Second-Order Linear ODEs

Chap. 2 Second-Order Linear ODEs Sec. 2.1 Homogeneous Linear ODEs of Second Order On pp. 45-46 we extend concepts defined in Chap. 1, notably solution and homogeneous and nonhomogeneous, to second-order ODEs; take a look into Secs. 1.1 and 1.5 before you continue. You will see in this section that a homogeneous linear ODE is of the form y px y qx 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 x 0. 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 y c 1 y 1 c 2 y 2 where y 1 and y 2 are such that they cannot be pooled together with just one arbitrary constant remaining; expressed technically, y 1 and y 2 are 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 52 5. General solution. Initial value problem. Substitution shows that x 2 and x 2 x 0 are solutions. (The simple algebraic derivation of such solutions will be shown in Sec. 2.5 on p. 70.) They are linearly independent (not proportional) on any interval not containing 0 (where x 2 is not defined). Hence y c 1 x 2 c 2 x 2 is a general solution of the given ODE. Set x 1 and use the initial conditions in y and y 2 c 1 x2c 2 x 3. This gives y1 c 1 c 2 11, y 1 2c 1 c 2 6. The solution is obtained by inspection or elimination, c 1 4, c 2 7. 9. Linear independence. e a x /e a x e 2ax is not constant, unless a 0. Hence y c 1 e a x c 2 e a x with a 0 can be a general solution of an ODE. You may verify that the ODE is y a 2 y 0. A derivation will be given in the next section. 11. Linear dependence. This follows by noting that ln x 2 2 ln x. The problem is typical of cases in which some functional relation is used to show linear dependence. Problem 13 is of the same kind. 21. Reduction to first order. The most general second-order ODE is of the form Fx, y, y, y 0. It can be reduced to first order if [Case (A)] x does not occur explicitly, or if [Case B] y does not occur explicitly. The ODE y y 3 sin y 0 is Case (A). To reduce it, set z y dy/dx and use y as the independent variable. This can be done by using the chain rule, namely, y dy dy dy dx dy dz dx dy z z3 sin y where the last equality follows by using the given ODE. Now divide byz 3 and separate variables to get dz sin y dy. z 2 Integration gives 1 z cos yc 1. Next use z dy/dx, hence 1/z dx/dy, and separate again, obtaining dx cos yc 1 dy. By integration, x sin yc 1 yc 2. This is given on p. A7. The derivation shows that the two arbitrary

Chap. 2 Second-Order Linear ODEs 11 constants result from the two integrations which finally gave us the answer. The trick of reducing the second derivative is worth remembering. 25. Another reduction. Use of initial conditions. The use of the two initial conditions is similar to the use of a single initial conditon in Chap. 1. The only difference is that you also have to calculate the first derivative of the general solution, and that you may have to solve a system of two equations in two unknowns by algebra. Problem 25 concerns the initial value problem y ky, y0 0, y 0 1 (with an unspecified constant of proportionality k) because the solution curve should pass through the origin and have slope 1 at the origin. The reduction to first order is called Case ( B) in Prob. 21 (see just before). It is simpler and more natural than Case (A). Set y z. The reduced problem is z kz, z0 1. Solution z e k x. By integration and use of the first initial condition, y z dx 1 k ekx c, y0 1 k c 0, c 1 k, y 1 k ekx 1. [Note that here we solved the two equations for the arbitrary constants individually, not as a system.] Sec. 2.2 Homogeneous Linear ODEs with Constant Coefficients Solving such an ODE amounts to solving the quadratic equation y a y b y 0 (a, b constant) (1) 2 ab 0. (3) From algebra you know that (3) may have two real roots 1, 2 or a real double rootor complex conjugate roots 1 ai, 1 ai with i 1 and b 1 2 2 4 a2. The corresponding general solution of (1) in these three cases are (6), (7), and (9). In (9) we have oscillations, harmonic if a 0 and damped (going to zero as x increases) if a 0. See Fig. 31 on p. 57. The key in the derivation of (9) is the Euler formula (11) with t x, that is, which you will also need later. Problem Set 2.2. Page 59 e ix cos xisin x 3. General solution. Problems 1-14 amount to solving quadratic equations. Observe that (3) and (4) refer to the standard form, namely, the case that y has the coefficient 1. Hence in Prob. 3 you have y 5y 25 y 0, so that the characteristic equation 4 2 5 25 4 5 2 has the double root 5/2. Hence the ODE has the general solution y c 1 c 2 x e 5x/2 given on p. A7. 7. Complex roots. The ODE y y 2.5y 0 has the characteristic equation 2 2.5 0, whose solutions are 2 0 1 2 1 4 5 2 1 2 1 2 9 1 2 This gives the real general solution 3 i.

