Mathematics, 20.10.2020 21:01 briarwilliams9668
Solve the differential equation y" + 3y' + 2y = P1 where, y" = d2y/dx2, y' = dy/dx, P1 = P1(x)) and the initial conditions are y(0) = 0, y'(0) = 0. The input function is given by P1() = { 0, < 0 1, 0 ≤ ≤ 1 0, > 1
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Me answer this question: -2/3p + 1/5 - 1 + 5/6p i think the simplified expression is 1/6p - 4/5 correct me if i'm wrong, and explain it! if i have it right, just tell me. you so
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Solve the differential equation y" + 3y' + 2y = P1 where, y" = d2y/dx2, y' = dy/dx, P1 = P1(x)) and...
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