#$&* course MTH-279 08/03 around 11:20 PM Query 25 Differential Equations*********************************************
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: Determine whether the set of solutions {y_1, y_2, y_3} is linearly independent, where y_1 = [ 1, sin^2(t), 0] y_2 = [ 0, 2 - 2 cos^2(t), -2] y_3 = [ 1, 0, 1] YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: psi(t) = [1, 0, 1; sin2t, 2-2cos^2(t), 0 ; 0, -2, 1] W(t) = det psi(t) = 1 [2-2cos^2(t), 0 ; -2, 1] + 0 [sin^2(t), 0 ; 0, 1] + 1 [sin^2(t), 2-2cos^2(t); 0, -2] 1 * [(2-scos^2(t)*1) - (-2*0)] + 1 * [(sin^2(t) * -2) - (0 * 2 - 2cos^2(t))] = 1 * [2 - 2cos^2(t)] + 1* [-2sin^2(t)] --- = 2 - 2cos^2(t) - 2 sin^2(t) -> -2(cos^2(t) + sin^2(t) - 1) -> -2 (1-1) = 0, not linearly independent Confidence rating: 2,3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: Determine whether there is a matrix P(t) such that y_1 = [ t^2, 0 ] y_2 = [ 2t, 1 ] is a fundamental set of solutions to the equation y ' = P(t) y. If so, find such a matrix P(t). Hint: The matrix psi(t) = [y_1, y_2 ] = [ t^2, 2 t; 0, 1 ] would need to satisfy psi ' (t) = P(t) psi(t). In standard notation we could write this as follows: satisfies YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: psi(t) = [t^2, 2t; 0, 1] note: inverse of 2x2 matrix is 1/det(A) [d, -b ; -c, a] --- y’ = p(t)y psi’(t) = p(t)psi(t) p(t) = psi’(t) * psi^-1(t) psi’(t) = [2t, 2; 0, 0] 1/psi(t) = 1/t^2 [ 1, -2t; 0, t^2] = [-1/t^2, -2t/t ; 0, 1] p(t) = psi’(t) * psi^-1(t) = [2t, 2; 0, 0] * [-1/t^2, -2t/t; 0, 1] = [2/t + 0, -4 + 2 ; 0 + 0, 0 + 0] = [2/t, -2; 0, 0] -> p(t) Confidence rating: 3
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Given Solution: Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: If the matrix psi(t) = [y_1, y_2] = [e^t, e^(-t); e^t, - e^(-t)]: What are the vector functions y_1 and y_2? Write out the system of two differential equations represented by the equation y ' = P(t) y with P(t) = [0, 1; 1, 0]. Show that y_1 and y_2 are both solutions of the equation y ' = P(t) y with P(t) = [0, 1; 1, 0]. Show that { y_1 , y_2} is a fundamental set for this equation. Show that the matrix psi(t) is a solution of the matrix equation psi ' = P(t) psi. Show that the matrix psi(t) is a fundamental matrix for the linear system of equations. Let psi_hat(t) = [ 2 e^t - e^(-t), e^t + 3 e^(-t); 2 e^t + e^(-t), e^t - 3e^(-t) ]. Find a constant matrix C such that psi_hat(t) = psi(t) * C. Based on your matrix C, is psi_hat(t) a solution matrix for the system? Based on your matrix C, is psi_hat(t) a fundamental matrix for the system? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Confidence rating: 2
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Given Solution: Did this on paper, quite a bit of work, but got answers and concepts. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: Given the system y ' = [ 1, 1; 0, -2 ] y verify that psi(t) = [ e^t, e^(-2 t); 0, e^(-2 t) ] is a fundamental matrix for the system. Find a matrix C such that psi_hat(t) = psi(t) * C is a solution matrix satisfying initial condition psi_hat(0) = I, where I is the identity matrix. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: W(t) = (e^t * e^-2t) - (0 - e^-2t) -> e^-t - 0 = e^-t, so is a fundamental set --- psi^(0) = psi(0) + C I = [1, 0 ; 0, 1] = [1, 1; 0, 1] * C ---rearrange C = psi^(0) + psi^-1(0) C = [1, 0; 0, 1] * 1/1 [1, -1; 0, 1] C = [1, 0; 0, 1] * [1, -1; 0, 1] = [1+0, -1+0 ; 0+0, 0+1] = [1, -1; 0, 1] = C Confidence rating: 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK Self-critique rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!