Unformatted text preview: th this we can determine a specific solution to the differential equation. A = 20 ! T (t ) = e Kt +C + 20
(0, 95) ! 95 = e K (0 )+C + 20
! e K = 75
! K = ln 75 " 4.3175
! T (t ) = e Kt +ln 75 + 20 = e Kt e ln 75 + 20 = 75e Kt + 20
(20, 70) ! 70 = 75e K (20 ) + 20 ! 50 = 75e K (20 )
2
! = e20 K
3
#2&
! ln % ( = 20 K
$ 3'
1 #2&
! K = ln % ( " )0.0202732554
20 $ 3 '
t #2&
ln% (
! T (t ) = 75e 20 $ 3 ' + 20 t !2$
ln# &
The particular solution T (t ) = 75e 20 " 3 % + 20 , with T in degrees centigrade and t in minutes for the given conditions, can now be used to answer a wide variety of questions about the situation. (a) Determine the object’s temperature 45 min...
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This document was uploaded on 03/05/2014 for the course MATH 146 at Illinois State.
 Summer '09
 Calculus, Equations

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