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...
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!
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?
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?
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...
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?
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?
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?
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...
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...
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.
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?