Study Guide of Applied Differential Equations

2 User Manual: (I). Make sure you can do all the 19 Problems in this Study Guide by yourself. (II). Under every important topic in each section, I list the Examples and Exercises from the textbook. Read the step-by-step solutions in these Examples provided by the book; then choose an Exercise in the list to practice. (Most of the Exercises listed in this Study Guide have been assigned as homework.) You may choose the odd-numbered exercises so you can compare your solution with the answer key in the back of the book. Remark. You are to solve a wide variety of differential equations. Always identify the type of the DEi and be clear about the method you use. Main types of DEs are given in the following, in which each DE can be incorporated with initial value problems. • Chapter 2. Autonomous, separable, and linear first order DEs. • Chapter 3. Linear models by first order linear DEs (e.g. mixture of salt solutions). • Chapter 4. Homogeneous and nonhomogeneous linear second order DEs. • Chapter 5. Linear models by second order linear DEs (e.g. spring/mass systems). • Chapter 7. Initial value problems by the Laplace transform. • Chapter 8. Systems of two linear first order DEs in the matrix form. Chapter 1. Introduction to Differential Equations Important topics in this chapter: (i). Order and linear vs nonlinear of a DE. (ii). Verify (explicit, implicit, or integral-defined) solutions of DEs. (iii). Find interval of solutions of DEs. 1.1. Definitions and Terminology. Give a DE. • Determine its order and whether it is linear or nonlinear. (Example 4 and Exercises 1-8) • Verify an explicit solution. (Example 5 and Exercises 11-18) • Verify an implicit solution. (Example 7 and Exercises 19-20) • Verify an integral-defined solution. (Exercises 25-28) 1.2. Initial Value Problems. Give an initial value problem (IVP) dy = f (x, y), y(x0 ) = y0 . dx • Determine the interval of definition of a solution. (Example 2 and Exercises 3-6) • Determine the region where the IVP has a unique solution:   ∂f 2 (x, y) ∈ R : f and are continuous ∂y (Examples 2, 4-5 and Exercises 3-6, 17-28) Problem 1 (Exercise 24 in §1.2). Find all (x0 , y0 ) such that the initial value problem (y − x)y 0 = y + x, y(x0 ) = y0 has a unique solution on a region with x near x0 . Justify your answer. Do not solve. Answer. The DE can be rewritten as dy y+x = . dx y−x 3 Inspection of the functions y+x ∂f 2x and =− y−x ∂y (y − x)2 shows that they are continuous when y − x 6= 0. So through a point (x0 , y0 ) in the regions f (x, y) = {y > x} or {y < x} the differential equation has a unique solution. Chapter 2. First Order Differential Equations The important topics in this chapter: (i). Autonomous DEs: dy = f (y). dx Find the critical points; sketch the general solutions; classify the critical points. (ii). Separable DEs: dy = g(x)h(y). dx Solve the DE by separating independent and dependent variables. (iii). Linear DEs: dy + p(x)y = f (x). dx R Solve the DE by inserting the integrating factor e P (x) . (iv). Use substitution to solve some DEs. Remark (Review of integrals). 1. Integrate polynomials,Rrational functions, exponential functions, logarithmic functions, and trigonoR 2 metric functions, e.g. 1/x = ln |x| + c and 1/(x + 1) = arctan x + c. 2. Use substitution to integrate. 3. Use integration by parts to integrate. 4. Use identities to simplify trigonometric functions, e.g. sin2 x + cos2 x = 1, sin(2x) = 2 sin x cos x, cos(2x) = 2 cos2 x − 1 = 1 − 2 sin2 x. 5. Use partial fractions to simplify rational functions, e.g. 1 1 1 1 = = + . 2 P −P P (1 − P ) P 1−P 2.1. Solution Curves Without a Solution. Give a DE dy = f (x, y). dx • Use the direction field to sketch the solution curves of the DE. (Examples 1-2 and Exercises 1-4) • Determine whether the DE is autonomous, that is, an autonomous DE has the form dy = f (y). dx (1). Find all the critical points by solving f (y) = 0. (2). Sketch the general solutions, i.e. draw the phase portrait. (3). Classify the critical points as asymptotically stable attractors, semi-stable critical points, or unstable repellers. 4 (Examples 3-5 and Exercises 21-30, 38-40) Problem 2 (Exercise 24 in §2.1). Find the critical points and phase portrait of dy = 10 + 3y − y 2 . dx Classify each critical point as asymptotically stable, unstable, or semi-stable. By hand, sketch typical solutions curves in the regions in the xy-plane determined by the graphs of the equilibrium solutions. Answer. Let f (y) = 10 + 3y − y 2 = (5 − y)(2 + y). The critical points are y = 5 and y = −2. In the region above the equilibrium solution y = 5, f (y) < 0 so y 0 (x) = f (y) < 0 and the function y(x) is decreasing. In the region below the equilibrium solution y = 5 and above the equilibrium solution y = −2, f (y) > 0 so y 0 (x) = f (y) > 0 and the function y(x) is increasing. In the region below the equilibrium solution y = −2, f (y) < 0 so y 0 (x) = f (y) < 0 and the function y(x) is decreasing. Therefore, the critical point 5 is an asymptotically stable attractor and the critical point −2 is a unstable repeller. The phase portrait (i.e. general solutions curves) is omitted. 