. (i) The spring constant [in Hooke’s law (2)] is k1 = 20 [nt/m]. Thus the desired frequency of vibration is w0 2π = k1/m 2π = √ 20/5 2π [Hz] = 1 π [Hz] = 0.3183 [Hz]. (ii) Here the frequency of vibration is √ 45/5/(2π) [Hz] =3/2π [Hz] =0.4775 [Hz]. (iii) Let K denote the modulus of the springs in parallel. Let F be some force that stretches the combination of springs by an amount s0. Then F = Ks0. Chap. 2 Second-Order Linear ODEs 19 Let k1s0 = F1, k2s0 = F2. Then F = F1 + F2 = (k1 + k2)s0. By comparison, K = k1 + k2 = 20 + 45 = 65 [nt/m] so that the desired frequency of vibrations is √ K/m 2π = √ 65/5 2π = √ 13 2π = 0.5738 [Hz]. 7. Pendulum. In the solution given on p. A7, the second derivative θ is the angular acceleration, hence Lθ is the acceleration and mLθ is the corresponding force. The restoring force is caused by the force of gravity −mg whose tangential component −mg sin θ is the restoring force and whose normal component −mg cos θ has the direction of the rod in Fig. 42, p. 63. Also ω2 0 = g/L is the analog of ω2 0 = k/m in (4) because the models are θ + g L θ = 0 and y + k m y = 0. 13. Initial value problem. The general formula follows from y = (c1 + c2t)e−αt , y = [c2 − α(c1 + c2t)]e−αt by straightforward calculation, solving one equation after the other. First, y(0) = c1 = y0 and then y(0) = c2 − αc1 = c2 − αy0 = v0, c2 = v0 + αy0. Together we obtain the answer given on p. A7. 15. Frequency. The binomial theorem with exponent 12 gives (1 + a)1/2 = 12 0 + 12 1 a + 12 2 a2 +· · · = 1 + 1 2 a + 12 −12 2 a2 +· · · . Applied in (9), it gives ω ∗ = k m − c2 4m2 1/2 = k m 1/2 1 − c2 4mk 1/2 ≈ k m 1/2 1 − c2 8mk . For Example 2, III, it gives ω ∗ = 3(1 − 100/(8 · 10 · 90)) = 2.9583 (exact 2.95833). Sec. 2.5 Euler–Cauchy Equations This is another large class of ODEs that can be solved by algebra, leading to single powers and logarithms, whereas for constant-coefficient ODEs we obtained exponential and trigonometric functions. 20 Ordinary Differential Equations (ODEs [Show Less]