#$&* course Mth 279 5/17 Query 25 Differential Equations*********************************************
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Determine whether the set of solutions {y_1, y_2, y_3} is linearly independent, where y_1 = [ 1, sin^2(t), 0] y_2 = [ 0, 2 - 2 cos^2(t), -2] y_3 = [ 1, 0, 1] YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: psi(t) = [ 1, 0, 1; sin^2(t), 2-2cos^2(t), 0; 0, -2, 1] W(t) = det[psi(t)] =(1)[ 2-2cos^2(t), 0; -2, 1] - (0)[sin^2(t), 0; 0, 1] + (1)[ sin^2(t), 2-2cos^2(t); 0, -2] = 2-2cos^2(t) - 2sin^2(t) = -2(-1+cos^2(t) + sin^2(t)) = -2(-1 + 1) = -2*0 = 0 Since W(t) = 0 this is not a fundamental set, which makes it linearly dependent. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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: Since psi ' (t) = P(t) psi(t) p(t) = psi ' (t) * psi^(-1)(t) psi^(-1)(t) = (1/t^2)*[ 1, -2t; 0, t^2] = [ 1/t^2, -2/t; 0, 1] psi ' (t) = [ (t^2)', (2t)'; (0)', (1)' ] = [ 2t, 2; 0, 0] psi ' (t) * psi^(-1)(t) = p(t) [ 2t, 2; 0, 0]*[ 1/t^2, -2/t; 0, 1] = [ 2t, -2; 0, 0] which satisfies psi ' (t) = p(t) * psi(t) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: If the matrix psi(t) = [y_1, y_2] = [e^t, e^(-t); e^t, - e^(-t)]: What are the vector functions y_1 and y_2? Write out the system of two differential equations represented by the equation y ' = P(t) y with P(t) = [0, 1; 1, 0]. Show that y_1 and y_2 are both solutions of the equation y ' = P(t) y with P(t) = [0, 1; 1, 0]. Show that { y_1 , y_2} is a fundamental set for this equation. Show that the matrix psi(t) is a solution of the matrix equation psi ' = P(t) psi. Show that the matrix psi(t) is a fundamental matrix for the linear system of equations. Let psi_hat(t) = [ 2 e^t - e^(-t), e^t + 3 e^(-t); 2 e^t + e^(-t), e^t - 3e^(-t) ]. Find a constant matrix C such that psi_hat(t) = psi(t) * C. Based on your matrix C, is psi_hat(t) a solution matrix for the system? Based on your matrix C, is psi_hat(t) a fundamental matrix for the system? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: p(t) = psi ' (t) * psi^(-1)(t) psi(t) = [ e^t, e^(-t); e^t, -e^(-t)] psi ' (t) = [ e^t, -e^(-t); e^t, e^(-t)] psi^(-1) = 1/(-2) * [ -e^(-t), -e^(-t); -e^t, e^t] = [(1/2)e^(-t), (1/2)e^(-t); (1/2)e^t, (-1/2)e^t] psi ' (t) * psi^(-1) = p(t) [ e^t, -e^(-t); e^t, e^(-t)]*[(1/2)e^(-t), (1/2)e^(-t); (1/2)e^t, (-1/2)e^t] = [ 0, 1; 1, 0] this proves that y1, and y2 are solutions. To Check: psi ' (t) = p(t)*psi(t) [ 0, 1; 1, 0]*[ e^t, e^(-t); e^t, -e^(-t)] = [ e^t, -e^(-t); e^t, e^(-t)] This proves to check out! psi(t) = [ e^t, e^(-t); e^t, -e^(-t)] W(t) = det[psi(t)] cannot be zero W(t) = -1 -1 = -2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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) = det[psi(t)] is not zero. = e^(-t) psi(0) = [ 1, 1; 0, 1] psi_hat(0) = I = psi(t) * C [1, 0; 0, 1] = [ 1, 1; 0, 1]*C psi_hat(0) * psi^(-1)(0) = C [1, 0; 0, 1]*[1, -1; 0, 1] = [1, -1; 0, 1] Check: psi(0) * C = psi_hat(0) [ 1, 1; 0, 1] * [1, -1; 0, 1] = [ 1, 0; 0, 1] Which checks out! confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:"