12 Ordinary Differential Equations (ODEs) Part A y e x/2 A cos 3 2 xbsin 3 2 x. This represents oscillations with an increasing amplitude. An oscillation with decreasing amplitude is obtained in Prob. 23. 35. Suitable abbreviated notations. The point of the problem is to show that in complicated calculations it is essential to choose suitable notations. This is a general rule. In the present case the suggestions given on p. A7 look in full detail as follows. y 1 E c, E e ax/2, E a 2 E, E a2 4 E s sin x, c cos x, c s, c 2 c y 1 E ce c a 2 c s E y 1 E c2 E c E c a 2 4 c2 a 2 s 2 c E. Insert these expressions into the ODE, drop the common factor E (the exponential function). This gives a 2 4 cas2 c a2 2 cas bc 0. The sine terms cancel. Also the sum of the cosine terms is zero, as you can see by remembering that 2 b 1 4 a2, namely, 1 4 a2 b 1 4 a2 1 2 a2 b 0. Sec. 2.3 Differential Operators Problem Set 2.3. Page 61 1. Differential operators. For e x you obtain D Ie x e x e x 0. For xe x you first have D I xe x D xe x xe x e x xe x xe x e x. HenceD I xe x e x. Applying D I again gives D I 2 xe x D I e x e x e x 0. Hence xe x is a solution only in the case of a double root, the ODE being For sin x you obtain Alternatively, y 2 y y D 2 2 D I y D I 2 y 0. D I 2 sin x D 2 2D I sin x sin x2 cos xsin x 2cos x. D I 2 sin x D Icos xsin x sin xcos xcos xsin x 2 cos x. 11. 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

Chap. 2 Second-Order Linear ODEs 13 D 2 4.1 D 3.1 Iy D ID 3.1I y y 4.1y 3.1y 0. From the two factors we conclude that a general solution is and because and y c 1 e 3.1 x c 2 e x D ID 3.1Ic 1 e 3.1 x c 2 e x 0 D Ie x e x e x 0 D 3.1 Ie 3.1 x 3.1e 3.1x 3.1e 3.1x 0. Sec. 2.4 Modeling: Free Oscillations (Mass-Spring System) Newton s law and Hooke s law give the model, namely, the ODE (3) if the damping is negligibly small over the time considered, and (5) if there is damping that cannot be neglected and the model contains 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. Eq. (4*) also shows the phase shift. Hence it gives a better impression than the sum (4) of sines and cosines. The derivation of (4*) suggested in the text begins with yt C cos 0 t Ccos 0 t cossin 0 t sin C coscos 0 tcsin sin 0 t A cos 0 tbsin 0 t. By comparing you see that A C cos, B C sin, hence and Problem Set 2.4. Page 68 A 2 B 2 C 2 cos 2 C 2 sin 2 C 2 tan sin cos C sin C cos B A. 3. Pendulum. In the solution given on p. A8, the second derivative is the angular acceleration, hence L is the acceleration and m L is the corresponding force. The restoring force is caused by the force of gravitymg whose tangential componentm g sin is the restoring force and whose normal component m g coshas the direction of the rod in Fig. 41. Also 02 g/l is the analog of 02 k/m in (4) because the models are g L 0 and y k m y 0. 9. Frequency. The binomial theorem with exponent 1/2 gives 1a 1/2 1/2 1/2 a 0 1 Applied in (9), it gives 1 1 2 a 1/21/2 2 1/2 2 a 2. a 2