2.2. Separable Equations. Give a DE dy = g(x)h(y). dx (1). Step 1. Determine if the DE is separable. If yes, then separate! Move all terms containing y to one side and move all terms containing x to the other side: dy = g(x)dx. h(y) (2). Step 2. Integrate both sides: Z Z dy = g(x)dx. h(y) (3). Step 3. Simplify the constant. (4). Step 4. We only do this step if the problem is an initial value problem (IVP). That is, if there is initial condition y(x0 ) = y0 , then plug (x0 , y0 ) into the general solution to find the constant. (Examples 3-5 and Exercises 1-28) Problem 3 (Exercise 23 in §2.2). Find an explicit solution to the initial value problem dx = 4(x2 + 1), x(π/4) = 1. dt Answer. The DE is separable. Separate the DE: dx = 4dt. 2 x +1 Integrate both sides: Z Z dx = 4dt. x2 + 1 5 Hence, arctan(x) = 4t + c and so x = tan(4t + c). Use the initial condition that x(π/4) = 1:  π  1 = tan 4 · + c = tan(π + c). 4 Hence, π 3 π + c = , and c = − π. 4 4 The solution to the initial value problem is   3 x = tan 4t − π . 4 2.3. Linear Equations. Give a DE. Determine whether it is linear. If yes, then change it into the standard form dy + p(x)y = f (x). dx R (1). Step 1. Multiply both sides of the DE by the integrating factor e P (x) : R R R dy + e P (x) p(x)y = e P (x) f (x). e P (x) dx Write the left-hand-side as an integral of a product: R d  R P (x)  e y = e P (x) f (x). dx (2). Step 2. Integrate both sides: Z R R P (x) e y = e P (x) f (x) dx. (3). Step 3. Simplify the constant. (4). Step 4. We only do this step if the problem is an initial value problem (IVP). That is, if there is initial condition y(x0 ) = y0 , then plug (x0 , y0 ) into the general solution to find the constant. (Examples 1-5 and Exercises 1-36) Problem 4 (Exercise 27 in §2.3). Solve the initial value problem xy 0 + y = ex , y(1) = 2. Answer. (1). Step 1. Dividing by x, the standard form of the given DE is 1 ex y0 + y = . x x From this form we identify P (x) = 1/x and f (x) = ex /x and further observe that P and f are continuous on x ∈ (0, ∞). Hence the integrating factor is R e 1/x = eln |x| = eln x = x. Multiply by x to the DE and rewrite dy x + y = ex dx and d [xy] = ex . dx 6 (2). Step 2. Integrate both sides: Z xy = ex dx = ex + c. (3). Step 3. Solve for y: ex c + . x x (4). Step 4. Use the initial condition that y(1) = 2: y= 2= e1 c + . 1 1 Solve for c: c = 2 − e. Hence, the solution to the IVP is ex 2 − e + . x x 2.5. Solutions by Substitutions. Give a DE. Use the substitution provided in the problem to solve the DE. You do not need to remember the substitution rules for different types of DEs! • In Example 1, use the substitution y = ux to solve y= (x2 + y 2 )dx + (x2 − xy)dy = 0. • In Example 2, use the substitution u = y −1 to solve dy x + y = x2 y 2 . dx • In Example 3, use the substitution u = −2x + y to solve dy = (−2x + y)2 − 7, y(0) = 0. dx (Examples 1-3 and Exercises 1-8, 11-12, 15-19, 21, 23-28) Chapter 3. Modeling with First Order Differential Equations The important models include (i). growth and decay of population, C-14, etc. (Examples 1-3 and Exercises 1-8, 11-12), (ii). Newton’s law of cooling/warming (Example 4 and Exercises 13-18), (iii). mixture of salt solutions (Examples 5-6 and Exercises 21-25, 27-28), (iv). series circuits (Example 7 and Exercises 29-30). 3.1. Linear Models. Give a model. (1). Translate the information in the problem and set up the DE. In this section, it is a first order linear DE. (2). Solve the DE. Read the background in the textbook and understand the problems! Notice that the numbers you encounter in Tests and Final are easy to calculate. You should keep the fractions and do not simplify them to decimals! Problem 5 (Exercise 13 in §3.1). A thermometer with initial temperature 70◦ is placed outside where the temperature is 10◦ . (a). Solve the differential equation for Newton’s law of cooling to derive the temperature of the thermometer T (t). (b). Graph the solution T (t). 7 Answer. (a). The differential equation for Newton’s law of cooling is dT = k(T − Tm ), dt in which Tm is the ambient temperature. Here, Tm = 10. So the initial value problem is dT = k(T − 10), T (0) = 70. dt The above differential equation is separable. Separate the DE: dT = kdt. T − 10 Integrating both sides, log |T − 10| = kt + c1 , which implies that |T − 10| = ekt+c1 = ec1 ekt . So T = ±c1 ekt + 10. Denote c = ±ec1 . Then T = cekt + 10. When t = 0, T = 70: 70 = c + 10, so c = 60. Thus, the temperature of the thermometer T = 60ekt + 10. (b). Graph is omitted here. You can consult Figure 3.1.6 on page 89 in the textbook. Notice that the graph for this problem starts at T = 70 when t = 0, i.e. the initial condition that T (0) = 70, and then converges to the asymptotic T = 10 as t → ∞, i.e. the temperature of the thermometer converges to the ambient temperature. Problem 6 (Exercise 23 in §3.1). A tank contains 500 gallons of water and no salt. Water containing 2 pounds of salt is pumping into the tank at a rate of 5 gallons per minute. Water drains out the tank at the same rate. (a). Set up a differential equation for the amount of salt in the tank, x(t). (b). Solve for x(t). (c). Graph the solution x(t). Answer. (a). The input rate of salt is 2 · 5 = 10 pounds per minute; the output rate of salt is x 1 ·5= x. 500 100 The differential equation for the amount of salt in the tank is therefore dx 1 = 10 − x, dt 100 with initial condition that x(0) = 0 (since initially there is no salt in the tank.) 8 (b). The above differential equation is linear and first order. Change it to the standard form: dx 1 + x = 10, dt 100 R 1 1 Multiply the DE by the integrating factor e 100 = e 100 t and rewrite 1 1 1 1 dx 1 d h 1 t i e 100 t + e 100 t x = 10e 100 t and e 100 x = 10e 100 t . dt 100 dx Integrate both sides: Z 1 1 1 t 100 e x = 10e 100 t = 1000e 100 t + c. Hence, 1 x = 1000 + ce− 100 t . When t = 0, x = 0: 0 = 1000 + c, so c = −1000. Thus, the amount of salt in the tank at time t is given by 1 x(t) = 1000 − 1000e− 100 t . (c). Graph is omitted here. You can consult Figure 3.1.6 on page 89 in the textbook. Notice that the graph for this problem starts at the origin since x(0) = 0, and then converges to the asymptotic x = 1000 as t → ∞. Problem 7 (Exercise 29 in §3.1). Suppose an RL circuit has inductance L = 0.1 H, resistance 50 Ω, and a constant voltage E(t) = 30 volt is applied. (a). Solve for the currant i(t), if i(0) = 0 Amp. (b). Sketch the current i(t). Answer. (a). The linear differential equation for the current i(t) is di L + Ri = E(t). dt With L = 0.1, R = 50, and E(t) = 30, we have that 1 di + 50i = 30. 