11 Linear homogeneous equations: Problem 9 (1 point) -92 are solutions to the differential equation It can be shown that

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answerhappygod
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11 Linear homogeneous equations: Problem 9 (1 point) -92 are solutions to the differential equation It can be shown that

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11 Linear Homogeneous Equations Problem 9 1 Point 92 Are Solutions To The Differential Equation It Can Be Shown That 1
11 Linear Homogeneous Equations Problem 9 1 Point 92 Are Solutions To The Differential Equation It Can Be Shown That 1 (30.65 KiB) Viewed 32 times
11 Linear Homogeneous Equations Problem 9 1 Point 92 Are Solutions To The Differential Equation It Can Be Shown That 2
11 Linear Homogeneous Equations Problem 9 1 Point 92 Are Solutions To The Differential Equation It Can Be Shown That 2 (61.28 KiB) Viewed 32 times
11 Linear homogeneous equations: Problem 9 (1 point) -92 are solutions to the differential equation It can be shown that yı = 3+ and y2 = e D'y+6Dy - 27y=0. W(y1, y2) = C1yı + c2y2 is the general solution to the equation on the interval
11 Linear homogeneous equations: Problem 10 (1 point) One of the following is a general solution of the homogeneous differential equation y' - *- 2y=0 y= ae" + be y=ae + be-22 y=ae y=ae-12 + be-22 -12 + be- -la and one of the following is a solution to the nonhomogeneous equation y" - y - 2y = 2e+ y=-3e* y=-4e* y=-2e y=-e By superposition, the general solution of the equation y' - y - 2y = 2e is y= Find the solution with y(0) = 5 7(0) = 2 y= Being the highly trained differential equations fighting machine that I am, I checked the Wronskian (using the solutions to the homogeneous equation without the coefficients a and b) for good measure, and found it to The fundamental theorem tells me that that this is the unique solution to the IVP on the interval be Note: You can earn partial credit on this problem.
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