#$&* course Mth 279 5/21 Query 28 Differential Equations*********************************************
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: Find the algebraic and geometric multiplicity of each eigenvalue and, if possible, diagonalize the matrix [7, -2, 2; 8, -1, 4; 8, 4, -1 ]. The characteristic equation of this matrix is (lambda - 3)^2 ( lambda + 1). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The algebraic multiplicity of Lambda_1 = 3 is two, and Lambda_2 is one. Which can be seen from the characteristic equation above. for Lambda_1 = 3 we solve (A - 3I)x_1 = 0 [ 4, -2, 2; 8, -4, 4; 8, 4, -4][ x_1; x_2; x_3] = [ 0; 0; 0] Using elementary row operations: -2R1 + R2 -> R2 Swap R2, R3 -2R1 + R2 -> R2 (1/4)R2 + R1 -> R1 (1/4)R1 -> R1 (1/8)R2 -> R2 [ 1, 0, 0; 0, 1, -1; 0, 0, 0][ x_1; x_2; x_3] = [ 0; 0; 0] Which results in simple equations: x_1 = 0 and x_2 - x_3 = 0 Solution would be x_1 = [ 0; 1; 1] With geometric Multipicity of 1 -------------------------------------------------------- for Lambda_2 = -1 we solve (A + I)x_2 = 0 [ 8, -2, 2; 8, 0, 4; 8, 4, 0][ x_1; x_2; x_3] = [ 0; 0; 0] Using elementary row operations puts the system into rref [ 1, 0, 0; 0, 1, 0; 0, 0, 1][ x_1; x_2; x_3] = [ 0; 0; 0] Which results in simple equations: x_1 = 0 x_2 = 0 x_3 = 0 Solution would be x_1 = [ 0; 0; 0] With geometric Multipicity of 0 -------------------------------------------------------------- Repeated Eigenvalue Lambda_1 = 3 v_1 = [ 0; 1; 1] y_1(t) = e^(3t) v_1 y_2(t) = te^(3t) v_1 + e^(3t) v_2 (A - lambda*I)v_2 = v_1 [ 4, -2, 2; 8, -4, 4; 8, 4, -4][ x_1; x_2; x_3] = [ 0; 1; 1] RREF [ 1, 0, 0; 0, 1, -1; 0, 0, 0][ x_1; x_2; x_3] = [ 0; 0; 1] yeilds equations: x_1 = 0 x_2 - x_3 = 0 v_2 = [ 0; 1; 1] with geometric multiplicity of 1 ---------------------------------------------------------------- y_1 = e^(3t) [ 0; 1; 1] = [ 0; e^(3t); e^(3t)] y_2 = te^(3t) [ 0; 1; 1] + e^(3t) [ 0, 1, 1] = [ 0; te^(3t) + e^(3t); te^(3t) + e^(3t)] Psi(t) = T = [ 0, 0, 0; 1, 1, 0; 1, 1, 0] since det[Psi(t)] = 0 cannot diagonalize confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: Find the algebraic and geometric multiplicity of each eigenvalue and, if possible, diagonalize the matrix [ 5, -1, 1; 14, -3, 6; 5, -2, 5 ]. The characteristic equation of this matrix is (lambda - 2)^2 ( lambda -3). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Lambda_1 = 2 which has an algebraic multiplicity of 2 and Lambda_2 = 3 which has an algebraic multiplicity of 1 This all can be seen from the charactersitic equation given. First for Lambda_1 = 2, we solve (A - 2I)x_1 = 0 [ 3, -1, 1; 14, -5, 6; 5, -2, 3][ x_1; x_2; x_3] = [ 0; 0; 0] RREF [ 1, 0, -1; 0, 1, -4; 0, 0, 0][ x_1; x_2; x_3] = [ 0; 0; 0] Which yeilds simple equations: x_1 - x_3 = 0 x_2 - 4(x_3) = 0 A simple solution to this set of equations