Study Guide of Differential Equations

STUDY GUIDE OF DIFFERENTIAL EQUATIONS Abstract. This Study Guide includes the important topics and problems that are featured in the Tests and the Final of Differential Equations. Under each topic, examples and exercises from the book by Zill (A first course in differential equations with modeling applications, 11th Edition) are listed for more information and practice. Topic 0. Classification of differential equations Given a DE, determine its order and whether it is linear or nonlinear. (Example 4 and Exercises 1, 3, 5, 7 in §1.1) Topic 1. First order differential equations To solve DEs, i.e., equations that involve derivatives, the skills of integration are indispensable, which include the following. • Integrate polynomials, exponential functions, logarithmic functions, and trigonometric functions, e.g., Z Z 1 1 dx = ln |x| + c and = arctan x + c. 2 x x +1 • Apply substitution. • Apply integration by parts (IBP). Moreover, the following skills are also important in DEs. 1 2 • Apply trigonometric identities, e.g., sin2 x + cos2 x = 1, sin(2x) = 2 sin x cos x, cos(2x) = 2 cos2 x − 1 = 1 − 2 sin2 x. • Apply partial fractions to break a large fraction into smaller ones. For example, 1 A B A(1 − P ) + BP (B − A)P + A 1 = + = = , = P − P2 P (1 − P ) P 1−P P (1 − P ) P (1 − P ) in which A and B are to be determined. Comparing the terms of P and constants in the numerators of both sides, we have that B − A = 0 and A = 1. So A = 1 and B = 1. Therefore, 1 1 1 . = + 2 P −P P 1−P Topic 1.A. Autonomous differential equations. Consider the DE dy = f (y). dx It is called an autonomous DE because y 0 depends only on y, i.e., a function of itself. (1). Find all the equilibrium solutions by solving f (y) = 0. (2). Sketch phase portrait of solutions curves. (3). Classify the equilibrium solutions as asymptotically stable attractors, semi-stable equilibrium solutions, or unstable repellers. (Examples 3-5 and Exercises 21, 23, 25, 27 in §2.1) Problem 1 (Exercise 24 in §2.1). Let dy = 10 + 3y − y 2 . dx (a). Find the equilibrium solutions. (b). Classify each equilibrium solution (asymptotically stable, semi-stable, or unstable). (b). Sketch the phase portrait. Answer. (a). Solving f (y) = 10 + 3y − y 2 = (5 − y)(2 + y) = 0, the equilibrium solutions are y = 5 and y = −2. (b). The equilibrium solutions y = 5 and y = −2 divide the whole xy-plane into three regions. In the region {y > 5}, f (y) < 0 so y 0 = f (y) < 0 and the function y(x) is decreasing. In the region {−2 < y < 5}, f (y) > 0 so y 0 = f (y) > 0 and the function y(x) is increasing. In the region {y < −2}, f (y) < 0 so y 0 = f (y) < 0 and the function y(x) is decreasing. Therefore, y = 5 is an asymptotically stable and y = −2 is unstable. (c). The phase portrait is omitted. 3 Topic 1.B. Separable differential equations. Consider the DE dy = g(x)h(y). dx It is called a separable DE because the variables x, y can be separated to the opposite sides of the DE as follows. (1). Separate the variables: dy = g(x)dx. h(y) (2). Integrate both sides: Z Z dy = g(x)dx. h(y) (3). Simplify the function and the constant to get the general solution. (4). If the problem is an initial value problem (IVP), that is, 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, 3, 5, 15, 17, 23, 25 in §2.2) Problem 2 (Exercise 17 in §2.2). Solve the differential equation dP = P − P 2. dt Answer. Separate the variables: dP = dt. P − P2 Integrate both sides: Z Z dP = dt = t + c1 . P − P2 Using partial fractions, Z Z Z dP 1 1 P . = dP + dP = ln |P | − ln |1 − P | = ln 2 P −P P 1−P 1−P Hence, P = t + c1 , ln 1−P and P = ±ec1 et = cet . 