$80.00 Differential Equations help
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- Posted on Oct 30, 2009 at 11:36:52PM
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$30.00 Please confirm to me, I can help you.
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- Posted on Oct 30, 2009 at 11:42:36PM
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The full tutorial is about 8 words long .
$32.00 please confirm me by buying this tutorial. I can send you solution with 100 % accuracy before deadline.
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- Posted on Oct 31, 2009 at 01:01:38PM
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The full tutorial is about 18 words long .
$20.00 solutions with great steps
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- Posted on Oct 31, 2009 at 06:20:41PM
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The full tutorial is about 4 words long plus attachments.
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$10.00 sure A+ answer
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- Posted on Oct 31, 2009 at 07:03:48PM
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The full tutorial is about 8 words long plus attachments.
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$15.00 FULL WORK WITH DETAILED STEPS SHOWN OF ALL QUESTIONS
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- Posted on Oct 31, 2009 at 07:37:05PM
A:
Preview: ... of c1 and c2 in eqn(1)
y(t) =(2/3) e^(3t) + (1/3) e^(-3t)
b) lim ( t-> ∞) y(t) =(2/3) e^(∞) +(1/3) e^(-∞)
=∞ +0
=∞
=PINF
7(a) y" + 5y' + 4y =0 , y(0) =6 , y'(0) =1
so (D^2 + 5D +4 )y =0
auxillary eqn is
m^2 + 5m + 4 =0
so (m+1)(m+4) =0
so m =-1 , -4
solution is y(t) = c1 e^(-t) + c2 e^(-4t) ---------eqn(1)
y' (t)= -c1 e^(-t) -4c2 e^(-4t) ------------eqn(2)
when t =0 , y =6 , y' =1
putting the above values in eqn(1) and(2)
6 =c1 + c2 -----------eqn(3)
1 = -c1 -4c2
so c1 + 4c2 =-1 --eqn(4)
eqn(3) -eqn(4) is
-3c2 =7
c2 = -7/3
c1 = 6 + 7/3
=25/3
puttng the values of c1 and c2 in eqn(1)
y(t) =(25/3) e^(-t) -(7/3) e^(-4t)
b) y'(t) = -25/3 e^(-t) +(28/3) e^(-4t)
put y'(t) =0
so (28/3) e^(-4t) = (25/3) e^-t
so 28 e^t =25 e^4t
so e^3t =28/25
so 3t = ln(28/25)
so t = (1/3) ln (28/25)
at t = (1/3) ln (28/25) , y"(t) is negative
so t coordinate of the absolute maximum of the solution is ...
The full tutorial is about 793 words long .