10 dt Multiply both sides by 10: di + 500i = 300. dt R Multiply the DE by the integrating factor e 500 = e500t and rewrite di e500t + e500t 500i = 300e500t . dt Integrate both sides: Z 3 500t e i = 300e500t dt = e500t + c. 5 Hence, 3 i = + ce−500t . 5 When t = 0, i = 0: 3 0 = + c, 5 9 so c = −3/5. Thus, the current 3 3 −500t − e . 5 5 (b). Graph is omitted here. Notice that the graph for this problem starts at the origin since i(0) = 0, and then converges to the asymptotic i = 3/5 as t → ∞. i(t) = Chapter 4. Higher Order Differential Equations The important topics in this chapter: (i). Give an n-th order linear DE. The general solution is a linear combination of fundamental set of solutions: y = c1 y1 + c2 y2 + · · · + cn yn . Here, we use Wronskian to check that y1 , ..., yn are linearly independent. (ii). Reduction of order: Given y1 as a solution to a second order linear DE, set y2 = uy1 to solve for y2 . (iii). Second order homogeneous linear DE with constant coefficients ay 00 + by 0 + cy = 0, where a, b, c are numbers. Use y = emx to deduce a quadratic equation and solve the DE in three cases: The quadratic equation am2 + bm + c = 0 has two distinct real roots (Case 1), two equal real roots (Case 2), or two conjugate complex roots (Case 3). (iv). Second order nonhomogeneous linear DE with constant coefficients ay 00 + by 0 + cy = g, where a, b, c are numbers. The general solution y = yc +yp , in which yc solves the associated homogeneous DE ay 00 +by 0 + cy = 0 and we deduce yp by methods of Undetermined Coefficients or Variation of Parameters. (v). Solve systems of first order linear DEs by elimination. 4.1. Preliminary Theory – Linear Equations. Give a linear n-th order DE dn y dn−1 y dx an (x) n + an−1 (x) n−1 + · · · + a1 (x) + a0 (x)y = g(x). dx dx dx • The general solution is a linear combination of fundamental set of solutions: y = c1 y1 + c2 y2 + · · · + cn yn . • There are n equations to define the initial value problem (IVP): y(x0 ) = y0 , y 0 (x0 ) = y1 , ..., y (n−1) (x0 ) = yn−1 . Determine the interval of solutions: The IVP has a unique solution in an interval I where an (x), an−1 (x), ..., a1 (x), a0 (x), g(x) are continuous and an (x) 6= 0. • Give a homogeneousi linear second order DE a2 (x)y 00 + a1 (x)y 0 + a0 (x)y = 0. The general solution is a linear combination of two functions. ◦ If we find two such functions y1 and y2 , then we can compute their Wronskian W (y1 , y2 ) = iHomogeneous y1 y2 . y10 y20 means the right-hand-side of the DE equals 0. 10 ◦ If the Wronskian W (y1 , y2 ) 6= 0 on an interval I, then we say the two functions y1 and y2 are linearly independent on I and they form the fundamental set of solutions to the DE on I. ◦ The general solution to the DE is a linear combination of y1 and y2 : y = c1 y 1 + c2 y 2 on the interval I. ◦ If there is initial condition y(x0 ) = y0 and y 0 (x0 ) = y1 , then use these information to the general solution to find the constants c1 and c2 . You need to solve a system of linear equations for c1 and c2 . • The nonhomogeneousi linear second order DE a2 (x)y 00 + a1 (x)y 0 + a0 (x)y = g(x) has general solution y = yc + yp = complementary function + any particular solution. Here, yc solves the homogeneous DE a2 (x)y 00 + a1 (x)y 0 + a0 (x)y = 0 and yp is a solution to the nonhomogeneous DE. See §4.4 Undetermined Coefficients and §4.6 Variation of Parameters for finding yp of second order linear DEs. (Example 9 and Exercises 1-4, 9-10, 25-26, 28-29) 4.2. Reduction of Order. Give a homogeneous linear second order DE a2 (x)y 00 + a1 (x)y 0 + a0 (x)y = 0. We expect the general solution is a linear combination y = c1 y1 + c2 y2 . Provide y1 . Solve for y2 : (1). normalized the DE to the form y 00 + P (x)y 0 + Q(x)y = 0. (2). use the formula Z y2 = y1 R e− P . y12 (Examples 1-2 and Exercises 9-12) Problem 8 (Exercise 10 in §4.2). Suppose that we know that y1 (x) = x2 solves x2 y 00 + 2xy 0 − 6y = 0 on (0, ∞). (a). Use reduction of order to find a second solution y2 (x) to x2 y 00 + 2xy 0 − 6y = 0. (b). Show that y1 (x) and y2 (x) are linearly independent. (c). Write the general solution to x2 y 00 + 2xy 0 − 6y = 0 on (0, ∞). Answer. (a). Divide the DE by x2 : 6 2 y 00 + y 0 − 2 y = 0. x x iNonhomogeneous means the right-hand-side of the DE does not equal 0. 11 Use reduction of order with P (x) = 2/x: Z y2 (x) = y1 R e− P y12 R e− 2/x = x2 (x2 )2 Z −2 ln |x| e = x2 x4 Z −2 x = x2 x4 Z = x2 x−6 Z 1 = − x−3 . 5 (b). To show that y1 (x) and y2 (x) are linearly independent, we compute the Wronskian: W (y1 , y2 ) = x2 − 51 x−3 y1 y2 = = x−2 6= 0 y10 y20 2x 35 x−4 on (0, ∞). (c). The general solution to x2 y 00 + 2xy 0 − 6y = 0 is y(x) = c1 y1 + c3 y2 = c1 x2 − c3 −3 x = c1 x2 + c2 x−3 , 5 replacing −c3 /5 by c2 . 4.3. Homogeneous Linear Equations with Constant Coefficients. Give a second order homogeneous linear DE with constant coefficients ay 00 + by 0 + cy = 0, where a, b, c are numbers. Set y = emx . So y 0 = memx and y 00 = m2 emx . The DE simplifies to