would be: x_1 = [ 1; 4; 1] With a Geometric Multiplicity of 1 --------------------------------------------------------------- Since this is a repeated Eigenvalue y_1 = e^(2t) v_1 v_1 = [ 1, 4, 1] y_2 = te^(2t) v_1 + e^(2t) v_2 (A - 2I)v_2 = v_1 [ 3, -1, 1; 14, -5, 6; 5, -2, 3][ x_1; x_2; x_3] = [ 1; 4; 1] RREF [ 1, 0, -1; 0, 1, -4; 0, 0, 0][ x_1; x_2; x_3] = [ 1; 4; 1] Which yeilds simple equations: x_1 - x_3 = 1 x_2 - 4(x_3) = 2 A simple solution to this set would be: v_2 = [ 2; 6; 1] ---------------------------------------------------------------- For lambda_2 = 3 we solve (A - 3I)x_2 = 0 [ 2, -1, 1; 14, -6, 6; 5, -2, 2][ x_1; x_2; x_3] = [ 0; 0; 0] RREF [ 1, 0, 0; 0, 1, -1; 0, 0, 0][ x_1; x_2; x_3] = [ 0; 0; 0] Which yeilds equations: x_1 = 0, and x_2 - x_3 = 0 A solution to this simple set: x_2 = [ 0, 1, 1] Which has a geometric multiplicity of 1 ----------------------------------------------------------------- T = [ 1, 2, 0; 4, 6, 1; 1, 1, 1] det(T) = (1)[ 6, 1; 1, 1] - (2)[ 4, 1; 1, 1] + (0)[ 4, 6; 1, 1] = 6 - 1 - 2(4) + (2)(1) + 0 = 8 - 9 = -1 T^(-1) = 1/det * [ adj(M)] ---------------------------------------------------------------- Finding the adj(M). M = [ 1, 2, 0; 4, 6, 1; 1, 1, 1] M^T = [ 1, 4, 1; 2, 6, 1, 0, 1, 1] -------------------------------------------- determinants of the Mini Matricies of M^T M_11 = [ 6, 1; 1, 1] = 5 M_12 = [ 2, 1; 0, 1] = 2 M_13 = [ 2, 6; 0, 1] = 2 M_21 = [ 4, 1; 1, 1] = 3 M_22 = [ 1, 1; 0, 1] = 1 M_23 = [ 1, 4; 0, 1] = 1 M_31 = [ 4, 1; 6, 1] = -2 M_32 = [ 1, 1; 2, 1] = -1 M_33 = [ 1, 4; 2, 6] = -2 Adj(M) = [ 5, 2, 2; 3, 1, 1; -2, -1, -2][ +, -, +; -, +, -; +, -, +] = [ 5, -2, 2; -3, 1, -1; -2, 1, -2] T^(-1) = -1*[ 5, -2, 2; -3, 1, -1; -2, 1, -2] = [ -5, 2, -2; 3, -1, 1; 2, -1, 2] Now to diagonalize: T^(-1)AT = D [ -5, 2, -2; 3, -1, 1; 2, -1, 2][ 5, -1, 1; 14, -3, 6; 5, -2, 5 ][ 1, 2, 0; 4, 6, 1; 1, 1, 1] = [ 2, 1, 0; 0, 2, 0; 0, 0, 3] confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: Solve the system y ' = [ -4, -6; 3, 5 ] y + [e^(2 t) - 2 e^t; -e^(2 t) + e^t] YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y_c(t) y' = [ -4, -6; 3, 5]y P(lambda) = det[ A - lambda_1*I] = [ (-4-lambda), -6; 3, (5 - lambda)] = (-4-lambda)(5 - lambda) - (-6)(3) = lambda^2 - 5*lambda + 4*lambda - 20 + 18 = lambda^2 - lambda -2 = (lambda - 2)(lambda + 1) Lambda_1 = 2 Lambda_2 = -1 -------------------------------------------------------------------------- for Lambda_1 = 2 we solve (A - 2I)x_1 = 0 [ -6, -6; 3, 3][ x_1; x_2] = [ 0; 0] RREF [ 1, 1; 0, 0][ x_1; x_2] = [ 0; 0] Which yeilds simple equation: x_1 + x_2 = 0 A solution would be x_1 = [ 1, -1] --------------------------------------- for Lambda_2 = -1 we solve (A + I)x_2 = 0 [ -3, -6; 3, 6][ x_1; x_2] = [ 0; 0] RREF [ 1, 2; 0, 0][ x_1; x_2] = [ 0; 0] Which yeilds simple equation: x_1 + 2(x_2) = 0 A solution