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# EDO 9th Boyce Cap9 Resolução

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429 CH A P T E R 9 Nonlinear Differential Equations and Stability 9.1 2.(a) Setting x= \u3be ert results in the algebraic equations( 5\u2212 r \u22121 3 1\u2212 r )( \u3be1 \u3be2 ) = ( 0 0 ) . For a nonzero solution, we must have det(A\u2212 r I) = r2 \u2212 6 r + 8 = 0 . The roots of the characteristic equation are r1 = 2 and r2 = 4 . For r = 2, the system of equa- tions reduces to 3\u3be1 = \u3be2. The corresponding eigenvector is \u3be(1) = (1 , 3)T . Substi- tution of r = 4 results in the single equation \u3be1 = \u3be2 . A corresponding eigenvector is \u3be(2) = (1 , 1)T . (b) The eigenvalues are real and positive, hence the critical point is an unstable node. 430 Chapter 9. Nonlinear Differential Equations and Stability (c,d) (a) x1 vs t (b) x1 vs t (c) x1 vs t (d) x1 vs t 3.(a) Solution of the ODE requires analysis of the algebraic equations( 2\u2212 r \u22121 3 \u22122\u2212 r )( \u3be1 \u3be2 ) = ( 0 0 ) . For a nonzero solution, we must have det(A\u2212 r I) = r2 \u2212 1 = 0 . The roots of the characteristic equation are r1 = 1 and r2 = \u22121. For r = 1, the system of equations reduces to \u3be1 = \u3be2 . The corresponding eigenvector is \u3be(1) = (1 , 1)T . Substitution of r = \u22121 results in the single equation 3 \u3be1 \u2212 \u3be2 = 0 . A corresponding eigenvector is \u3be(2) = (1 , 3)T . 9.1 431 (b) The eigenvalues are real, with r1 r2 < 0 . Hence the critical point is a saddle. (c,d) (a) x1 vs t (b) x1 vs t (c) x1 vs t (d) x1 vs t 5.(a) The characteristic equation is given by\u2223\u2223\u2223\u22231\u2212 r \u221251 \u22123\u2212 r \u2223\u2223\u2223\u2223 = r2 + 2 r + 2 = 0 . The equation has complex roots r1 = \u22121 + i and r2 = \u22121\u2212 i. For r = \u22121 + i, the components of the solution vector must satisfy \u3be1 \u2212 (2 + i)\u3be2 = 0 . Thus the corresponding eigenvector is \u3be(1) = (2 + i , 1)T . Substitution of r = \u22121\u2212 i results 432 Chapter 9. Nonlinear Differential Equations and Stability in the single equation \u3be1 \u2212 (2\u2212 i)\u3be2 = 0 . A corresponding eigenvector is \u3be(2) = (2\u2212 i , 1)T . (b) The eigenvalues are complex conjugates, with negative real part. Hence the origin is a stable spiral. (c,d) (a) x1 vs t (b) x1 vs t (c) x1 vs t (d) x1 vs t 6.(a) Solution of the ODEs is based on the analysis of the algebraic equations( 2\u2212 r \u22125 1 \u22122\u2212 r )( \u3be1 \u3be2 ) = ( 0 0 ) . 9.1 433 For a nonzero solution, we require that det(A\u2212 r I) = r2 + 1 = 0. The roots of the characteristic equation are r = ±i . Setting r = i , the equations are equivalent to \u3be1 \u2212 (2 + i)\u3be2 = 0 . The eigenvectors are \u3be(1) = (2 + i , 1)T and \u3be(2) = (2\u2212 i , 1)T . (b) The eigenvalues are purely imaginary. Hence the critical point is a center. (c,d) (a) x1 vs t (b) x1 vs t 7.(a) Setting x= \u3be ert results in the algebraic equations( 3\u2212 r \u22122 4 \u22121\u2212 r )( \u3be1 \u3be2 ) = ( 0 0 ) . For a nonzero solution, we require that det(A\u2212 r I) = r2 \u2212 2r + 5 = 0. The roots of the characteristic equation are r = 1 ± 2i . Substituting r = 1 \u2212 2i , the two equations reduce to (1 + i)\u3be1 \u2212 \u3be2 = 0 . The two eigenvectors are \u3be(1) = (1 , 1 + i)T and \u3be(2) = (1 , 1\u2212 i)T . (b) The eigenvalues are complex conjugates, with positive real part. Hence the origin is an unstable spiral. 434 Chapter 9. Nonlinear Differential Equations and Stability (c,d) (a) x1 vs t (b) x1 vs t (c) x1 vs t (d) x1 vs t 8.(a) The characteristic equation is given by\u2223\u2223\u2223\u2223\u22121\u2212 r \u221210 \u22121/4\u2212 r \u2223\u2223\u2223\u2223 = (r + 1)(r + 1/4) = 0 , with roots r1 = \u22121 and r2 = \u22121/4. For r = \u22121, the components of the solution vector must satisfy \u3be2 = 0 . Thus the corresponding eigenvector is \u3be(1) = (1 , 0)T . Substitution of r = \u22121/4 results in the single equation 3\u3be1/4 + \u3be2 = 0 . A corre- sponding eigenvector is \u3be(2) = (4 ,\u22123)T . 9.1 435 (b) The eigenvalues are real and both negative. Hence the critical point is a stable node. (c,d) (a) x1 vs t (b) x1 vs t 9.