Second Order Equations-Reduction of Order: Problem 2 (1 point) Consider the differential equation which is linear with constant coefficients. First we will work on solving the corresponding homogeneous equation. The complementary homogenous equation (using m as your variable) is d'y dy 10- +25y = x dr2 da So, plugging y into the left side of the differential equation, and reducing, we get Because this is a repeated root, we don't have much choice but to use the exponential function corresponding to this root: to do reduction of order. y = ue5x Then (using the prime notation for the derivatives) y = Therefore the general solution is y = = 0 which has root y" - 10y + 25y = So now our equation is e5u". To solve for u we need only integrate ze-5 twice, using a as our first constant of integration and b as the second we get U= We knew from the beginning that e5* was a solution of the complementary homogeneous equation. We have worked out is that ze5* is another solution to the homogeneous equation, which is generally the case when we have multiple roots. Then 2+5 is the particular solution to the nonhomogeneous equation, and the general solution we derived is pieced together using superposition.

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Second Order Equations-Reduction of Order: Problem 2 (1 point) Consider the differential equation which is linear with constant coefficients. First we will work on solving the corresponding homogeneous equation. The complementary homogenous equation (using m as your variable) is d'y dy 10- +25y = x dr2 da So, plugging