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246_pdfsam_math 54 differential equation solutions odd

# 246_pdfsam_math 54 differential equation solutions odd -...

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Chapter 4 where we take s = 1 because r = 3 i is a simple root of the auxiliary equation. Computing the derivatives y ( t ) = A cos 3 t + B sin 3 t + t ( 3 A sin 3 t + 3 B cos 3 t ) , y ( t ) = 6 B cos 3 t 6 A sin 3 t + t ( 9 A cos 3 t 9 B sin 3 t ) , and substituting y ( t ) and y ( t ) into the original equation, we get 6 B cos 3 t 6 A sin 3 t + t ( 9 A cos 3 t 9 B sin 3 t ) + 9 t ( A cos 3 t + B sin 3 t ) = 2 cos 3 t 6 B cos 3 t 6 A sin 3 t = 2 cos 3 t A = 0 , B = 1 / 3 . So, y p ( t ) = (1 / 3) t sin 3 t , and y ( t ) = C 1 cos 3 t + C 2 sin 3 t + (1 / 3) t sin 3 t is a general solution. To satisfy the initial conditions, we solve y (0) = C 1 = 1 , y (0) = 3 C 2 = 0 C 1 = 1 , C 2 = 0 . So, the solution to the given initial value problem is y ( t ) = cos 3 t + 1 3 t sin 3 t . The graph of
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Unformatted text preview: 5. (a) The corresponding homogeneous equation, my + ky = 0, is the equation of a simple harmonic motion, and so its general solution is given by y h ( t ) = C 1 cos ωt + C 2 sin ωt, ω = p k/m . Since γ 6 = ω , we look for a particular solution of the form y p ( t ) = A cos γt + B sin γt ⇒ y p ( t ) = − Aγ sin γt + Bγ cos γt ⇒ y p ( t ) = − Aγ 2 cos γt − Bγ 2 sin γt. 242...
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