% 2250-notes-S2008-ch7.txt ========================================================================= # 7.1: 6, 18 ========================================================================= 7.1-6: ========================================================================= Make the position-velocity substitution u1=x(t), u2=x'(t), u3=y(t), u4=y'(t) This changes the coupled system of two second order DE into a first order system u' = Au for some 4x4 matrix A. The book's answer is u'=Au in scalar form using x1,x2,x3,x4 instead of u1,u2,u3,u4. 7.1-18: Skip the direction field work. Submit only the solution for the general solution and the evaluation of initial conditions. Start by writing the system as u'=Au. [x(t)] [0 -1] u = [ ], A = [ ] [y(t)] [10 -7] The characteristic equation of A is r^2 + 7r + 10 = 0 with roots -2,-5. The theory says that u1 and u2 are solutions of the second order equation obtained from this characteristic equation [namely w''+7w'+10w=0] and therefore u1=x(t) is a linear combination of the atoms exp(-2t) and exp(-5t). This gives the answer x(t) = c1 exp(-2t) + c2 exp(-5t) y(t) = -x'(t) from the DE system = 2c1 exp(-2t) + 5c2 exp(-5t) ========================================================================= # 7.2: 8, 16 ========================================================================= 7.2-8: ========================================================================= Define u1=x(t), u2=y(t), u3=z(t). Then u' = Au for the 3x3 matrix [2 -3 0] A = [1 1 2] [0 5 -7] 7.2-16: ========================================================================= The general solution of the system is x = c1 x1 + c2 x2. Independence of x1, x2 is determined by the value of the wronskian of x1 and x2, which is the determinant of the matrix [ exp(3t) 2 exp(-2t)] W = [ ] [ -exp(3t) -2 exp(2t) ] Evaluate det(W) and show it is nonzero. The two requested answer checks are for the two differential equations x1' = Ax1 and x2' = Ax2 where x1, x2 are the columns of matrix W and [ 4 1] A = [ ]. [-2 1] The two answer checks can be done at once by verifying W' = AW. The latter is done by expanding the LHS and the RHS, showing LHS=RHS. ========================================================================= # 7.3: 10, 22 ========================================================================= 7.3-10: ========================================================================= The system can be written u'=Au where [x1] [-3 -2] u =[ ], A = [ ] [x2] [ 9 3] The characteristic equation of A is r^2 + 9 = 0 with roots 3i, -3i. The corresponding eigenvectors are [i-1] [-i-1] v1 =[ ], v2 = [ ] [ 3 ] [ 3 ] These should be found from one frame sequence starting with B = A - (3i)I and ending with rref(B), by applying the last frame algorithm. The second eigenvector v2 is found from the THEOREM: v2 = conjugate(v1), which applies to conjugate eigenvalues only. The solution by the eigenvalue method [Edwards Penney naming convention] is then u = c1 v1 exp(3ti) + c2 v2 exp(-3ti). This formula is expected to be translated to its real form as in Example 3, section 7.3 of the textbook. The method is to focus on one complex solution v1 exp(3ti) = v1(cos 3t + i sin 3t) and then take real and imaginary parts to get two vector solutions w1, w2 which are independent. Then a REAL solution is the formula u = c1 w1 + c2 w2. Your problem is to find formulas for vectors w1 and w2 from the identity v1 exp(3ti) = v1(cos 3t + i sin 3t) = w1 + i w2. 7.3-16: ========================================================================= The system can be written u'=Au where [x1] [-50 20] u =[ ], A = [ ] [x2] [100 -60] The characteristic equation of A is r^2 + 110r + 1000 = 0 with roots -10, -100. The corresponding eigenvectors are [ 1 ] [ 2 ] v1 =[ ], v2 = [ ] [ 2 ] [ -5 ] These should be found from two frame sequences starting with B = A - (lambda)I [using lambda = -10 then lambda = -100] and ending with rref(B), by applying the last frame algorithm. The solution by the eigenvalue method [Edwards Penney naming convention] is then u = c1 v1 exp(-10t) + c2 v2 exp(-100t). ========================================================================= # 7.4: 6 7.4-6: ========================================================================= This problem follows exactly Example 1 in the textbook, section 7.4. An answer is provided in the answer section of the textbook. The second order matrix differential system is [1 0] [-(2+4) 4 ] [-6 4] M x'' = K x, M = [ ], K = [ ] = [ ] [0 2] [ 4 -(4+4)] [ 4 -8] Then x'' = Ax with A=inverse(M)K: [-6 4] A = [ ] [ 2 -4] The eigenpairs of A are found from two frame sequences. Show the details. [-2] [1] (-8, [ ]), (-2, [ ]) [ 1] [1] The NATURAL FREQUENCIES are sqrt(-lambda): omega1 = sqrt(8), omega2 = sqrt(2) The NATURAL MODES x1(t) and x2(t) are the two vector terms in (14), Example 1 of section 7.4, with changes made for this problem. Describe the two modes with the same language as used in the textbook. ================ LAPLACE THEORY. This problem can also be solved by Laplace theory applied to the matrix differential system x'' = Ax, x(0)=c2, x'(0)=c1. However, the solution x(t) is not in a form which identifies the natural modes. What is useful is the Laplace model itself, which identifies the natural frequencies as the roots of the denominator delta [see below] in all the Laplace entries of the resolvent. Step 1. Transform the system to s^2 L(x) = A L(x)+ c1 + c2 s. Step 2. Use the resolvent of s^2 I - A, which is inverse(s^2I-A), to write L(x) = inverse(s^2 I - A)(c1 + c2 s) delta = s^4+10*s^2+16 = (s^2+8)(s^2+2) 1 [s^2+4 4 ] resolvent = ----- [ ] delta [ 2 s^2+6] Then the resolvent formula for L(x) gives 1 [s^2+4 4 ] 1 [s^s+4s 4s ] L(x) = c1 ----- [ ] + c2 ----- [ ] delta [ 2 s^2+6] delta [ 2s s^3+6s] The rest of the problem is partial fractions, which can be done with maple assist using the function convert(). =========================================================================