{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Ordinary Diff Eq Exam Review Solutions 25

# Ordinary Diff Eq Exam Review Solutions 25 - 1 Solutions 27...

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

1 Solutions 27 and simplifying gives y = (5+ t )+ c (5+ t ) 1 . The initial condition y (0) = 0 implies c = 25 so y = (5+ t ) 25(5+ t ) 1 ; for 0 t 45 . At t = T = 45 we get y (45) = 50 25 50 = 49 : 5 lbs salt. Once the tank is full, the inflow and outflow rates will be equal and the brine in the tank will (in the limit as t ! 1 ) stabilize to the concentration of the incoming brine, i.e., 0 : 5 lb/gal. Since the tank holds 100 gal, the total amount present will approach 0 : 5 100 = 50 lbs. Thus lim t !1 y ( t ) = 50 . 31. input rate: input rate = rc output rate: output rate = r P ( t ) V Let P 0 denote the amount of pollutant at time t = 0 . Since P 0 = input rate output rate it follows that P ( t ) is a solution of the initial value problem P 0 = rc rP ( t ) V ; P (0) = P 0 : Rewriting this equation in standard form gives the differential equation P 0 + r V P = rc . The coefficient function is p ( t ) = r=V and the integrating factor is ( t ) = e rt=V . Thus ( e rt V P ) 0 = rce rt V . Integrating and simplifying gives P ( t ) = cV + ke rt V ; where k is the constant of integration. The initial condition P (0) = P 0 implies c = P 0 cV so P ( t ) = cV + ( P 0 cV ) e rt V : (a) lim t !1 P ( t ) = cV: (b) When the river is cleaned up at t = 0 we assume the input con- centration is c = 0 . The amount of pollutant is therefore given by P ( t ) = P
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}