247_pdfsam_math 54 differential equation solutions odd

247_pdfsam_math 54 differential equation solutions odd -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Exercises 4.9 Substitution into the original equation yields m ( 2 cos γt 2 sin γt ) + k ( A cos γt + B sin γt )= F 0 cos γt A ( 2 + k ) cos γt + B ( 2 + k ) sin γt = F 0 cos γt A = F 0 / ( k 2 ) , B =0 y p ( t )= F 0 k 2 cos γt. Therefore, a general solution to the original equation is y ( t )= C 1 cos ωt + C 2 sin ωt + F 0 k 2 cos γt. With the initial conditions, y (0) = y 0 (0) = 0, we get y (0) = C 1 + F 0 / ( k 2 )=0 , y 0 (0) = ωC 2 =0 C 1 = F 0 / ( k 2 ) , C 2 =0 . Therefore, y ( t )= F 0 k 2 cos ωt + F 0 k 2 cos γt, which can also be written in the form y ( t )= F 0 k 2 (cos γt cos ωt )= F 0 m ( ω 2 γ 2 ) (cos γt cos ωt ) . (b) Here one can apply the “diFerence-to-product” identity
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

Ask a homework question - tutors are online