14 Ordinary Differential Equations (ODEs) Part A k m c2 4 m 2 1/2 k m 1/2 1 c2 4 m k 1/2 k m 1/2 1 c2 8 m k For Example 2, III, it gives 31100/8 10 90 2.9583 (exact 2.95833.. 15. Initial value problem. The general formula follows from y c 1 c 2 t e t, y c 2 c 1 c 2 te t by straightforward calculation, solving one equation after the other. First, y0 c 1 y 0 and then y 0 c 2 c 1 c 2 y 0 v 0, c 2 v 0 y 0. Together you obtain the answer given on p. A8. 17. Overdamping. c 1 e t c 2 e t 0 gives c 1 /c 2 e t e 2t. Since the exponential function is nonnegative and you have a minus sign on the right, you see that c 1 /c 2 must be negative to have a solution. 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 you obtained exponential and trigonometric functions. Three cases appear, as for those other ODEs, and Fig. 47 gives an idea of what kind of solution you can expect. In some cases x 0 must be excluded (when you have a power with a negative exponent), and in other cases the solutions are restricted to positive values for the independent variable x; this happens when a logarithm or a root appears (see Example 1). Note further that the auxiliary equation for determining exponents m in y x m is mm 1 a m b 0, thus m 2 a1 m b 0, with a 1 as the coefficient of the linear term; here the ODE is written x 2 y axy b y 0, (1) which is no longer in the standard form with y as the first term. Whereas constant-coefficient ODEs are basic in mechanics and electricity, Euler-Cauchy equations are less important. A typical application is shown on p. 72. Problem Set 2.5. Page 72 1. General solution. Problems 1-10 are solved as explained in the text by determining the roots of the auxiliary equation (3). For the ODE x 2 y 6y 0 you obtain mm 1 6 m 2 m 6 m 3m 2 0 and from this the general solution y c 1 x 3 c 2 x 2 valid for all x 0. 3. Double root (Case II). The ODE x 2 y 7 xy 16y 0 has the auxiliary equation mm 1 7 m 16 m 2 8 m 16 m 4 2 0. According to (7), a general solution for positive x is y c 1 c 2 ln x x 4. 5. Complex roots. The ODE x 2 y xy 2 y 0 has the auxiliary equation m 2 2 m 2 m 1 im 1i 0. A basis of complex solutions is x 1i, x 1i. From it you obtain real solutions by a trick that introduces exponential functions, namely, by first writing (Euler s formula!)

Chap. 2 Second-Order Linear ODEs 15 x 1i x 1 x i xe i ln x xcosln x i sin ln x and then taking linear combinations to obtain a real basis of solutions x cosln x, and x sin ln x for positive x or writing ln x as in the solution on p. A8 if you want to admit all x 0. 13. Initial value problem. Initial values cannot be prescribed at x 0 because the coefficients of an Euler-Cauchy equation in standard form y a/x y b/x 2 y 0 are infinite at x 0. Choosing x 1 makes the problem simpler than other values would do because ln 1 0. The given ODE has the auxiliary equation x 2 y 2 xy 100.25y 0 m 2 m 100.25 m 0.510im 0.510i 0. Solutions are0.5 10i. A general solution for positive x, involving the corresponding real solutions, is y x 0.5 A cos10 ln x B sin 10 ln x. You need the derivative (chain rule!) y 0.5x 1.5 A cos10 ln x B sin10 ln x x 0.5 A sin 10 ln x B cos10 ln x 10 x. Hence y1 A 2 since the cosine gives 1 and the sine gives 0. Similarly, with A 2 inserted, y 1 10010B 11, hence B 1. This yields the answer given on p. A8. Sec. 2.6 Existence and Uniqueness of Solutions. Wronskian The Wronskian Wy 1, y 2 of two solutions y 1 and y 2 of an ODE is defined by (6), Sec. 2.6. It is conveniently written as a second-order determinant (but this is not essential for using it; you need not be familiar with determinants here). It serves for checking linear independence or dependence, which is important in obtaining bases of solutions. The latter are needed, for instance, in connection with initial value problems, where a single solution will generally not be sufficient for satisfying two given initial conditions. Of course, two functions are linearly independent if and only if their quotient is not constant. To check this, you would not need Wronskians, but we discuss them here in the simple case of second-order ODEs as a preparation for Chapter 3 on higher order ODEs, where Wronskians will show their power and will be very useful. Problem Set 2.6. Page 77 1. Wronskian. The solutions form a basis for an ODE with constant coefficients, which is obtained from the characteristic equation 0.50.5 2 0.25 0, namely, y 0.25y 0. The Wronskian is We 0.5 x, e 0.5 x e 0.5x e 0.5 x 0.5e 0.5x 0.5e 0.5 x e 0.5x 0.5e 0.5x e 0.5x 0.5e 0.5 x 0.50.5 1. If a lengthy calculation gives a simple result, you may suspect that there may be a simpler way. In the present case, you can verify that, by the quotient rule, W y 2 y 1 y 1 2 y 2 y 1 y 2 y 1 y 1 2 y 1 2 y 2 y 1 y 2 y 1. (A)

