- #1
WendysRules
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Homework Statement
## y''+\lambda y = 0 ; y(0) = 0, y(\pi)-y'(\pi) = 0##
Homework Equations
The Attempt at a Solution
So, we have to test when lambda is equal to, less than and greater than 0.
Let ## \lambda = 0## thus, the ODE becomes ## y'' = 0 ## which implies solutions of the form ## y(t) = C_1t+C_2 ## which would make the derivative ## y'(t) = C_1 ##. When we apply our boundary conditions we see that ## y(0) = C_2 = 0 ## and ##y(\pi)-y'(\pi) = C_1 \pi - C_1= 0## therefore ##C_1 = 0## which are trivial solutions.
Let ## \lambda < 0 ## thus, the ODE becomes ## y''- \lambda y = 0 ## which implies solutions of the form ## y(t) = C_1e^{\sqrt{\lambda}t}+C_2e^{-\sqrt{\lambda}t} ## which would make the derivative ## y'(t) = \sqrt{\lambda}C_1e^{\sqrt{\lambda}t}-\sqrt{\lambda}C_2e^{-\sqrt{\lambda}t} ##. When we apply our boundary conditions we see that ## y(0) = C_1+C_2 = 0 ## therefore ## C_2=-C_1 ## and ## y(\pi)-y'(\pi) = C_1e^{\sqrt{\lambda}\pi}+C_2e^{-\sqrt{\lambda}\pi}-[\sqrt{\lambda}C_1e^{\sqrt{\lambda}\pi}-\sqrt{\lambda}C_2e^{-\sqrt{\lambda}\pi}].## If i make my substitution from the first boundary condition, we see this become ## C_1e^{\sqrt{\lambda}\pi}-C_1e^{-\sqrt{\lambda}\pi}-\sqrt{\lambda}C_1e^{\sqrt{\lambda}\pi}-\sqrt{\lambda}C_1e^{-\sqrt{\lambda}\pi}## The two positive square root terms cancel out and we get ## C_1[-e^{-\sqrt{\lambda}\pi}-e^{-\sqrt{\lambda}\pi} = 0 ## Since the exponentials can't ever be zero, it implies that ## C_1 = 0 ## this, these solutions are trivial.
But this is wrong, so I must've done something wrong in this step. If anyone could point out my error, I'd appreciate it. Am I allowed to say ##\sqrt{\lambda}C_1 = C_1##? Maybe I should convert to hyperbolic trig functions? That would be my guess, but I don't want to do more work if I just made a silly algebra error. I also got the next part wrong.
Let ## \lambda > 0 ## thus, the ODE becomes ## y''+\lambda y = 0 ## which implies solutions of the form ## y(t) = C_1\cos{\sqrt{\lambda}t}+ C_2\sin{\sqrt{\lambda}t}. ##Which would make the derivative ## y'(t) = -\sqrt{\lambda}C_1\sin{\sqrt{\lambda}t} + \sqrt{\lambda}C_2\cos{\sqrt{\lambda}t} ##. When I apply my boundary conditions, we see that ## y(0) = C_1 = 0 ## and ## y(\pi)-y'(\pi) = 0-C_2 \cos \pi \sqrt{\lambda} = 0 ## Now we solve for ## \lambda ## to see that ## \sqrt{\lambda} pi = \frac{npi}{2} ## which makes ## \lambda = \frac{n^2}{4} ## which would make our solutions ## y(t) = C \sin {\frac{n}{2}} t ## which is wrong, but I don't quite understand the answer for this one.
The solution I'm given for ## \lambda > 0 ## is ## y(t) = c \sin \sqrt{\lambda_n} t ## where ##\tan \sqrt{\lambda_n} \pi = \lambda_n ##
Should I start just using determinant to get the solutions to the system of equations? Or am I able to continue down the path as above?
THanks for any help.