1−P in which we denote by c = ±ec1 . Then P = cet (1 − P ) = cet − cet P, which implies that (1 + cet )P = cet . So the general solution to the DE is cet . 1 + cet Problem 3 (Exercise 23 in §2.2). Solve the initial value problem dx = 4(x2 + 1), x(π/4) = 1. dt P = 4 Answer. Separate the variables: dx = 4dt. +1 Z Z dx = 4dt. x2 + 1 x2 Integrate both sides: Hence, arctan(x) = 4t + c, so the general solution to the DE is x = tan(4t + c). Use the initial condition that x(π/4) = 1:  π  1 = tan 4 · + c = tan(π + c). 4 Hence, 3 π π + c = , and c = − π. 4 4 The solution to the initial value problem is   3 x = tan 4t − π . 4 Topic 1.C. Linear differential equations. Consider the first order linear DE dy + p(x)y = f (x). dx R (1). Multiply both sides of the DE by the integrating factor e p : 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). Integrate both sides: Z R e p(x) y= e R p(x) f (x) dx. (3). Simplify the function and the constant. (4). If the problem is an IVP with initial condition y(x0 ) = y0 , then plug (x0 , y0 ) into the general solution to find the constant. (Examples 1-5 and Exercises 3, 5, 7, 9, 11, 27, 31 in §2.3) Problem 4 (Exercise 27 in §2.3). Solve the initial value problem xy 0 + y = ex , Answer. y(1) = 2. 5 (1). Divide by x to normalize the DE: 1 ex y0 + y = . x x From this form we identify the integrating factor R e 1/x = eln |x| = eln x = x. Here, we take x ∈ (0, ∞) since the initial condition is at x = 1. Multiply by x to the DE and rewrite dy d x + y = ex and [xy] = ex . dx dx (2). Integrate both sides: Z xy = ex dx = ex + c. (3). Solve for y: y= ex c + . x x (4). Use the initial condition that y(1) = 2: 2= e1 c + . 1 1 Solve for c: c = 2 − e. Hence, the solution to the IVP is ex 2 − e + . x x Topic 1.D. Models by first order differential equations. The models include • Newton’s law of cooling/warming (Example 4 and Exercises 13, 17 in §3.1), • mixture of salt solutions (Examples 5-6 and Exercises 21, 27, 29 in §3.1), y= Problem 5 (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). Use differential equation to find the number of pounds of salt A(t) at time t. (b). Sketch the solution. Answer. (a). The input rate of salt is 2 · 5 = 10 pounds per minute, whereas the output rate of salt is A 1 ·5= A. 500 100 The DE for the amount of salt in the tank is therefore 1 dA = 10 − A, dt 100 with initial condition that A(0) = 0 since initially there is no salt in the tank. The above DE is linear and first order. Change it to the standard form: dA 1 + A = 10, dt 100 6 R 1 1 Multiply the DE by the integrating factor e 100 dt = e 100 t and rewrite 1 1 1 1 dA 1 d h 1 t i + e 100 t A = 10e 100 t and e 100 A = 10e 100 t . e 100 t dt 100 dA Integrate both sides: Z 1 1 1 t 100 e A = 10e 100 t = 1000e 100 t + c. Hence, 1 A = 1000 + ce− 100 t . When t = 0, A = 0: 0 = 1000 + c, so c = −1000. Thus, the amount of salt in the tank at time t is given by 1 A = 1000 − 1000e− 100 t . (b). The solution curve starts from the origin and converges to A = 1000 as t → ∞. Topic 2. Second order linear differential equations with constant coefficients Topic 2.A. Homogeneous differential equations. Consider ay 00 + by 0 + cy = 0, in which a, b, c are real numbers. It is said to be homogeneous because the right-hand-side is zero. Set y = emx . Then 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. Depending on the roots, the general solution to the DE can be represented as follows. • Case 1. Two distinct real roots m1 and m2 . Then y = c1 em1 x + c2 em2 x . • Case 2. Repeated root, i.e., m1 = m2 = m. Then y = c1 emx + c2 xemx . • Case 3. Two complex and conjugate roots m1 = α + iβ and m2 = α − iβ, in which α, β ∈ R. Then y = c1 eαx cos(βx) + c2 eαx sin(βx). 