am2 emx + bmemx + cemx = emx am2 + bm + c = 0. Solve the quadratic equation am2 + bm + c = 0. ◦ Case 1. If there are two distinct real roots m1 and m2 , then the fundamental set of solutions is em1 x and em2 x . ◦ Case 2. If the two roots are equal, i.e. m1 = m2 = m, then the fundamental set of solutions is emx and xemx . ◦ Case 3. If there are two complex and conjugate roots m1 = α + iβ and m2 = α − iβ, then the fundamental set of solutions is eαx cos(βx) and eαx sin(βx). (Examples 1-4 and Exercises 1-24, 29-32, 49-58) 12 4.4. Undetermined Coefficients. Give a nonhomogeneous second order linear DE ay 00 + by 0 + cy = g, where a, b, c are numbers. Attention: Use the method of Undetermined Coefficients only for constant coefficients (i.e. a, b, c are numbers) and g contains polynomials, exponential, and cos or sin. If a, b, c are not numbers or g contains rational, log, tan, sin2 , etc, then use the method of Variation of Parameters in §4.6. (1). Step 1. Solve the associated homogeneous linear DE ay 00 + by 0 + cy = 0 by §4.3. The complimentary solution is yc = c1 y1 + c2 y2 . (2). Step 2. Use undetermined coefficients to find one solution yp to the nonhomogeneous DE. To find one solution yp to the nonhomogeneous DE, depending on the nonhomogenous term g(x), we choose different forms of yp . ◦ If g contains polynomial, then yp should contain polynomial (with possibly higher degree). ◦ If g contains exponential ejx , then yp should contain ejx . ◦ If g contains cos(kx) or sin(kx), then yp should contain both cos(kx) and sin(kx). (3). Step 3. The general solution to the nonhomogeneous DE is y = yc + yp = c1 y1 + c2 y2 + yp . (4). Step 4. We only do this step if the problem is an initial value problem (IVP). That is, if there is initial condition y(x0 ) = y0 and y 0 (x0 ) = y1 , then plug them into the general solution to find the constants c1 and c2 . (Table 4.4.1, Examples 1-11, and Exercises 1-21, 27-30) Problem 9. Solve the initial value problem by undetermined coefficients. y 00 − y = 2e−x − 2x2 , y(0) = 0, y 0 (0) = 1. Answer. (1). Step 1. Solve the associated homogeneous linear DE y 00 − y = 0 The associated quadratic equation is m2 − 1 = 0. The two roots are m = 1 and m = −1. So the complementary solution yc = c1 ex + c2 e−x . (2). Step 2. Set yp = Axe−x + Bx2 + Cx + D. (Here, the nonhomogeneous term contains e−x so we set yp to contain Axe−x ; the nonhomogeneous term also contains −2x2 so we set yp to contain Bx2 + Cx + D.) Compute that yp0 = Ae−x − Axe−x + 2Bx + C, and yp00 = −Ae−x − Ae−x + Axe−x + 2B = −2Ae−x + Axe−x + 2B. 13 Hence, plug in the DE y 00 − y = 2e−x − 2x2 : yp00 − yp = −2Ae−x + Axe−x + 2B − (Axe−x + Bx2 + Cx + D) = −2Ae−x − Bx2 − Cx + (2B − D) = 2e−x − 2x2 . So −2A = 2, −B = −2, C = 0, 2B − D = 0 C = 0, D = 4, imply that A = −1, B = 2, and yp = −xe−x + 2x2 + 4. (3). Step 3. The general solution to the nonhomogeneous DE is y = yc + yp = c1 ex + c2 e−x − xe−x + 2x2 + 4. (4). Step 4. Using the initial condition that y(0) = 0: c1 + c2 + 4 = 0, using the initial condition that y 0 (0) = 1: y 0 (x) = c1 ex − c2 e−x − e−x + xe−x + 4x, and y 0 (0) = c1 − c2 − 1 = 1. Solve for c1 and c2 : c1 = −1 and c2 = −3. Hence, the solution is y = −ex − 3e−x − xe−x + 2x2 + 4. 4.6. Variation of Parameters. Give a nonhomogeneous second order linear DE y 00 + P (x)y 0 + Q(x)y = f (x). If the DE is not in this form, change it. (1). Step 1. Solve the associated homogeneous linear DE y 00 + P (x)y 0 + Q(x)y = 0. The complimentary solution is yc = c1 y1 + c2 y2 . You may be provided y1 and y2 as in Exercise 27-28. (2). Step 2. Use variation of parameters to find one solution yp to the nonhomogeneous DE. That is, set yp = u1 y1 + u2 y2 . Then we have that ( u01 y1 + u02 y2 = 0, u01 y10 + u02 y20 = f (x). Solve for u01 and u02 . We can use either of the following. ◦ Solve the system of equations directly by elimination or substitution. ◦ Formulas: y2 f (x) y1 f (x) u01 = − and u02 = . W (y1 , y2 ) W (y1 , y2 ) Here, W (y1 , y2 ) is the Wronskian of y1 and y2 . Attention: Integrate u01 and u02 to solve for u1 and u2 so yp is solved. 14 (3). Step 3. The general solution to the nonhomogeneous DE is y = yc + yp = c1 y1 + c2 y2 + yp . (4). Step 4. We only do this step if the problem is an initial value problem (IVP). That is, if there is initial condition y(x0 ) = y0 and y 0 (x0 ) = y1 , then plug them into the general solution to find the constants c1 and c2 . (Examples 1-3 and Exercises 1-15, 23-28) Problem 10 (Exercise 27 in §4.6).
1 1 The homogeneous equation x2 y 00 +xy 0 − x2 − 41 y = 0 has fundamental solution set {x− 2 cos x, x− 2 sin x} on (0, ∞). (a). Use variation of parameters to find a particular solution to the nonhomogeneous equation   3 1 2 00 0 2 x y + xy + x − y = x2 . 4 (b). Write the general solution. Answer. (a). We first set the DE in the standard form by dividing by x2 :   1 1 0 1 00 y + y + 1 − 2 y = x− 2 . x 4x Set the particular solution yp = u1 y1 + u2 y2 , in which 1 1 y1 = x− 2 cos x and y2 = x− 2 sin x. Then we have that ( u01 y1 + u02 y2 = 0, u01 y10 + u02 y20 = f (x), 1 in which f (x) = x− 2 , 1 1 1 3 1 3 y10 = − x− 2 cos x − x− 2 sin x and y20 = − x− 2 sin x + x− 2 cos x. 2 2 The system of equations becomes ( 0 −1 1 2 u x + u02 x− 2 sin x = 0,  1 x cos  1 1 − 32 1 − 32 − 12 − 21 0 − 2 x cos x − x sin x u1 + − 2 x sin x + x cos x u02 = x− 2 . 1 Multiply x 2 to the first and second equations: ( u01 cos x + u02 sin x = 0,