16 Ordinary Differential Equations (ODEs) Part A For the present problem, y 2 /y 1 e 0.5x e 0.5 x e x and e x e x. Multiplication by y 1 2 e x gives1. 3. Wronskian. This is the case of a double root. The characteristic equation is 2 2 kk 2 k 2 0. The ODE is y 2 ky k 2 y 0. The Wronskian can be obtained by the formula (A) above or by direct calculation. By that formula, y 2 /y 1 xe k x /e k x x, hence y 2 /y 1 x 1, W 1 y 1 2 e k x 2 e 2kx. The direct calculation by (6) is more complicated. Sincee k x ke k x and xe k x e k x kxe k x, you obtain W e k x xe k x xe k x e k x e 2 k x 1 kx kx e 2k x. 9. Euler-Cauchy equation. This concerns an Euler-Cauchy equation with characteristic equation m 1.5m 0.5 m 2 m 0.75 mm 1 0.75 0. Hence the ODE is x 2 y 0.75 y 0. The Wronskian is W x 1.5 0.5x 1.5 x 0.5 1.5x 0.5 0.51.5 2. The formula in Prob. 3 gives y 2 /y 1 x 2, y 2 /y 1 2 x 3, so that multiplication by y 12 x 3 gives the previous result W 2. 13. Damped oscillations. You see from the given solutions that the roots of the characteristic equation are complex conjugates, namely, 1 10.8i and 2 10.8i. Hence, using the familiar formula abab a 2 b 2, you obtain the characteristic equation 1 2 10.8i10.8i 1 2 0.8 2 2 21.64 0. The Wronskian is obtained from (6), or you can calculate it more simply from the general formula in Prob. 3 (see the solution above), namely, and y 1 e x cos 0.8x, y 2 e x sin 0.8 x, y 2 /y 1 tan 0.8 x, y 2 /y 1 1/cos 2 0.8x 0.8 so that multiplication by y 12 (see the formula) gives W 0.8e 2x. 17. Double root. From the form of the given functions you see that they are solutions of a linear ODE with constant coefficients, and the x in the second solution shows you that it is the case of a double root of the characteristic equation, the root being 3.8. Hence the characteristic equation is 3.8 2 2 7.6 14.44 2 0. Hence the ODE is y 7.6 y 14.44 2 y 0. For the Wronskian W you need y 1 e 3.8x, y 2 xe 3.8x and the derivatives Hence the Wronskian is y 1 3.8e 3.8x, y 2 1 x3.8e 3.8x.