7 (Examples 1-4 and Exercises 1, 3, 5, 7, 9, 11, 29, 31 in §4.3) Topic 2.B. Non-homogeneous differential equations by undetermined coefficients. Consider ay 00 + by 0 + cy = g(x), in which a, b, c are real numbers. It is said to be non-homogeneous because the right-hand-side is non-zero. (1). Solve the associated homogeneous DE ay 00 + by 0 + cy = 0. The solution yc = c1 y1 + c2 y2 is called the complimentary solution. (2). Use undetermined coefficients to find one solution yp to the non-homogeneous DE. To this end, set yp according to the non-homogeneous term g(x). ◦ If g(x) is a polynomial, then set yp as a polynomial. ◦ If g(x) is an exponential function, then set yp to contain the same exponential function. ◦ If g(x) contains cos(kx) or sin(kx), then yp should contain both cos(kx) and sin(kx). Moreover, compare yp with the complimentary solution yc . Make necessary modifications to the setup if yp is contained in yc . (3). The general solution to the non-homogeneous DE is y = yc + yp = c1 y1 + c2 y2 + yp . (4). If the problem is an initial value problem (IVP), that is, 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, 3, 5, 9, 11, 13, 17, 31 in §4.4) Problem 6. Solve the initial value problem by undetermined coefficients. y 00 − y = 2e−x − 2x2 , y(0) = 0, y 0 (0) = 1. Answer. (1). Solve the associated homogeneous DE y 00 − y = 0 Set y = emx . Then m2 − 1 = 0. The two roots are m = 1 and m = −1. So the complementary solution yc = c1 ex + c2 e−x . (2). Set yp = Axe−x + Bx2 + Cx + D. (Here, the non-homogeneous term contains e−x so we set yp to contain xe−x , considering that e−x is already contained in the complimentary solution yc ; the non-homogeneous term also contains −2x2 so we set yp to contain a polynomial of degree two, i.e., 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. 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 . 8 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). The general solution to the non-homogeneous DE is y = yc + yp = c1 ex + c2 e−x − xe−x + 2x2 + 4. (4). Using the initial condition that y(0) = 0: c1 + c2 + 4 = 0. 0 Using the initial condition that y (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 to the IVP is y = −ex − 3e−x − xe−x + 2x2 + 4. Topic 2.C. Non-homogeneous differential equations by variation of parameters. Consider ay 00 + by 0 + cy = g(x), in which a, b, c are real numbers. Normalize the DE: b c g(x) y 00 + y 0 + y = . a a a Denote by f (x) = g(x)/a. (1). Solve the associated homogeneous DE ay 00 + by 0 + cy = 0. The complimentary solution is yc = c1 y1 + c2 y2 . (2). Set a particular solution yp = u1 y1 + u2 y2 . Then Z Z y2 f (x) y1 f (x) u1 = − dx and u2 = dx. W (y1 , y2 ) W (y1 , y2 ) Here, W (y1 , y2 ) is the Wronskian of y1 and y2 : W (y1 , y2 ) = y1 y2 = y1 y20 − y2 y10 . y10 y20 (3). The general solution to the non-homogeneous DE is y = yc + yp = c1 y1 + c2 y2 + yp . (4). If the problem is an initial value problem (IVP) with 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, 5, 11, 13, 19, 27 in §4.6) 9 Problem 7 (Exercise 27 in §4.6). The homogeneous equation x2 y 00 + xy 0 − x2 − 4 y = 0 has 1 1 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 non-homogeneous equation   3 1 2 00 0 2 x y + xy + x − y = x2 . 4
