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Solution 10112017

Content type
User Generated
Subject
Mathematics
Type
Homework
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1. Let the concentrate of the incoming solution be c (in unit of lb/gallon, to be determined), and
the amount of salt in the tank be y(t), the rate of change dy(t)/dt equals the amount of salt
adding to the tank per minutes at t, which is 5c, minus the amount of salt drained from the tank,
which equals to y(t)/100 *5 =y/20, thus we have
dy/dt = 5c- y(t)/20
or
dy/dt+ y /20 =5c
The initial condition is y (0) = 50.
The particular solution to the equation is y
p
= 100c, the general solution to the homogeneous
equation dy/dt + y/20 =0 is
y = Ce
-t/20
where C is arbitrary constant.
the general solution to the equation is
y= Ce
-t/20
+100c
the initial condition y(0) = 100c requires that
y(0) = 50 = C+100c
C = 50-100c
y= (50-100c)e
-t/20
+100c
=50 e
-t/20
+100c(1- e
-t/20
)

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t= 27 minutes, y = 233 lb requires that
233 =50 e
-27/20
+100c(1- e
-27/20
) = 12.96+100C*(1-0.259)
=12.96+74.08c
c = (233-12.96)/74.08
=2.97 lb/gallon
2. The general solution to the homogeneous equation
y’+y=0
y= Ce
-t
and a particular solution of the original equation can be found by method of variation of the
constant, let the particular solution be y = C(t)e
-t
, plugging it into the equation
C’(t) e
-t
C(t)e
-t
+C(t)e
-t
= tsint
C’(t) = te
t
sint
)coscossin(
2
1
sin)(
ttttte
tdttetC
t
t
+=
=
Thus the particular solution is
)coscossin(
2
1
ttttty +=
and the general solution to the equation y’+y= tsint is
t
Cettttty
++= )coscossin(
2
1
y(0) = 3/2 requires that
3/2 = ½ +C
C=1
The solution is
t
etttty
+= )cos)1(sin(
2
1
The graph of the solution:

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1. Let the concentrate of the incoming solution be c (in unit of lb/gallon, to be determined), and the amount of salt in the tank be y(t), the rate of change dy(t)/dt equals the amount of salt adding to the tank per minutes at t, which is 5c, minus the amount of salt drained from the tank, which equals to y(t)/100 *5 =y/20, thus we have dy/dt = 5c- y(t)/20 or dy/dt+ y /20 =5c The initial condition is y (0) = 50. The particular solution to the equation is yp = 100c, the general solution to the homogeneous equation dy/dt + y/20 =0 is y = Ce-t/20 where C is arbitrary constant. the general solution to the equation is y= Ce-t/20+100c the initial condition y(0) = 100c requires that y(0) = 50 = C+100c C = 50-100c y= (50-100c)e-t/20+100c =50 e-t/20+100c(1- e-t/20) t= 27 minutes, y = 233 lb requires that 233 =50 e-27/20+100c(1- e-27/20) = 12.96+100C*(1-0.259) =12.96+74.08c c = (233-12.96)/74.08 ...
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