− 12 x−1 cos x − sin x u01 + − 12 x−1 sin x + cos x u02 = 1. Solve for u01 and u02 : u01 = − sin x and u02 = cos x. Hence, Z u1 = Z − sin x = cos x and u2 = cos x = sin x. The particular solution 1 1 1 1 yp (x) = u1 y1 + u2 y2 = cos x · x− 2 cos x + sin x · x− 2 sin x = x− 2 (sin2 x + cos2 x) = x− 2 . 15 (b). The general solution 1 1 1 y(x) = c1 x− 2 cos x + c2 x− 2 sin x + x− 2 . Remark. Alternatively, in the above problem, one can use the formulas to derive u01 and u02 . W (y1 , y2 ) = y1 y2 y10 y20 1 1 x− 2 sin x x− 2 cos x . = 1 1 3 3 − 21 x− 2 cos x − x− 2 sin x − 12 x− 2 sin x + x− 2 cos x     1 1 1 1 −3 1 −3 − 21 − − − = x cos x − x 2 sin x + x 2 cos x − x 2 sin x − x 2 cos x − x 2 sin x 2 2
−1 2 2 = x cos x + sin x = x−1 . With f (x) = x−1/2 , we have that 1 u01 1 y2 f (x) x− 2 sin x · x− 2 =− =− = − sin x, W (y1 , y2 ) x−1 and 1 1 x− 2 cos x · x− 2 y1 f (x) = = cos x. = W (y1 , y2 ) x−1 One then resume the solution as in above. u02 4.9. Solving Systems of Differential Equations by Elimination. (Examples 1 and Exercises 1-4, 21-22) Problem 11 (Exercise 4 in §4.9). Solve the system of DEs by elimination. ( Answer. (1). Step 1. Rewrite the system to ( dx dt dy dt − 4y = 1, + x = 2. 1 Dx − 4y = 1, ○ 2 Dy + x = x + Dy = 2. ○ Here, D = d/dt. Notice that x and y are dependent variables and t is the independent variable. (2). Step 2. 1 − D × ○. 2 ◦ Eliminate x to get a DE of y: ○ (Dx − 4y) − D(x + Dy) = 1 + D(2). 2 is 0 since it is constant. We have that Notice that D on the right-hand side of ○ −4y − D2 y = −y 00 − 4y = 1. Solve the above second order nonhomogeneous linear DE (by methods of Undermined Coefficients or Variation of Parameters): 1 y(t) = yc (t) + yp (t) = c1 cos(2t) + c2 sin(2t) − . 4 16 1 + 4 × ○. 2 ◦ Eliminate y to get a DE of x: D × ○ D(Dx − 4y) + 4(x + Dy) = D(1) + 4 × 2. 2 is 0 since it is constant. We have that Notice that D on the right-hand side of ○ D2 x + 4x = x00 + 4x = 8. Solve the above second order nonhomogeneous linear DE (by methods of Undermined Coefficients or Variation of Parameters): x(t) = xc (t) + xp (t) = c3 cos(2t) + c4 sin(2t) + 2. (3). Step 3. Plug x(t) and y(t) into any equation in the system to reduce the four constants c1 , c2 , c3 , c4 to two constants. That is, apply 1 x(t) = c3 cos(2t) + c4 sin(2t) + 2 and y(t) = c1 cos(2t) + c2 sin(2t) − 4 to the first equation in the system dx − 4y = 1. dt We have that dx = −2c3 sin(2t) + 2c4 cos(2t) dt and   dx 1 − 4y = −2c3 sin(2t) + 2c4 cos(2t) − 4 c1 cos(2t) + c2 sin(2t) − dt 4 = (−2c3 − 4c2 ) sin(2t) + (2c4 − 4c1 ) cos(2t) + 1 = 1. This forces −2c3 − 4c2 = 0 and 2c4 − 4c1 = 0. So c3 = −2c2 and c4 = 2c1 . Hence, the general solution to the system of linear system 1 x(t) = −2c2 cos(2t) + 2c1 sin(2t) + 2 and y(t) = c1 cos(2t) + c2 sin(2t) − . 4 (4). Step 4. We only do this step if the problem is an initial value problem (IVP). That is, if there is initial condition x(t0 ) = x0 and y(t0 ) = y0 , then plug them into the general solution to find the constants c1 and c2 . Chapter 5. Modeling with Higher Order Differential Equations The important topic in this chapter is to use second order linear DE to solve the spring/mass model. 17 5.1. Linear Models: Initial Value Problems. Give a spring/mass problem. We determine the motion of the mass under the influence of four possible forces: • Gravitational force. • Restoring force of the spring. • Damping force of the environment (e.g. air, fluid, etc). • Driven/external force. (1). Translate the information in the problem. Determine one of the following cases depending on the forces in consideration. ◦ Free undamped motion, i.e. gravitational force and restoring force are considered. See §5.1.1. (Examples 1 and Exercises 1-6) ◦ Free damped motion, i.e. gravitational force, restoring force, and damping force are considered. See §5.1.2. (Examples 3-5 and Exercises 21-31) ◦ Driven motion, i.e. gravitational force, restoring force, damping force, and driven force are considered. See §5.1.3. (Examples 6-8 and Exercises 33-37, 41-42) (2). Set up the DE. Denote x(t) for the position of the mass. So the DE has x as the dependent variable and t as the independent variable. Let m be the mass, k be the spring constant, β be the damping constant, and f (t) be the driven force. These constants or functions are provided in the problem. ◦ Free undamped motion: A second order homogeneous linear DE dx2 m 2 + kx = 0. dt p Here, ω = k/m is the frequency and T = 2π/ω is the period. ◦ Free damped motion: A second order homogeneous linear DE dx dx2 = 0. m 2 + kx + β dt dt Here, there are three cases: overdamped (two distinct real roots), critically damped (two equal roots), and underdamped (two conjugate and complex roots). ◦ Driven motion: A second order nonhomogeneous linear DE dx2 dx m 2 + kx + β = f (t). dt dt Here, x(t) = xc (t) + xp (t), where xc (t) is the transient term and xp (t) is the steady-state term. (3). Solve the DE with initial conditions x(t0 ) = x0 (initial position) and x0 (t0 ) = x1 (initial velocity). Notice that x(t) is positive below the equilibrium position and is negative above the equilibrium position; x0 (t) is positive when moving downward and is negative when moving upward. Read the background in the textbook and understand the problems! Notice that the numbers you encounter in Tests and Final are in the metric system and easy to calculate. You should keep the fractions and do not simplify them to decimals! Problem 12. A ball on a spring has mass m = 2 kg. Suppose the spring has spring constant k = 2, with resistance 4 times the velocity. (a). Suppose the ball starts at equilibrium with no initial velocity, and a force f (t) = 2 sin t is applied. Set up an initial value problem for the position of the ball. (b). Solve the initial value problem. (c). Identify the transient solution. 