1 (b). Write the general solution. Answer. (a). Normalize the DE by dividing by x2 :   1 1 1 0 y + y + 1 − 2 y = x− 2 . x 4x 00 Compute that W (y1 , y2 ) = y1 y2 y10 y20 1 1 x− 2 sin x x− 2 cos x . 3 1 3 1 − 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 . Set a particular solution yp = u1 y1 + u2 y2 . Then Z u1 = − and Z u2 = 1 1 x− 2 sin x · x− 2 dx = − x−1 1 1 x− 2 cos x · x− 2 dx = x−1 Z sin x dx = cos x, Z cos x dx = sin x. Therefore, 1 1 1 1 yp = u1 y1 + u2 y2 = cos x · x− 2 cos x + sin x · x− 2 sin x = x− 2 (sin2 x + cos2 x) = x− 2 . (b). The general solution 1 1 1 y = c1 x− 2 cos x + c2 x− 2 sin x + x− 2 . Topic 2.D. The spring-mass model. The motion of a mass attached to a spring (which satisfies the Hooke’s law) is governed by the following DE. mx00 + βx0 + kx = f (t). Here, m is the mass (in kilogram), β is the damping constant (in newton-second/meter), k is the spring constant (in newton/meter), and f (t) is the driven force (in newton). (Examples 1-8 and Exercises 6, 7, 27, 37 in §5.1) Problem 8. A mass weighing 1 kg is attached to a spring whose spring constant is 5 n/m. The damping force equals 2 times the instantaneous velocity. Initially, the mass is released from the equilibrium position with a downward velocity of 2 m/s. 10 (a). Determine the equation of the motion. (Set up the IVP and find the solution.) (b). Sketch the solution. Answer. (a). The IVP reads x00 + 2x0 + 5x = 0, x(0) = 0, x0 (0) = 2. The general solution is x(t) = c1 e−t cos(2t) + c2 e−t sin(2t). Using the initial condition that x(0) = 0: c1 = 0. Using the initial condition that x0 (0) = 2: x0 (t) = −c1 e−t cos(2t) − 2c1 e−t sin(2t) − c2 e−x sin(2t) + 2c2 e−x cos(2t), and x0 (0) = −c1 + 2c2 = 2. Solve for c1 and c2 : c1 = 0 and c2 = 1. Hence, the solution to the IVP is x = e−t sin(2t). (b). The solution curve has magnitude e−t → 0 as t → ∞ and oscillates at a period of π. Topic 3. The Laplace transform Topic 3.A. The Laplace transform by integrals. Let f be a function on (0, ∞). Define the Laplace transform of f as Z ∞ L {f } = e−st f (t) dt. 0 (Examples 1-6 and Exercises 1, 3, 5, 13, 15, 19, 21, 23, 25, 27, 29, 31 in §7.1) Some formulas of the Laplace transforms: • 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 11 • L {eat } = • 1 , s−a s > a. k , + k2 s L {cos(kt)} = 2 , s + k2 L {sin(kt)} = • • s2 s > 0. s > 0. L {eat f } = L {f }(s − a). Topic 3.B. Initial value problems by the Laplacian transform. 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). Take Laplace transform on both sides of the DE. The DE then is transformed to an algebraic equations of Y (s). Solve for Y (s) for the algebraic equation. (2). Take inverse Laplace transform of Y (s) to get y(t) as the solution to the IVP. (Example 4-5 and Exercises 1, 3, 7, 9, 11, 13, 15, 17, 19, 29, 35, 37, 39, 41 in §7.2) Problem 9 (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. (a). Take Laplace transform on both sides of the DE: √ √ L {y 00 } + L {y} = 2L {sin( 2t)}. Write Y = L {y}. Then L {y 00 } = s2 Y − 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 − 10s + Y = 2 = 2 . 2 2 s +2 s + ( 2) Solve for Y in the above algebraic equation 2 10s Y = 2 + . (s + 1)(s2 + 2) s2 + 1 (b). Take inverse Laplace transform of Y (s) to get y(t) as the solution to the IVP. That is,   10s 2 −1 −1 y = L {Y } = L + . (s2 + 1)(s2 + 2) s2 + 1 Use partial fractions: 2 A B A(s2 + 2) + B(s2 + 1) (A + B)s2 + 2A + B = + = = . (s2 + 1)(s2 + 2) s2 + 1 s2 + 2 (s2 + 1)(s2 + 2) (s2 + 1)(s2 + 2) 12 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 = L + (s2 + 1)(s2 + 2) s2 + 1       2 −2 10s −1 −1 −1 = L +L +L s2 + 1 s2 + 2 s2 + 1 ) ( √     √ −1 2 s 1 −1 −1 √ − 2L + 10L = 2L s2 + 1 s2 + 1 s2 + ( 2)2 √ √ = 2 sin(t) − 2 sin( 2t) + 10 cos(t). Topic 4. First order linear differential systems Consider ( dx dt dy dt = ax + by, = cx + dy. Here, a, b, c, d are real numbers. Write the system in matrix form:  0    x a b x = . 