Chap. 2 Second-Order Linear ODEs 17 W y 1 y 2 y 2 y 1 e 3.8x 1 3.8x e 3.8x xe 3.8x 3.8 e 3.8x e 7.6x 1 3.8xx 3.8 e 7.6x. Much more simply you obtain by the general formula in Prob. 1 W y 2 y 1 y 12 x y 12 1 e 3.8x 2 e 7.6x. The simplification is due to the fact that the quotient of the two solutions is x and its derivative is 1. Can you see that this is typical of all ODEs with constant coefficients that have a double root of the characteristic equation? Can you see that also the cases of two real roots and of complex roots must lead to substantial simplifications? In the last formula you used thate a 2 e 2a, where a 3.8x. Sec. 2.7 Nonhomogeneous ODEs This section and problem set concern nonhomogeneous linear ODEs y px y qx y rx (1) where rx is not identically zerorx 0. The new task is the determination of a particular solution y of (1). For this you can use the method of undetermined coefficients. Because of the Modification Rule it is necessary to first determine a general solution of the homogeneous ODE since the form of differs depending on whether or not the function (or a term of it) on the right side of the ODE is a solution of the homogeneous ODE. If you forget to take this into account, you will not be able to determine the coefficients; in this sense the method will warn you that you made a mistake. Problem Set 2.7. Page 83 1. General solution. The characteristic equation of the homogeneous ODE is 2 32 12 0. You see that it has the solutions1 and 2. Hence a general solution of the homogeneous ODE y 3y 2y 0 is y h c 1 e x c 2 e 2x. The function 30 e 2x on the right is not a solution of the homogeneous ODE. Hence you do not apply the Modification Rule. Table 2.1 requires that you start from C e 2x. Two differentiations give 2 C e 2x and 4 C e 2x. Substitution of y and its derivatives into the given ODE yields 4 C e 2x 3 2 C e 2x 2Ce 2x 12Ce 2x 30e 2x. Hence 12 C 30, C 30/12 2.5. This gives the answer (a general solution of the given ODE; see page A8) y c 1 e x c 2 e 2x 2.5e 2x. 11. Modification rule. The homogeneous ODE y 1.44y 0 has the characteristic equation 2 1.44 0. Its roots are 1.2i and 1.2i. Hence a real general solution of the homogeneous ODE is y h c 1 cos 1.2xc 2 sin 1.2x. You now see that the right side of the nonhomogeneous ODE 24 cos 1.2x is a solution of the homogeneous ODE. Hence you need the Modification Rule for a simple root. That is, you multiply your choice function K cos 1.2xMsin 1.2 x in Table 2.1 by x, obtaining

18 Ordinary Differential Equations (ODEs) Part A xk cos 1.2xMsin 1.2 x. By differentiation, K cos 1.2xMsin 1.2 xx1.2k sin 1.2 x1.2m cos 1.2x 21.2 K sin 1.2 x1.2m cos 1.2x x1.44k cos 1.2x1.44Msin 1.2 x. You now substitute and into y 1.44y 24 cos 1.2x. Then 1.44 and x in cancel and you are left with 21.2 K sin 1.2 x1.2m cos 1.2x 24 cos 1.2x. Equating the cosine terms gives 2.4M 24, hence M 10. There is no sine term on the right. Hence K must be zero. You thus obtain the answer given on p. A8 y c 1 cos 1.2xc 2 sin 1.2 x10x sin 1.2 x. 19. Initial value problem. The homogeneous ODE y y 12y 0 has the characteristic equation 2 12 0. The usual formula for the solutions of a quadratic equation gives 1 2 1 4 12 1 2 49 1 4 2 7, thus 3 and 4. 2 Hence a general solution of the homogeneous ODE is y h c 1 e 3x c 2 e 4x. You now see that the terms on the right side of the given ODE are not solutions of the homogeneous ODE. Hence you do not apply the Modification Rule, but simply start from the choice function Its derivatives are Substitution into the given ODE yields K 3 x 3 K 2 x 2 K 1 xk 0. 3 K 3 x 2 2 K 2 xk 1 6 K 3 x2 K 2. 6 K 3 x2 K 2 3 K 3 x 2 2 K 2 xk 1 12K 3 x 3 K 2 x 2 K 1 xk 0 144x 3 12.5. Now compare powers of x on both sides. The x 3 -terms give12k 3 144, so that K 3 12. Furthermore, the x 2 -terms give3 K 3 12K 2 0 since there is no x 2 -term on the right. That is, 312 12 K 2 0 and K 2 3. The x-terms give 6 K 3 2 K 2 12K 1 0 since there is no x-term on the right. Thus, 612 2 3 12K 1 and K 1 6.5. Finally, from the constant terms on both sides you obtain 2K 2 K 1 12 K 0 12.5, thus 6 6.512K 0 12.5, K 0 0. You have obtained a general solution of the given ODE, y y h c 1 e 3x c 2 e 4x 12x 3 3 x 2 6.5x0. Only now can you consider the initial conditions. (Why not earlier?) The first condition is y0 5 and gives y0 c 1 c 2 5, hence c 2 5c 1. For the second initial condition y 0 0.5 you need the derivative and its value y 3 c 1 e 3 x 4 c 2 e 4x 36x 2 6x6.5 y 0 3 c 1 4 c 2 6.5 3 c 1 45 c 1 6.5 7 c 1 13.5. This must equal0.5. Hence c 1 2 and thus c 2 5 c 1 3. With the values of the constants thus

