278_pdfsam_math 54 differential equation solutions odd

278_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
Chapter 5 “output A ” consists of two flows: one is going out of the system and the other one is going to the tank B. So, output A = x ( t ) 100 kg / L · (4 + 3) L / min = 7 x ( t ) 100 kg / min , and the Frst equation in (5.17) becomes dx dt =1 . 2+ y 100 7 x 100 . Similarly, the second equation in (5.17) can be written as dy dt = 3 x 100 3 y 100 . Rewriting this system in the operator form, we obtain ( D +0 . 07)[ x ] 0 . 01 y =1 . 2 , 0 . 03 x +( D +0 . 03)[ y ]=0 . (5.18) Eliminating y yields { ( D +0 . 07)( D +0 . 03) ( 0 . 01)( 0 . 03) } [ x ]=( D +0 . 03)[1 . 2] = 0 . 036 , which simpliFes to ( D 2 +0 . 1 D +0 . 0018 ) [ x ]=0 . 036 . (5.19) The auxiliary equation, r 2 +0 . 1 r +0 . 0018 = 0, has roots r 1 =
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online