18 (d). Identify the steady-state solution. Answer. (a). The differential equation reads dx2 dx + f (t), = −kx − β dt2 dt in which m = 2, k = 2, β = 4, and f (t) = 2 sin t. Thus, m 2 dx2 dx = −2x − 4 + 2 sin t, 2 dt dt that is 2x00 + 4x0 + 2x = 2 sin t. The initial conditions are x(0) = 0 and x0 (0) = 0. (b). Solve the associated homogeneous linear DE 2x00 + 4x0 + 2x = 0. The complementary solution xc = c1 te−t + c2 e−t . Set xp = A sin t + B cos t. (Here, the nonhomogeneous term is 2 sin t so we set yp to contain both sin t and cos t.) Compute that x0p = A cos t − B sin t, and x00p = −A sin t − B cos t. Hence, plug in the DE 2x00 + 4x0 + 2x = 2 sin t: 2x00p + 4x0p + 2xp = 2(−A sin t − B cos t) + 4(A cos t − B sin t) + 2(A sin t + B cos t) = (−2A − 4B + 2A) sin t + (−2B + 4A + 2B) cos t = −4B sin t + 4A cos t = 2 sin t. Therefore, −4B = 2 and 4A = 0. So B = −1/2 and A = 0. The particular solution 1 xp = − cos t. 2 The general solution is 1 x = xc + cp = c1 te−t + c2 e−t − cos t. 2 Using the initial condition that x(0) = 0: 1 c2 − = 0, 2 0 using the initial condition that y (0) = 1: 1 x0 (t) = c1 e−t − c1 te−t − c2 e−t + sin t, 2 and x0 (0) = c1 − c2 = 0. 19 Solve for c1 and c2 : 1 2 c1 = 1 and c2 = . 2 Hence, the solution is 1 1 1 x = te−t + e−t − cos t. 2 2 2 (c). The transient solution is 1 −t 1 −t te + e . 2 2 (d). The steady-state solution is 1 − cos t. 2 Chapter 7. The Laplace Transform The important topics in this chapter: (i). The Laplace transform of polynomials, exponentials, sin and cos, and piecewise functions. (ii). The Laplace transform of derivatives and integrals: Let Y (s) = L {y}. Then L {y 0 } = sY (s) − y(0), and Z L {y 00 } = s2 Y (s) − sy(0) − y 0 (0), t  Y (s) . s 0 (iii). The inverse Laplace transform of rational functions. Partial fractions are needed here. (iv). Use Laplace transform to solve IVPs. Here, we need to use the Laplace transforms and inverse Laplace transforms of the above functions or formulas provided in the problem. (See Sections 7.2 and 7.4.) L f (τ ) dτ = 7.1. Definition of the Laplace Transform. Give a function f . Define the Laplace transform of f as Z ∞ e−st f (t) dt. L {f } = 0 To evaluate the Laplace transform of a function, we need (1). integrate by parts, if necessary, (2). notice that e−st → 0 as t → ∞ when s > 0, (3). specify the domain of s. (Examples 1-4 and Exercises 11-16, 19-32, 37-38) The important Laplace transforms are • For n = 0, 1, 2, ..., n! L {tn } = n+1 , s > 0. s For example, 1 1 L {1} = and L {t} = 2 . s s • 1 L {eat } = , s > a. s−a • k L {sin(kt)} = 2 , s > 0. s + k2 • s L {cos(kt)} = 2 , s > 0. s + k2 20 • Piecewise functions. (Example 6 and Exercises 1-10) 7.2. Inverse Transforms and Transforms of Derivatives. The important inverse Laplace transforms are • For n = 0, 1, 2, ...,   n! −1 L = tn . sn+1 For example,     1 1 −1 −1 L = 1 and L = t. s s2 •   1 −1 = eat . L s−a •   k −1 L = sin(kt). s2 + k 2 •   s −1 L = cos(kt). s2 + k 2 (Example 1-2 and Exercises 1-30) In order to use Laplace transform to solve IVPs, we need to evaluate the Laplace transforms of derivatives: Write L {y} = Y (s). Then L {y 0 } = sY (s) − y(0), and L {y 00 } = s2 Y (s) − sy(0) − y 0 (0). Give an IVP of y(t). (1). Step 1. Take Laplace transform on both sides of the DE. The DE then is transformed to an algebraic equations of Y (s). Here, we need to use the above two formulas and the initial conditions. (2). Step 2. Solve for Y (s) for the algebraic equation. (3). Step 3. Take inverse Laplace transform of Y (s) to get y(t) as the solution to the IVP. We usually need to use partial fractions here. See the following example. (Example 4-5 and Exercises 35-42, 45-46) Remark. • Laplace transform can be used to solve IVPs only, i.e. differential equations with initial conditions. Whereas, the methods of Undetermined Coefficients and Variation of Parameters apply to differential equations with or without initial conditions. • The initial conditions are applied in Step 1. Taking the Laplace transform. So we do NOT apply these conditions in the end. Problem 13 (Exercise 41 in §7.2). Consider √ √ y 00 + y = 2 sin( 2t), y(0) = 10, y 0 (0) = 0. (a). Take Laplace transform of the above, and solve for Y (s). (b). Use Y (s) to solve for y(t). Answer. 21 (a). Take Laplace transform on both sides of the DEs: √ √ L {y 00 } + L {y} = 2L {sin( 2t)}. Denote Y (s) = L {y}. Then L {y 00 } = s2 Y (s) − sy(0) − y 0 (0) = s2 Y (s) − 10s, by the initial conditions that y(0) = 10 and y 0 (0) = 0. So the DE becomes √ √ 2 2 2 √ s Y (s) − 10s + Y (s) = 2 . = 2 2 2 s +2 s + ( 2) Solve for Y (s) in the above algebraic equation 2 10s Y (s) = 2 + 2 . 