Chap. 2 Second-Order Linear ODEs 19 determined you obtain the answer of the initial value problem given on p. A9, y 2 e 3x 3 e 4x 12x 3 3 x 2 6.5x. Sec. 2.8 Modeling: Forced Oscillations. Resonance In the solution a, b of (4) in the formula before (5) (the formula without number) the denominator is the coefficient determinant; furthermore, for a the numerator is the determinant Similarly for b, by Cramer s rule or by elimination. Problem Set 2.8. Page 90 F 0 c F 0km 2. 0 km 2 1. Steady-state solution. Because of the function r 130 cos 3t you have to choose By differentiation, K cos 3tMsin 3t. 3 K sin 3 t3 M cos 3t 9 K cos 3t9 M sin 3 t. Substitute this into the ODE y 6y 8y 130 cos 3t. To get a simple formula, use the abbreviations C cos 3t and S sin 3t. Then 9 K C 9 MS 63 K S 3 MC 8K C MS 130C. Collect the C-terms and equate their sum to 130. Collect the S-terms and equate their sum to 0 because there is no sine term on the right side of the ODE. You obtain 9 K 18M8 K K 18M 130 9 M18K 8 M 18K M 0. From the second equation, M 18 K. Then from the first equation, K 18 18K 130, K 130/325 0.4. Hence M 18K 7.2. The steady-state solution is (cf. p. A9) y 0.4 cos 3t7.2 sin 3 t. 9. Transient solution. The homogeneous ODE y 2 y 0.75y 0 has the characteristic equation 2 20.75 1 2 3 0. 2 Hence the roots are1/2 and 3/2. A general solution of the homogeneous ODE is y h c 1 e t/2 c 2 e 3t/2. To obtain a general solution of the given ODE you need a particular solution. According to the method of undetermined coefficients (Sec. 2.7) set Differentiate to get K cos tmsin t. K sin tmcos t K cos tmsin t. Substitute this into the given ODE. Abbreviate C cos t, S sin t. You obtain K C MS 2K S MC 0.75K C MS 13S. Collect the C-terms and equate their sum to 0 (why?)

20 Ordinary Differential Equations (ODEs) Part A K 2 M0.75 K 0, thus 1 K 2M 0, K 8M. 4 Collect the S-terms and equate their sum to 13: M2 K 0.75M 13. Since K 8 M, conclude that this gives M16 M0.75 M 1 65 4 M64M3 M 4 4 M 13, M 4 5. Hence K 8M 32/5. This confirms the answer on p. A9 that the transient solution is y y h c 1 e t/2 c 2 e 3 t/2 32 5 cos t 4 5 sin t. 17. Initial value problem. The characteristic equation 2 68 42 0 of the homogeneous ODE y 6 y 8y 0 has the roots4 and 2, so that a general solution of the homogeneous ODE is y h c 1 e 4 t c 2 e 2 t. You also need a particular solution of the given ODE. Using the method of undetermined coefficients, set Differentiate to get K cos 2tMsin 2t. 2 K sin 2 t2 M cos 2t 4 K cos 2t4 M sin 2 t. Substitute this into the given ODE, abbreviating C cos 2t, S sin 2t : 4 K C 4 MS 62 K S 2 MC 8K C MS 4 S. Collect the cosine terms and equate the sum of their coefficients to 0: 4 K 12 M8 K 4 K 12M 0, thus K 3 M. Collect the sine terms and equate the sum of their coefficients to 4 because the right side of the ODE is 4 sin 2t 4 S: 4 M12 K 8 M 4 M36M8 M 40M 4, M 0.1. Now K 3 M (see before), so that K 0.3. You thus have the general solution of the given ODE yt c 1 e 4t c 2 e 2t 0.3 cos 2t0.1 sin 2 t. You can now determine c 1 and c 2 from the initial conditions y0 0.7, y 0 11.8. The first condition gives (note that cos 0 1, sin 0 0) y0 c 1 c 2 0.3 0.7, thus c 1 c 2 1. For the second condition you need the derivative Thus, y t 4c 1 e 4t 2 c 2 e 2t 0.6 sin 2 t0.2 cos 2t. y 0 4c 1 2 c 2 00.2 11.8, hence 4 c 1 2 c 2 12. Since c 1 c 2 1, you have c 2 1c 1, so that 4c 1 2 c 2 4c 1 21c 1 2 c 1 2 12, c 1 5, c 2 1 c 1 4. You thus obtain the answer (cf. p. A9) yt 5 e 4 t 4e 2t 0.3 cos 2t0.1 sin 2 t.

