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Finding the constants in a general solution 
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#1
Nov2010, 09:27 PM

P: 38

I have
x(t) = C(sub1) sin(16t) + C(sub2) cos(16t) Given: initial position x(0) = 1/6 so: 1/6 = C(sub1) sin(0) + C(sub2) cos(0) 1/6 = C(sub2) but how do i find C(sub1)? im not given initial velocity 


#2
Nov2010, 09:36 PM

P: 150

Well you have:
C1*sin(16t)+(1/6)cos(16t)=0 It's possible to isolate C1 here with simple algebra. What do you get when you try this? 


#3
Nov2010, 09:45 PM

P: 52

No, you don't have enough information to determine [tex]C_1[/tex].



#4
Nov2010, 09:50 PM

P: 38

Finding the constants in a general solution
C1 = (1/6)*(cos(16t)/sin(16t))
im a allowed to do that (set the eq. to zero)? also, the book answer is x(t) = (1/6)*cos(16t) so C1 must be zero, but i cannot solve for C1(cant plug in initial position 0 because that would be dividing by zero. 


#5
Nov2010, 09:59 PM

P: 52

C1 is not a constant if it depends on 1/tan(16t).
Besides, that comes from assuming that x(t) is zero everywhere, which you did not state in the problem. You have only presented one equation to extract information from: x(0)=1/6. It is not possible to determine both constants from one piece of information. Any value for [tex]C_1[/tex] is consistent with the information you have given us. Is there more? 


#6
Nov2010, 10:14 PM

P: 38

sooo sorry. I've been reading the problem over and over (it's actually a mass on spring problem). It uses the term "from rest", meaning initial velocity is 0. Now when i solve for the constants i get
0 = 16C1 and C1 = 0 Wow, an hour wasted because i missed that part 


#7
Nov2010, 10:34 PM

P: 52

Good start on any problem to count the number of unknowns and see what information you need to hunt 'em down :)



#8
Nov2110, 08:38 AM

Math
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PF Gold
P: 39,490

I once watched a calculus lesson on a local educational station. The problem was the typical "a rock is dropped from height....". In the middle of the problem the instructor said "We are not told the initial speed so we will take that to be 0." I nearly threw a brick through the television screen!



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