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y = C₁ cos(3x) + C₂ sin(3x) is a two-parameter family of solutions of the second-order DE y" + 9y = 0. If possible, find a solution of the differential equation that satisfies the given side conditions. The conditions specified at two different points are called boundary conditions. (If not possible, enter IMPOSSIBLE.) y(0) = 1, y'(π) = 7 y =

y = C₁ cos(3x) + C₂ sin(3x) is a two-parameter family of solutions of the second-order DE y" + 9y = 0. If possible, find a solution of the differential equation that satisfies the given side conditions. The conditions specified at two different points are called boundary conditions. (If not possible, enter IMPOSSIBLE.) y(0) = 1, y'(π) = 7 y =

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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y = C₁ cos(3x) + C₂ sin(3x) is a two-parameter family of solutions of the second-order DE y" + 9y = 0. If possible, find a solution of the differential equation that satisfies the given side conditions. The conditions specified at two different points are called boundary conditions. (If not possible, enter IMPOSSIBLE.) y(0) = 1, y'(π) = 7 y =
Transcribed Image Text:y = C₁ cos(3x) + C₂ sin(3x) is a two-parameter family of solutions of the second-order DE y" + 9y = 0. If possible, find a solution of the differential equation that satisfies the given side conditions. The conditions specified at two different points are called boundary conditions. (If not possible, enter IMPOSSIBLE.) y(0) = 1, y'(π) = 7 y =
Solve equation (10) from Section 7.4 1 -= /[^/(2T) ( i(t) dt = E(t) (10) с subject to i(0) = 0 with L, R, C, and E(t) as given. L = 0.1 h, R = 3 , C = 0.05 f, E(t) = 100 [U(t − 1) – U(t − 2)] Ju(t-1) + ( i(t) L. + Ri(t) + dt = Jau (+-
Transcribed Image Text:Solve equation (10) from Section 7.4 1 -= /[^/(2T) ( i(t) dt = E(t) (10) с subject to i(0) = 0 with L, R, C, and E(t) as given. L = 0.1 h, R = 3 , C = 0.05 f, E(t) = 100 [U(t − 1) – U(t − 2)] Ju(t-1) + ( i(t) L. + Ri(t) + dt = Jau (+-
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