Chap. 2 Second-Order Linear ODEs 21 Sec. 2.9 Modeling: Electric Circuits Problem Set 2.9. Page 97 5. LC-circuit. Modeling of the circuit is the same as for the RLC-circuit. Thus, By differentiation, L I Q/C Et. L I I/C E t. Here L 0.2, 1/C 20, Et sin t, E t cos t, so that 0.2 I 20I cos t, thus I 100I 5 cos t. The characteristic equation 2 100 0 has the roots10i, so that a real solution of the homogeneous ODE is I h c 1 cos 10tc 2 sin 10 t. Now determine a particular solution I p of the nonhomogeneous ODE by the method of undetermined coefficients in Sec. 2.7, starting from Substitute this and the derivatives I p K cos tmsin t. I p K sin tmcos t I p K cos tmsin t into the nonhomogeneous ODE I 100 I 5 cos t, obtaining K cos tmsin t 100K cos tmsin t 5 cos t. Equating the cosine terms on both sides, you get The sine terms give K 100 K 5, hence K 5/99. M100M 0, hence M 0. You thus have the general solution of the nonhomogeneous ODE I I h I p c 1 cos 10tc 2 sin 10 t 5 99 Now use the initial conditions I0 0 and I 0 0. You obtain cos t. Differentiation gives I0 c 1 5 99 0, thus c 1 5 99. I t 10 c 1 sin 10 t10c 2 cos 10t 5 sin t. 99 Since sin 0 0, this gives I 0 10 c 2 0, so that c 2 0. You thus obtain the answer (cf. p. A9) It 5 cos tcos 10t. 99 13. Transient current. You must find a general solution of the nonhomogeneous ODE. Since R 6, L 0.2, 1/C 40, E 110 sin 10t, E 1100 cos 10t, this ODE is 0.2 I 6 I 40I E 1100 cos 10t. Multiply by 5 to get the standard form with I as the first term, From the characteristic equation I 30I 200I 5500 cos 10t. 2 30200 2010 0

22 Ordinary Differential Equations (ODEs) Part A you see that a general solution of the homogeneous ODE is I h c 1 e 20 t c 2 e 10 t. This will go to zero as t increases, regardless of initial conditions (which are not given in this problem). You also need a particular solution I p of the nonhomogeneous ODE; this will be the steady-state solution. You obtain it by the method of undetermined coefficients, starting from By differentiation you have I p K cos 10tMsin 10 t. I p 10K sin 10 t10m cos 10t I p 100K cos 10t100M sin 10 t. Substitute all this into the nonhomogeneous ODE in standard form, abbreviating C cos 10t, S sin 10t. You obtain 100 KC 100 MS 3010KS 10MC 200KC MS 5500C. Equate the sum of the S-terms to zero, 100 M300 K 200M 0, 100M 300K, M 3 K. Equate the sum of the C-terms to 5500 (the right side), 100 K 300 M200 K 5500, 100K 300 3 K 1000K 5500. You see that K 5.5, M 3K 16.5, and have the solution (cf. p. A9) I I h I p c 1 e 20t c 2 e 10t 5.5 cos 10t16.5 sin 10 t. Sec. 2.10 Solution by Variation of Parameters This method is general, in contrast to the method of undetermined coefficients, which is restricted to constant-coefficient ODEs with special right sides. Nevertheless, the method of undetermined coefficients is more important to the engineer and physicist because it can take care of cases of the usual periodic driving forces (electromotive forces). The present method reduces the problem of solving a linear ODE to that of the evaluation of two integrals. Hence it is an extension of the solution method for linear first-order ODEs in Sec. 1.5. Problem Set 2.10. Page 101 3. General solution. The solution formula (2) was obtained for the standard form of an ODE. In the present problem, divide the given nonhomogeneous Euler-Cauchy equation x 2 y 2 xy 2 y x 3 cos x by x 2 in order to see what rx in (2) is. You obtain y 2 x y 2 y x cos x. 2 x The auxiliary equation for the homogeneous ODE is needed to determine y 1 and y 2. It is mm 1 2 m 2 m 2 3 m 2 m 2m 1 0. The roots are 1 and 2. Hence you have the basis of solutions y 1 x, y 2 x 2. You also need the corresponding Wronskian W x x 2 1 2 x 2x 2 x 2 x 2. Now use (2) (and integrate barts in the first integral), obtaining