(a) Solution of the ODEs is based on the analysis of the algebraic equations( 3\u2212 r \u22124 1 \u22121\u2212 r )( \u3be1 \u3be2 ) = ( 0 0 ) . For a nonzero solution, we require that det(A\u2212 r I) = r2 \u2212 2 r + 1 = 0. The single root of the characteristic equation is r = 1 . Setting r = 1 , the components of the solution vector must satisfy \u3be1 \u2212 2 \u3be2 = 0 . A corresponding eigenvector is \u3be = (2 , 1)T . (b) Since there is only one linearly independent eigenvector, the critical point is an unstable, improper node. 436 Chapter 9. Nonlinear Differential Equations and Stability (c,d) (a) x1 vs t (b) x1 vs t 10.(a) The characteristic equation is given by\u2223\u2223\u2223\u22231\u2212 r 2\u22125 \u22121\u2212 r \u2223\u2223\u2223\u2223 = r2 + 9 = 0 . The equation has complex roots r1,2 = ± 3i. For r = \u22123i, the components of the solution vector must satisfy 5 \u3be1 + (1\u2212 3i)\u3be2 = 0 . Thus the corresponding eigenvec- tor is \u3be(1) = (1\u2212 3i ,\u22125)T . Substitution of r = 3i results in 5 \u3be1 + (1 + 3i)\u3be2 = 0 . A corresponding eigenvector is \u3be(2) = (1 + 3i ,\u22125)T . (b) The eigenvalues are purely imaginary, hence the critical point is a center. 9.1 437 (c,d) (a) x1 vs t (b) x1 vs t 11.(a) The characteristic equation is (r + 1)2 = 0 , with double root r = \u22121 . It is easy to see that the two linearly independent eigenvectors are \u3be(1) = (1 , 0)T and \u3be(2) = (0 , 1)T . (b) Since there are two linearly independent eigenvectors, the critical point is a stable proper node. (c,d) 438 Chapter 9. Nonlinear Differential Equations and Stability (a) x1 vs t (b) x1 vs t 12.(a) Setting x= \u3be ert results in the algebraic equations( 2\u2212 r \u22125/2 9/5 \u22121\u2212 r )( \u3be1 \u3be2 ) = ( 0 0 ) . For a nonzero solution, we require that det(A\u2212 r I) = r2 \u2212 r + 5/2 = 0. The roots of the characteristic equation are r = 1/2 ± 3i/2 . Substituting r = 1/2 \u2212 3i/2 , the equations reduce to (3 + 3i)\u3be1 \u2212 5 \u3be2 = 0 . Therefore the two eigenvectors are \u3be(1) = (5 , 3 + 3i)T and \u3be(2) = (5 , 3\u2212 3i)T . (b) Since the eigenvalues are complex, with positive real part, the critical point is an unstable spiral. (c,d) (a) x1 vs t (b) x1 vs t 9.1 439 14. Setting x \u2032 = 0 , that is, (\u22122 1 1 \u22122 ) x = ( 2 \u22121 ) , we find that the critical point is x0 = (\u22121, 0)T . With the change of dependent variable, x=x0+u , the differential equation can be written as du dt = (\u22122 1 1 \u22122 ) u. The critical point for the transformed equation is the origin. Setting u= \u3be ert results in the algebraic equations(\u22122\u2212 r 1 1 \u22122\u2212 r )( \u3be1 \u3be2 ) = ( 0 0 ) . For a nonzero solution, we require that det(A\u2212 r I) = r2 + 4r + 3 = 0. The roots of the characteristic equation are r = \u22123 , \u22121 . Hence the critical point is a stable node. 15. Setting x \u2032 = 0 , that is, (\u22121 \u22121 2 \u22121 ) x = ( 1 \u22125 ) , we find that the critical point is x0 = (\u22122, 1)T . With the change of dependent variable, x=x0+u , the differential equation can be written as du dt = (\u22121 \u22121 2 \u22121 ) u. The characteristic equation is det(A\u2212 r I) = r2 + 2r + 3 = 0, with complex conju- gate roots r = \u22121 ± i\u221a2 . Since the real parts of the eigenvalues are negative, the critical point is a stable spiral. 16. The critical point x0 satisfies the system of equations( 0 \u2212\u3b2 \u3b4 0 ) x = (\u2212\u3b1 \u3b3 ) . It follows that x0 = \u3b3/\u3b4 and y0 = \u3b1/\u3b2 . Using the transformation, x=x0+u , the differential equation can be written as du dt = ( 0 \u2212\u3b2 \u3b4 0 ) u. The characteristic equation is det(A\u2212 r I) = r2 + \u3b2 \u3b4 = 0. Since \u3b2 \u3b4 > 0 , the roots are purely imaginary, with r = ± i\u221a\u3b2\u3b4 . Hence the critical point is a center. 20. The system of ODEs can be written as dx dt = ( a11 a12 a21 a22 ) x. 440 Chapter 9. Nonlinear Differential Equations and Stability The characteristic equation is r2 \u2212 p r + q = 0. The roots are given by r1,2 = p ± \u221a p2 \u2212 4q 2 = p ± \u221a\u2206 2 . The results can be verified using Table 9.1.1. 21.(a) If q > 0 and p < 0, then the roots are either complex conjugates with negative real parts, or both real and negative. (b) If q > 0 and p = 0, then the roots are purely imaginary. (c) If