- #1
Xyius
- 508
- 4
Homework Statement
[tex]y^{(4)}+\lambda y=0[/tex]
[tex]y(0)=y'(0)=0[/tex]
[tex]y(L)=y'(L)=0[/tex]
Homework Equations
The hint says...
[tex]let \lambda = -\mu ^4, \mu >0 or \lambda = 0[/tex]
The Attempt at a Solution
Listening to the hint, I got
[tex]r=\pm\mu[/tex] With multiplicity 2 of each. So that means..
[tex]y=c_1 e^{\mu t}+c_2te^{\mu t}+c_3e^{-\mu t}+c_4te^{-\mu t}[/tex]
Finding the derivative and solving the system of the 4x4 matrix (Thank god for the TI-89!) I find that all constants are equal to zero. When lambda equals zero I also get zero for all the constants.
The answer in the back of the book says..
for [tex]\lambda = -\mu^4[/tex] and
[tex]cos(\mu_n L)cosh(\mu_n L)=1[/tex]
[tex]y_n=c_n[(cos(\mu_n L)-cosh(\mu_n L))(sin(\mu_n x)-sinh(\mu_n x))-(sin(\mu_n L)-sinh(\mu_n L))(cos(\mu_n x)-cosh(\mu_n x))][/tex]
I do not understand where they got this. It looks like there are only 2 cases and each case has the trivial solution. Can anyone help me out?