2 (s + 1)(s + 2) s + 1 (b). Take inverse Laplace transform of Y (s) to get y(t) as the solution to the IVP. That is,   2 10s −1 −1 y(t) = L {Y (s)} = L . + (s2 + 1)(s2 + 2) s2 + 1 Use partial fractions: 2 A B A(s2 + 2) + B(s2 + 1) (A + B)s2 + 2A + B = 2 + = = . (s2 + 1)(s2 + 2) s + 1 s2 + 2 (s2 + 1)(s2 + 2) (s2 + 1)(s2 + 2) Hence, A + B = 0 and 2A + B = 2. Solve for A and B: A = 2 and B = −2. So 2 2 −2 = 2 + 2 , 2 2 (s + 1)(s + 2) s +1 s +2 and   2 10s −1 y(t) = L + (s2 + 1)(s2 + 2) s2 + 1       2 −2 10s −1 −1 −1 = L +L +L s2 + 1 s2 + 2 s2 + 1 ) ( √     √ −1 1 2 s −1 −1 √ = 2L − 2L + 10L s2 + 1 s2 + 1 s2 + ( 2)2 √ √ = 2 sin(t) − 2 sin( 2t) + 10 cos(t). 7.3. Operational Properties I: Unit Step Functions. Define the unit step function ( 0, 0 ≤ t < a; U (t − a) = 1, t ≥ a. 22 Then ∞ Z e−st U (t − a) dt L {U (t − a)} = Z0 ∞ = e−st dt a e−st t=∞ = −s t=a e−as . = s (Examples 6 and Exercises 61-62) Problem 14. Consider   0, 0 ≤ t < 2; f (t) = 3, 2 ≤ t < 5;  0, t ≥ 5. (a). Find F (s) = L {f } using the definition of Laplace transform. (Evaluate an integral.) (b). Find F (s) = L {f } using the unit step functions. Answer. (a). Evaluate an integral: ∞ Z e−st f (t) dt L {f } = 0 5 Z 3e−st dt = 2 3e−st t=5 = −s t=2 3e−2s 3e−5s = − . s s (b). Represent f (t) by unit step functions: f (t) = 3U (t − 2) − 3U (t − 5). Then 3e−2s 3e−5s − . s s 7.4. Operational Properties II: Laplace Transforms of Integrals. Let Y (s) = L {y}. Then Z t  Y (s) L f (τ ) dτ = . s 0 L {f } = 3L {U (t − 2)} − 3L {U (t − 5)} = (Exercises 45, 49-52) Problem 15 (Exercise 49 in §7.4). Consider 0 Z y (t) = 1 − sin t − t y(τ ) dτ, y(0) = 0. 0 (a). Take Laplace transform of the above, and solve for Y (s). (b). Use Y (s) to solve for y(t). 23 You may need L {t sin(kt)} = 2ks . (s2 + k 2 )2 Answer. (a). Take Laplace transform on both sides of the DEs: Z 0 L {y } = L {1} − L {sin t} − L t  y(τ ) dτ . 0 Denote Y (s) = L {y}. Then L {y 0 } = sY (s) − y(0) = sY (s), by the initial conditions that y(0) = 0. So the DE becomes 1 1 Y (s) sY (s) = − 2 − . s s +1 s Solve for Y (s) in the above algebraic equation 1 s Y (s) = 2 − 2 . s + 1 (s + 1)2 (b). Take inverse Laplace transform of Y (s) to get y(t) as the solution to the IVP. That is,     1 s −1 −1 y(t) = L −L s2 + 1 (s2 + 1)2   2s 1 = sin t − L −1 2 2 (s + 1)2 1 = sin t − t sin t. 2 Here, we use the formula provided in the problem (k = 1): 2s L {t sin t} = 2 . (s + 1)2 7.5. The Dirac Delta Function. The Laplace transform of the Dirac delta function L {δ(t − a)} = e−as . (Example 1 and Exercises 1-12.) Problem 16 (Exercise 3 in §7.5). Consider y 00 + y = δ(t − 2π), y(0) = 0, y 0 (0) = 1. (a). Take Laplace transform of the above, and solve for Y (s). (b). Use Y (s) to solve for y(t). You may need L {f (t − a) · U (t − a)} = L {f }e−as . Answer. (a). Take Laplace transform on both sides of the DEs: L {y 00 } + L {y} = L {δ(t − 2π)}. Denote Y (s) = L {y}. Then L {y 00 } = s2 Y (s) − sy(0) − y 0 (0) = s2 Y (s) − 1, 24 by the initial conditions that y(0) = 0 and y 0 (0) = 1. So the DE becomes s2 Y (s) − 1 + Y (s) = e−2πs . Solve for Y (s) in the above algebraic equation 1 e−2πs Y (s) = 2 + . s + 1 s2 + 1 (b). Take inverse Laplace transform of Y (s) to get y(t) as the solution to the IVP. That is,   −2πs   1 e −1 −1 y(t) = L +L 2 s +1 s2 + 1 = sin t + sin(t − 2π) · U (t − 2π) = sin t + sin(t) · U (t − 2π) Here, we use the formula provided in the problem (f (t) = sin t and a = 2π): L {sin(t − 2π)U (t − 2π)} = e−2πs . s2 + 1 Chapter 8. Systems of Linear First Order Differential Equations Important topics in this chapter: (i). Use Wronskian to check two vector functions are linearly independent. (ii). Use matrix form to solve homogeneous systems of linear DEs. 8.1. Preliminary Theory – Linear Systems. Give two vector functions     → − → − x1 x2 X1 = and X 2 = . y1 y2 Their Wronskian → − → − x x W ( X 1, X 2) = 1 2 . y1 y2 → − → − → − → − If the Wronskian W ( X 1 , X 2 ) 6= 0, then we say the two vector functions X 1 and X 2 are linearly independent. (Exercises 17-18) 8.2. Homogeneous Linear Systems. Give a homogeneous linear system of first order DEs with constant coefficients ( dx = ax + by, dt dy = cx + dy. dt Here, a, b, c, d are numbers. (1). Step 1. Write the system in matrix form.  0    x a b x = . y0 c d y Denote Then → − X = → −0 → − X = AX ,   x . y   a b . in which the matrix A = c d 25 (2). Find the eigenvalues and eigenvectors. The eigenvalues are the solutions to the characteristic equation of the matrix A: det(A − λI) = 0. → − → − ◦ Case 1. A has two distinct real eigenvalues λ1 and λ2 with eigenvectors K 1 and K 2 , respectively. The eigenvectors are the solutions to → − → − (A − λ1 I) K 1 = 0 and (A − λ2 I) K 2 = 0. Then the general solution → − → − → − X = c1 eλ1 t K 1 + c2 eλ2 t K 2 . (Example 1 and Exercises 1-6, 13) → − ◦ Case 2. A has two equal real eigenvalues λ (i.e. repeated eigenvalues) with eigenvector K → − and generalized eigenvector P . → − → − → − (A − λI) K = 0 and (A − λI) P = K . Then the general solution  → − → − → − → − X = c1 eλt K + c2 teλt K + eλt P . (Example 4 and Exercises 21-24, 31) ◦ Case 3. A has two complex conjugate eigenvalues λ1 = α + iβ and λ2 = α − iβ. The → − → − eigenvalue λ1 = α + iβ has real column vectors B 1 and B 2 from a complex eigenvector. → − That is, find the eigenvector K that → − (A − λ1 I) K = 0. We have that → − → − → − K = B 1 + i B 2. Notice that we need to use λ1 = α + iβ (instead of λ2 = α − iβ) to find the eigenvectors to write the general solution h h → − → − → − i → − → − i X = c1 eαt cos(βt) B 1 − sin(βt) B 2 + c2 eαt cos(βt) B 2 + sin(βt) B 1 . (Example 6 and Exercises 35-40, 48) (3). Step 3. We only do this step if the problem is an initial value problem (IVP). That is, if there → − → − is initial condition X (t0 ) = X 0 , then plug them into the general solution to find the constants c1 and c2 . Problem 17 (Exercise 1 in §8.2). Consider the system of ODE’s ( dx dt dy dt = x + 2y, = 4x + 3y. → − → − Write as a matrix equation X 0 = A X . Find det(A − λI). Find the eigenvalues and eigenvectors of A. Use the above to find the general to the system.  