Chap. 2 Second-Order Linear ODEs 23 y x x2 x cos x x 2 dxx 2 x x cos x x 2 x x cos x dxx 2 cos x dx dx xcos xxsin x x 2 sin x x cos x. Using this particular solution, you obtain the general solution given on p. A9 y c 1 xc 2 x 2 x cos x. You would obtain this solution directly if you added constants of integration when you evaluated the integrals. 5. General solution. The solution on p. A9 shows that undetermined coefficients would not help. It also shows that the basis of solutions y 1 cos x, y 2 sin x has been used. The corresponding Wronskian is simply W cos x Hence you have to evaluate the terms in 2 and sin x sin x cos x cos 2 xsin 2 x 1. cos xsin x tan x dx cos xsin xln secxtan x c 1 sin xcos x tan x dx sin x sin x dx sin xcos xc 2. You see that cos x sin x appears twice, with opposite sign; hence these terms cancel and you are left with the general solution y c 1 cos xc 2 sin xcos xsin xln sec xtan x sin x cos x. (We denoted one of the arbitrary constants byc 1 to havec 1 in the final solution.) 11. Choice of method. y 4 y cosh 2 x can be solved more easily by undetermined coefficients because the integrals occurring in the variation of parameters are somewhat complicated. Derive that For start from and use y h c 1 cos 2xc 2 sin 2x. K cosh 2xMsinh 2 x 4K cosh 2x4M sinh 2 x. Substitute this into the ODE to get 4K cosh 2 x4m sinh 2 x4k cosh 2xMsinh 2 x 8K cosh 2x8M sinh 2 x cosh 2x. Compare both sides to get K 1/8 and M 0. You thus have the general solution y c 1 cos 2xc 2 sin 2 x 1 8 cosh 2x (not 1 8 x cosh 2 x; misprint in the first printing). 17. General solution. The homogeneous ODE is x 2 y xy x 2 1 y 0. 4

24 Ordinary Differential Equations (ODEs) Part A Use the hint y u x 1/2 and differentiate: y u x 1/2 1 2 ux3/2 Substitute this into the homogeneous ODE to get y u x 1/2 u x 3/2 3 4 ux5/2. u x 3/2 u x 1/2 3 4 u x1/2 u x 1/2 1 2 u x1/2 x 2 1 4 ux1/2 0. The second term cancels the fourth term. The third term plus the fifth term cancel - 1 4 u x1/2. Divide the remaining ODE by x 3/2, obtaining u u 0 and the general solution u c 1 cos xc 2 sin x, so that y h ux 1/2 c 1 cos xc 2 sin x/ x. Now obtain a particular solution of the nonhomogeneous equation. For this you need You also need y 1 x 1/2 cos x, y 2 x 1/2 sin x. r x 1/2 sin x obtained by dividing the given ODE by x 2, to get the standard form. Calculate the Wronskian W Thus, by the solution formula (2), x 1/2 cos x 1 2 x3/2 cos xx 1/2 sin x x 1/2 sin x 1 2 x3/2 sin xx 1/2 cos x x 1/2 cos x x1/2 sin xx 1/2 sin x 1/x x 1/2 sin x x1/2 cos xx 1/2 sin x 1/x x 1/2 cos xsin 2 x dx x 1/2 sin x cos x sin x dx dx dx 1 x. x 1/2 cos x 1 2 sin x cos x 1 2 x x 1/2 sin x 1 2 x1/2 cos 2 x 1 2 x1/2 cos x because the terms 1 2 x1/2 cos 2 x sin x cancel each other. (Drop the second term in in the answer on p. A9 erroneously given in the first printing.)