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172_pdfsam_math 54 differential equation solutions odd

172_pdfsam_math 54 differential equation solutions odd -...

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Chapter 4 Writing y ( t ) in the form y ( t ) = 5 2 5 sin 3 t + 1 5 cos 3 t = 5 sin(3 t + γ ) , where γ = arctan(1 / 2), we conclude that | y ( t ) | = 5 | sin(3 t + γ ) | , and so max | y ( t ) | = 5 (since max | sin(3 t + γ ) | = 1). 5. We differentiate y ( t ) twice and obtain y ( t ) = e 2 t sin( 2 t ) y ( t ) = e 2 t [( 2) sin( 2 t ) + 2 cos( 2 t )] y ( t ) = e 2 t ( 2) 2 sin( 2 t ) + ( 2) 2 cos( 2 t ) + ( 2) 2 cos( 2 t ) ( 2) 2 sin( 2 t ) = e 2 t 2 sin( 2 t ) 4 2 cos( 2 t ) . Substituting these functions into the differential equation, we get my + by + ky = y + 4 y + 6 y = e 2 t 2 sin( 2 t ) 4 2 cos( 2 t ) +4 e 2 t [( 2) sin( 2 t ) + 2 cos( 2 t )] + 6 e 2 t sin( 2 t ) = e 2 t (2 8 + 6) sin( 2 t ) + ( 4 2 + 4 2) cos( 2 t ) = 0 . Therefore, y = e 2 t sin( 2 t ) is a solution. As t + , e 2 t 0 while sin( 2 t ) remains bounded. Therefore, lim t + y ( t ) = 0. 7. For y = A cos 5 t + B sin 5 t , y = 5 A sin 5 t + 5 B cos 5
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