would be x_1 = [ 2, -1] ------------------------------------------- Solutions: y_1 = e^(2t) [ 1; -1] = [ e^(2t); -e^(2t)] y_2 = e^(-t) [ 2, -1] = [ 2e^(-t); -e^(-t)] y_c(t) = [ e^(2t), 2e^(-t); -e^(2t), -e^(-t)][ c_1; c_2] --------------------------------------------------------------------------- Since g(t) = [ e^(2t), -2e^(t); -e^(2t), e^t] g_1(t) = e^(2t) [1, -1] g_2(t) = e^t [ -2, 1] u' = [ -4, -6; 3, 5]u + e^(2t) [ 1, -1] v' = [ -4, -6; 3, 5]v + e^t [ -2, 1] U_p(t) = e^(2t) a 2e^(2t) a = [ -4, -6; 3, 5] (e^(2t)*a) + e^(2t) [1, -1] [2, 0; 0, 2]a = [ -4, -6; 3, 5]a + [ 1, -1] [ 6, 6; -3, -3]a = [ 1, -1] RREF [ 1, 1; 0, 0][ a_1; a_2] = [ 0, 1] Which yeilds equation: a_1 + a_2 = 0 a = [ 1, -1] Up(t) = e^(2t)a = [ e^(2t); -e^(2t)] --------------------------------------------------- Vp(t) = a*e^t a*e^t = [ -4, -6; 3, 5] (a*e^t) + e^t [ -2, 1] a = [ -4, -6; 3, 5]a + [ -2, 1] [1, 0; 0, 1]a = [ -4, -6; 3, 5]a + [ -2, 1] [ 5, 6; -3, -4]a = [ -2, 1] RREF [1, 0; 0, 1][ a_1; a_2] = [ -1; (1/2)] Which yeilds equations: a_1 = -1 a_2 = (1/2) a = [ -1; (1/2)] Vp(t) = a*e^t = [ -e^t; (1/2)e^t] ---------------------------------------------------- y_p(t) = u(t) + v(t) = [ e^(2t) - e^t; -e^(2t) + (1/2)e^t] General solution: y(t) = y_c(t) + y+p(t) =[ e^(2t), 2e^(-t); -e^(2t), -e^(-t)][c_1; c_2] + [ e^(2t) - e^t; -e^(2t) + (1/2)e^t] confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: Solve x '' = [ 6, 7; -15, -16] x YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x'' + Ax = 0 A = [ -6, -7; 15, 16] P(lambda) = det[A -lambda*I] = 0 = [(-6-lambda), -7; 15, (16 - lambda)] = (-6-lambda)(16 - lambda) - (-7)(15) = lambda^2 - (16)lambda + (6)lambda - 96 + 105 = 0 = lambda^2 - (10)lambda + 9 = (lambda - 9)(lambda - 1) Lambda_1 = 9 Lambda_2 = 1 ----------------------------------------------------- for lambda_1 = 9 we solve (A - 9I)x_1 = 0 [ -15, -7; 15, 7][x_1; x_2] = [0, 0] RREF [ 1, (7/15); 0, 0][x_1; x_2] = [0, 0] Which results in simple equations: x_1 + (7/15)x_2 = 0 A solution to this may be x_1 = [ 7, -15] ------------------------------------------------------ for lambda_2 = 1 we solve (A - I)x_2 = 0 [ -7, -7; 15, 15][x_1; x_2] = [0, 0] RREF [ 1, 1; 0, 0][x_1; x_2] = [0, 0] Which results in simple equations: x_1 + x_2 = 0 A solution to this may be x_1 = [ 1, -1] ----------------------------------------------------- T = [ 7, 1; -15, -1] x(t) = Tw(t) Tw(t)'' + ATw(t) = 0 w'' + Dw = 0 Now to diagonalize T^(-1)AT = D det[T] = (7)(-1) - (1)(-15) = 8 T^(-1) = (1/det(T)) [d, -b; -c, a] = (1/8) [ -1, -1; 15, 7] = [ (-1/8), (-1/8); (15/8), (7/8)] T^(-1)AT = D [ (-1/8), (-1/8); (15/8), (7/8)][ -6, -7; 15, 16][ 7, 1; -15, -1] = [9, 0; 0, 1] w'' + Dw = 0 becomes: (w_1)'' + 9(w_1) = 0 and (w_2)'' + (w_2) = 0 w_1 = Asin(3t) + Bcos(3t) w_2 = Asin(t) + Bcos(t) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!