Recent content by aeroguy2008

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    Diffusion Problem (Conduction)

    Thank you for trying to help me. I have tried a lil bit but I got confused so I guess I wanna make sure that what I am doing is correct. X(x)=A*cos(lambda*x)+B*sin(lambda*x) X(x=0)=A*cos(0)+B*sin(0)=>A=0 X(x=2)=0=>B*sin(2*lambda)=0 Lambda=((n*Pi)/L)^2 so the eigenfunction EF...
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    Solve y''+9y=delta(t-pi)

    Yes now I understand what you mean. The answer to this problem is thus a/b=-1/2 :) This problem is now officially solved :D Defennder, thank you so much for your help. I learned alot!
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    Solve y''+9y=delta(t-pi)

    Thanks for all the help. I would say u(t-pi) goes to zero? So I will be left with y=1/3sin(t)+cos(3t) ... I cant get an exact fraction I get this to 1.05...how do you get exact fractions?
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    Solve y''+9y=delta(t-pi)

    So then my u(t-a)f(t) is equal to f(t) as Pi is smaller than (14*Pi)/9. Now I will only have: t=((14*pi)/9)=4.886921906 into cos(3t) & 1/3sin(3t)... cos(3t)=-0.5 1/3sin(3t)=-0.32827 f(t-a)=1.745329 y=1.745329-0.5-0.32827=0.917059 Is this correct?
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    Solve y''+9y=delta(t-pi)

    Ok I have read your message 20 times now...and I'm still confused. I follow until the fact that f(t-a) is the inverse Laplace transform of 1/(s^2+9) shifted by -a. So then f(t-a)=1/3sin(3(t-a))u(t-a)? Is a=pi? How do I get the fraction a/b out of all this? I'm gonna plug in some numerics and see...
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    Solve y''+9y=delta(t-pi)

    Can't anyone shed more light on this problem?
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    Solve y''+9y=delta(t-pi)

    I plugged in t=((14*pi)/9) into cos(3t) & 1/3sin(3t)... cos(3t)=-0.5 1/3sin(3t)=-0.32827 I still don't know what to to about the heaviside-term. How can I evaluate that numerically? And how do I get a fraction a/b from all this? Any ideas or suggestions?
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    Calculus question

    You did the same mistake I did when I started asking questions here and initially didn't show my attempt for a solution. As I did notice it myself later on...it actually helps if you show at least one of your attempts - even for self-reflection about the problem. Have you tried to cross-multiply?
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    Solve y''+9y=delta(t-pi)

    I see...is this what you mean? u(t-a)f(t-a)*(1/3)*sin(3t-a) Is the a-value for the shift pi? If this is true, will this be the expression for y: y=u(t-pi)f(t-pi)1/3sin(3t-pi)+1/3sin(3t)+cos(3t) How will I evaluate the first term in the expression above numerically?
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    Solve y''+9y=delta(t-pi)

    After some thinking about the Heaviside term I came to the conclusion that the last term in the attachment above should have an inverse Laplace transform of: u(t)f(t-pi)*(1/3)*sin(3t) If this is correct (please object if it is not) as I add all three terms I should get...
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    Solve y''+9y=delta(t-pi)

    I have the following... http://img71.imageshack.us/img71/4016/mathprob1na5.jpg" [Broken] But I'm not sure about that Heaviside function term. I really don't know what to do with (s^2+9) in denominator? Isn't there a jump (discontinuity in Heaviside functions)? I have not y yet. If I add these...
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    Solve y''+9y=delta(t-pi)

    Oh yeah and Laplace Transform of f(t-a)u(t-a) is exp(-a*s)F(s) is that correct now?
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    Solve y''+9y=delta(t-pi)

    The Laplace Transform of cos(at) is s/(s^2+a^2) The Laplace Transform of sin(at) is a/(s^2+a^2) and Laplace Transform of f(t-a)u(t-a) is exp(-Ts)F(s) Now I will try to inverse-transform them individually.
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    Solve y''+9y=delta(t-pi)

    Thanks for letting me know...I hope this works:) http://img57.imageshack.us/my.php?image=mathprobkv7.jpg"
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    Solve y''+9y=delta(t-pi)

    I just don't understand how these three fractions can give me y at the end. Is the approach of inverse transform incorrect? Cause, just by looking at s/(s^2+9) I can transform that to cos(3t) but what do I do with the rest. How can these three "add up" to give me the fraction a/b?
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