0 y c d y Denote ~ = X   x . y Then   a b ~ X = AX, in which the matrix A = . c d The eigenvalues are the solutions to the characteristic equation of the matrix A: ~0 det(A − λI) = 0. ~ 1 and K ~ 2 , respectively. • Case 1. Two distinct real eigenvalues λ1 and λ2 with eigenvectors K The eigenvectors are the solutions to ~ 1 = 0 and (A − λ2 I)K ~ 2 = 0. (A − λ1 I)K Then the general solution ~ 1 + c2 eλ2 t K ~ 2. ~ = c1 eλ1 t K X ~ and generalized eigenvector P~ . The • Case 2. Repeated eigenvalues λ with eigenvector K generalized eigenvector is the solution to ~ (A − λI)P~ = K. Then the general solution   ~ = c1 eλt K ~ + c2 teλt K ~ + eλt P~ . X 13 • Case 3. Two complex conjugate eigenvalues λ1 = α + iβ and λ2 = α − iβ. The eigenvalue λ1 = α + iβ has complex eigenvector ~ =B ~ 1 + iB ~ 2. K ~ 1 and B ~ 2 are both real vectors. Then the general solution in which B h i h i ~ = c1 eαt cos(βt)B ~ 1 − sin(βt)B ~ 2 + c2 eαt cos(βt)B ~ 2 + sin(βt)B ~1 . X (Example 1-6 and Exercises 1, 3, 5, 13, 21, 23, 31, 35, 37, 39, 48 in §8.2) Depending on the eigenvalues of A, the behaviors of the solution curves are drastically different. The following figure lists all the possible phase portraits. Figure 1. Phase portraits of systems of two linear DEs with constant coefficients Problem 10 (Exercise 1 in §8.2). Consider ( dx dt dy dt (a). (b). (c). (d). (e). = x + 2y, = 4x + 3y. ~ 0 = AX. ~ Write as a matrix equation X Find det(A − λI). Find the eigenvalues and eigenvectors of A. Use the above to find the general solution to the system. Sketch the phase portrait. Answer. (a). The matrix equation is  0    x 1 2 x = . y0 4 3 y 14 (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 + 1)(λ − 5) = 0. So the two eigenvalues are λ1 = −1 and λ2 = 5. ~ The eigenvector K1 corresponding to the eigenvalue λ1 = −1 solves ~ 1 = 0, (A − λ1 I)K that is,       1 − λ1 2 p1 2 2 p1 = = 0. 4 3 − λ1 p2 4 4 p2 Hence, ( 2p1 + 2p2 = 0, 4p1 + 4p2 = 0. Set p2 = 1. Then p1 = −1 and the eigenvector   −1 ~ K1 = . 1 ~ 2 corresponding to the eigenvalue λ2 = 5 solves The eigenvector K ~ 2 = 0, (A − λ2 I)K that is,  1 − λ2 2 4 3 − λ2      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   ~2 = 1 . K 2 (d). The general solution to the system is ~ = c1 e X −t     −1 5t 1 + c2 e . 1 2 (e). Refer to the phase portrait when λ1 < 0 < λ2 in Figure 1. Problem 11 (Exercise 31 in §8.2). Consider ( dx = 2x + 4y, dt dy = −x + 6y. dt (a). (b). (c). (d). ~ 0 = AX. ~ Write as a matrix equation X Find det(A − λI). Find the eigenvalues and eigenvectors of A. Use the above to find the general solution to the system. 15 (e). Sketch the phase portrait. Answer. (a). The matrix equation is    0  2 4 x x = . −1 6 y y0 (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. ~ corresponding to the eigenvalue λ = 4 solves The eigenvector K ~ = 0, (A − λI)K 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 ~ = K   2 . 1 The generalized eigenvector P~ corresponding to the eigenvalue λ = 4 solves ~ (A − λI)P~ = K, that is,       p1 p1 −2 4 2−λ 4 ~ = K. = p2 −1 6 − λ p2 −1 2 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 2 4t 2 4t 1 ~ X = c1 e + c2 te +e . 1 1 1 (e). Refer to the phase portrait when 0 < λ1 = λ2 in Figure 1. 16 Problem 12 (Exercise 48 in §8.2). Consider ( dx = 6x − y, dt dy = 5x + 4y. dt (a). (b). (c). (d). (e). ~ 0 = AX. ~ Write as a matrix equation X Find det(A − λI). Find the eigenvalues and eigenvectors of A. Use the above to find the general solution to the system. Sketch the phase portrait. Answer. (a). The matrix equation is    0  6 −1 x x = . 0 5 4 y y (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. ~ corresponding to the eigenvalue λ1 = 5 + 2i solves The eigenvector K ~ = 0, (A − λI)K that is,       6 − λ1 −1 r1 1 − 2i −1 r1 = = 0. 5 4 − λ1 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 ~1 = ~2 = 0 . B and B 1 −2 (d). The general solution to the system is           1 0 0 1 5t 5t ~ = c1 e cos(2t) X − 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). Refer to the phase portrait when α ± βi, α > 0 in Figure 1.