#$&* course MTH 279 Query 27 Differential Equations*********************************************
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: Suppose that i + 1 is an eigenvalue of a matrix A and [-1 + i, i ] is a corresponding eigenvector. Find a fundamental set of real solutions to the equation y ' = A y. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: λ_1 = 1 + i v_1 = [-1+i; i] y(t) = e^((1+t)t)*[-1+i; i] + X We don’t need the second part of this solution, because we’ll break down the first eigenpair solution to two separate solutions. Using Euler’s Formula, we get: y(t) = e^(t)(cos(t) + i*sin(t))[-1+i; i] = [-e^(t)(cos(t)+sin(t)) + i*e^(t)(cos(t)-sin(t)); -e^(t)*sin(t) + i*e^(t)*cos(t)] This can be broken down into y(t) = [-e^(t)(cos(t) + sin(t)); -e^(t)sin(t)] + i[e^(t)cos(t)-e^(t)sin(t); e^(t)cos(t)]. Both matrix parts of this are real solutions to the original DE. We now have y(t) = c_1*[-e^(t)(cos(t) + sin(t)); -e^(t)sin(t)] + c_2*[e^(t)cos(t)-e^(t)sin(t); e^(t)cos(t)]. We can even check that this is a fundamental set of solutions by computing the Wronskian. W(t) = -e^(2t) ≠ 0 for all values of t. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: Solve the equation y ' = [0, -9; 1, 0] y with initial condition y(0) = [6, 2]. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A = [0, -9; 1, 0], Assume the solution will take the form y(t) = e^(λt)x, so y’ = λe^(λt)x, λe^(λt)x = Ae^(λt)x, e^(λt)(λx) = e^(λt)(Ax), λx = Ax, (A-λI)x = 0, x ≠ 0. det([-λ, -9; 1, -λ]) = 0, λ^2+9=0, λ = ±3i. (A-λI) = 0. For λ = -3i: [3i, -9; 1, 3i]. Using row reduction, we get: [i, -3; 0, 0]. This translates to x= -3iy. In vector form, we get v_1=[3;i]. y_1 = e^(-3it)*[3;i] = (cos(3t) - i*sin(3t))[3;i] = [3cos(3t); sin(3t)] + i*[-3sin(3t); cos(3t)] = u + iv. Both u(t) and v(t) are solutions of the system. We check that they are a fundamental set: W(t) = [u,v] = [3cos(3t), -3sin(3t); sin(3t), cos(3t)] = 3 ≠ 0 for all values of t. This shows that u and v do form a fundamental set. y(t) = [3cos(3t), -3sin(3t); sin(3t), cos(3t)][c_1;c_2]. [3c_1cos(3t) - 3c_2sin(3t); c_1sin(3t) + c_2cos(3t)] = [6;2]. 3c_1 = 6 c_1 = 2. c_2 = 2. y(t) = 2[3cos(3t); sin(3t)] + 2[-3sin(3t);cos(3t)] confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: Find all values of mu such that any fundamental set [ y_1, y_2 ] of the system y ' = [1, 3; mu, -2] y have the property that the limit of the expression (y_1(t))^2 + (y_2(t))^2, as t -< infinity, is zero. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: lim(e^(-(t^2))) as t ∞ = 0. As long as the solution takes a form of e^(-(t^2)), the limit will be 0. In order to get a solution that will take that form when squared, we’re looking for complex eigenvalues, since (±i)^2 = -1. It doesn’t really matter if we have the positive or negative eigenvalue, as long as it is complex. We find the eigenvalues as normal, det([1-λ, 3; µ, -2-λ]) = 0, λ^2 + λ - (3µ + 2) = 0. Using the quadratic formula, we get: -1/2 ± sqrt(9+12µ)/2. We know that we need the sqrt()/2 term to equal a multiple of i. Knowing this, we set up an equality: sqrt(9+12µ)/2 = ni. Solving, we get µ = -n^2/3 - 3/4. This means that n can be any real value. In that regard, µ can take any value such that -12µ is greater than 9. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: A particle moves in an unspecified force field in such a way that its position vector r(t) = x(t) i + y(t) j and the corresponding velocity vector v(t) = r ' (t) satisfy the equation v ' = 2 k X v Write this condition as a system v ' = A v, with v = [v_x; v_y]. If the particle starts at position r(0) = 2 i + j, v(0) = i + 2 j, find its position at t = 3 pi / 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I’m a bit stumped by this problem. The X in v’ = 2kXv is throwing me off. I’m not sure if that’s supposed to indicate the cross product, as in “2k crossed with v”, or if X is intended to be a separate matrix, or if 2k is even supposed to represent 2 units in the k direction. Since you’ve used i and j to denote directions in r(t), I’m assuming that k is the k direction. But the X is still throwing me off. I’ll just assume that it’s “2k crossed with v.”
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