solution  → − 0 (e). Find the solution with X (0) = . 3 (a). (b). (c). (d). Answer. 26 (a). The matrix equation is    0  1 2 x x = . y0 4 3 y (b). 1−λ 2 = (1 − λ)(3 − λ) − 8 = λ2 − 4λ − 5. 4 3−λ (c). The eigenvalues are the solutions to det(A − λI) = det(A − λI) = λ2 − 4λ − 5 = (λ − 5)(λ + 1) = 0. So the two eigenvalues are λ1 = 5 and λ2 = −1. → − The eigenvector K 1 corresponding to the eigenvalue λ1 = 5 solves → − (A − λ1 I) K = 0, that is,  1 − λ1 2 4 3 − λ1      r1 −4 2 r1 = = 0. r2 4 −2 r2 Hence, ( −4r1 + 2r2 = 0, 4r1 − 2r2 = 0. Set r1 = 1. Then r2 = 2 and the eigenvector   → − 1 K1 = . 2 → − The eigenvector K 2 corresponding to the eigenvalue λ2 = −1 solves → − (A − λ2 I) K = 0, that is,       1 − λ2 2 p1 2 2 p1 = = 0. 4 3 − λ2 p2 4 4 p2 Hence, ( 2p1 + 2p2 = 0, 4p1 + 4p2 = 0. Set p2 = 1. Then p1 = −1 and the eigenvector   → − −1 K2 = . 1 (d). The general solution to the system is     → − 5t 1 −t −1 X = c1 e + c2 e . 2 1 (e). Using the initial condition at t = 0,       1 −1 0 c1 + c2 = . 2 1 3 27 That is, ( c1 − c2 = 0, 2c1 + c2 = 3. Solve for c1 and c2 : c1 = 1 and c2 = 1. Hence, the solution is       5t → − e − e−t 5t 1 −t −1 . X =e +e = 2e5t + e−t 2 1 Problem 18 (Exercise 31 in §8.2). Consider the system of ODE’s ( dx dt dy dt = 2x + 4y, = −x + 6y. → − → − Write as a matrix equation X 0 = A X . Find det(A − λI). Find the eigenvalues and eigenvectors of A. Use the above to find the general to the system.  solution  → − −1 (e). Find the solution with X (0) = . 6 (a). (b). (c). (d). Answer. (a). The matrix equation is  0    x 2 4 x = . y0 −1 6 y (b). 2−λ 4 = (2 − λ)(6 − λ) + 4 = λ2 − 8λ + 16. −1 6 − λ (c). The eigenvalues are the solutions to det(A − λI) = det(A − λI) = λ2 − 8λ + 16 = (λ − 4)2 = 0. So the repeated eigenvalues are λ = 4. → − The eigenvector K corresponding to the eigenvalue λ = 4 solves → − (A − λI) K = 0, that is,       2−λ 4 r1 −2 4 r1 = = 0. −1 6 − λ r2 −1 2 r2 Hence, ( −2r1 + 4r2 = 0, −r1 + 2r2 = 0. Set r2 = 1. Then r1 = 2 and the eigenvector   → − 2 K= . 1 28 → − The generalized eigenvector P corresponding to the eigenvalue λ = 4 solves → − → − (A − λI) P = K , that is,       → − 2−λ 4 p1 −2 4 p1 = = K. −1 6 − λ p2 −1 2 p2 Hence, ( −2p1 + 4p2 = 2, −p1 + 2p2 = 1. Set p2 = 1. Then p1 = 1 and the generalized eigenvector   → − 1 P = . 1 (d). The general solution to the system is        → − 4t 1 4t 2 4t 2 . +e + c2 te X = c1 e 1 1 1 (e). Using the initial condition at t = 0,       2 1 −1 c1 + c2 = . 1 1 6 That is, ( 2c1 + c2 = −1, c1 + c2 = 6. Solve for c1 and c2 : c1 = −7 and c2 = 13. Hence, the solution is          → − 26te4t − e4t 4t 2 4t 2 4t 1 X = −7e + 13 te +e = . 1 1 1 13te4t + 6e4t Problem 19 (Exercise 48 in §8.2). Consider the system of ODE’s ( dx dt dy dt = 6x − y, = 5x + 4y. → − → − Write as a matrix equation X 0 = A X . Find det(A − λI). Find the eigenvalues and eigenvectors of A. Use the above to find the general to the system.  solution  → − −2 (e). Find the solution with X (0) = . 8 (a). (b). (c). (d). Answer. (a). The matrix equation is  0    x 6 −1 x = . y0 5 4 y 29 (b). 6 − λ −1 = (6 − λ)(4 − λ) + 5 = λ2 − 10λ + 29. 5 4−λ (c). The eigenvalues are the solutions to det(A − λI) = det(A − λI) = λ2 − 10λ + 29 = (λ − 5)2 + 4 = 0. So the two complex eigenvalues are λ1 = 5 + 2i and λ2 = 5 − 2i. → − The eigenvector K corresponding to the eigenvalue λ1 = 5 + 2i solves → − (A − λI) K = 0, that is,  6 − λ1 −1 5 4 − λ1      r1 1 − 2i −1 r1 = = 0. r2 5 −1 − 2i r2 Hence, ( (1 − 2i)r1 − r2 = 0, 5r1 − (1 + 2i)r2 = 0. Set r1 = 1. Then r2 = 1 − 2i and the eigenvector       → − 1 1 0 K= = +i . 1 − 2i 1 −2 The two real column vectors corresponding to λ1 = 5 + 2i are     → − → − 1 0 B1 = and B 2 = . 1 −2 (d). The general solution to the system is           → − 1 0 0 1 5t 5t X = c1 e cos(2t) − sin(2t) + c2 e cos(2t) + sin(2t) 1 −2 −2 1     cos(2t) sin(2t) = c1 e5t + c2 e5t . cos(2t) + 2 sin(2t) −2 cos(2t) + sin(2t) (e). Using the initial condition at t = 0,       1 0 −2 c1 + c2 = . 1 −2 8 That is, ( c1 = −2, c1 − 2c2 = 8. Solve for c1 and c2 : c1 = −2 and c2 = −5. Hence, the solution is     → − cos(2t) sin(2t) 5t 5t X = −2e − 5e cos(2t) + 2 sin(2t) −2 cos(2t) + sin(2t)   −2 cos(2t) − 5 sin(2t) = e5t . 8 cos(2t) − 9 sin(2t) 30 Appendix: Table of Formulas Laplace Transforms. Write F = L {f } as the Laplace transform of f in the following formulas. • 1 L {1} = . s • n! L {tn } = n+1 . s • ω L {sin(ωt)} = 2 . s + ω2 • s . L {cos(ωt)} = 2 s + ω2 • 1 . L {eat } = s−a • n! L {tn eat } = . (s − a)n+1 • ω L {eat sin(ωt)} = . (s − a)2 + ω 2 • s−a L {eat cos(ωt)} = . (s − a)2 + ω 2 • L {eat f (t)} = F (s − a). • 1 L {U (t − a)} = e−as . s • L {f (t − a)U (t − a)} = F (s)e−as . • L {f (t)U (t − a)} = L {f (t + a)}e−as . •  n  d L f (t) = sn F (s) − sn−1 f (0) − · · · − sf (n−2) (0) − f (n−1) (0). dtn Linear Models. • Newton’s Law of cooling/warming. (Example 4 in §3.1) dT = k(T − Tm ). dt • Series circuits. (Example 7 in §3.1) di L + Ri = E(t). dt • Air resistance. (Exercise 35 in §3.1) dv m = mg − kv. dt • Spring/mass systems. dx2 dx m 2 +β + kx = F (t). dt dt