#$&* course MTH 279 Query 26 Differential Equations*********************************************
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: Find the solutions to y ' = A y when A = [ 4,2,0; 0,1,3; 0,0, -2 ] and y(0) = [-1;0;3]. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Assume y(t) = e^(λt)x y’ = λe^(λt)x Ay = Ae^(λt)x = e^(λt)(Ax) y’ = Ay e^(λt)(λx) = e^(λt)(Ax) λx = Ax (A-λI)x = 0 det(A-λI) = 0 det([4-λ, 2, 0; 0, 1-λ, 3; 0, 0, -2-λ]) = -λ^3 + 3λ^2 + 6λ - 8 = 0 λ = -2, 1, 4 For λ = -2: [4+2, 2, 0; 0, 1+2, 3; 0, 0, -2+2] = [6, 2, 0; 0, 0, 3; 0, 0, -3] Using row reduction, we get [3,1,0; 0,1,1; 0, 0,0]. We interpret this as x= -y/3, and y = -z. Relating this into a vector, we get v_1=[1;-3;3]. For λ = 1: [3,2,0; 0,0,3;0,0,-3]. Using row reduction, we get [3,2,0;0,0,3;0,0,0]. We interpret this, and put it into vector form to get v_2=[-2,3,0]. λ = 4: [0,2,0; 0,-3,3; 0,0,-6]. Using row reduction, we get [0,0,0;0,1,0;0,0,1]. We interpret this into algebra, and find that both y and z are 0, with x unrestricted. We can choose any value of x, because it is the same vector, just different lengths. We choose x=1 for simplicity. v_3 = [1;0;0]. y(t) = [e^(-2t), -2e^(t), e^(4t); -3e^(-2t), 3e^(t), 0; 3e^(-2t), 0, 0][c_1;c_2;c_3] Imposing the initial conditions, setting up y(t) as an augmented matrix, and using row reduction methods, we get: c_1 = 1, c_2 = 1, c_3 = 0. y(t) = e^(-2t), -2e^(t), e^(4t); -3e^(-2t), 3e^(t), 0; 3e^(-2t), 0, 0][1;1;0] confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: Three tanks, each full of a solution and all of equal volume V, are each connected to each of the others by two pipes. Each tank also has a third pipe through which pure water flows into it, and a fourth through which water exits. The flow rate r through every pipe is the same. Write the system of equations that relates the quantities Q_1, Q_2 and Q_3 representing the amount of solute in each tank as a function of time. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Each tank has an input of pure water, which contributes no solution, so does not show in the system. Each tank has two inputs, contributing solution from the other 2 tanks, which has concentration Q/V for each respective tank. Each tank also has a total of 3 outlets, which take concentration away at Q/V rate. When this is translated into a system of equations, we have Q_1’ = rQ_2/v_2 + rQ_3/v_3 - 3rQ_1/v_1 for example. We write the other two equations and set up a system in matrix form as below. [Q_1’;Q_2’;Q_3’] = [-3r/v_1, r/v_2, r/v_3; r/v_1, -3r/v_2, r/v_3; r/v_1, r/v_2, -3r/v_3][Q_1;Q_2;Q_3] confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK"