Question

find eigen values and functions:

Answer 1

\[y''+\lambda=0\]

Answer 2

y'(0)=0, y(L)=0

Answer 3

\[m^2=\pm\sqrt{\lambda}i\]

Answer 4

m= imean

Answer 5

\[y=c_1cos(\sqrt{\lambda}t)+c_2sin(\sqrt{\lambda}t)\]

Answer 6

\[y'(t)=(\sqrt{\lambda})-c_1sin(\sqrt{\lambda}t)+\sqrt{\lambda}c_2cos(\sqrt{\lambda}t)\]

Answer 7

y'(0)=0

Answer 8

\[c_2=0\]

Answer 9

since \[c_2=0\]

\[y=c_1cos(\sqrt{\lambda}t)\]

Answer 10

\[0=c_1cos(\sqrt{\lambda}t)\]

Answer 11

only happens when c1= 0 which is a trivial solution

Answer 12

or

Answer 13

\[\sqrt{\lambda}L=\frac{\pi}{2}+\pi n\]

Answer 14

\[\frac{\pi+2\pi n}{2}=L\sqrt{\lambda}\]

Answer 15

for some reason they get 2n-1)pi

Answer 16

but isn't thta the same as 2n+1)pi

Answer 17

only one is starting backwards and the other is forward

Answer 18

\[y''+\lambda=0\]\[m^2+\lambda=0\]\[(m+i\sqrt{\lambda})(m-i\sqrt{\lambda})=0\]\[m=\pm i\sqrt\lambda\]
{you already have this}

Answer 19

\[y_p=c_1\cos(\sqrt{\lambda}t)+c_2\sin(\sqrt{\lambda}t)\]\[y_p^\prime=-c_1\sqrt\lambda \sin(\sqrt{\lambda}t)+c_2\sqrt\lambda \cos(\sqrt{\lambda}t)\]\[y_P^{\prime\prime}=-c_1\lambda \cos\left(\sqrt{\lambda}t\right)- c_2\lambda\sin\left(\sqrt{\lambda}t\right)\]

Answer 20

\[y_p^\prime(t)=-c_1\sqrt\lambda \sin(\sqrt{\lambda}t)+c_2\sqrt\lambda \cos(\sqrt{\lambda}t)\]\[y_p^\prime(0)=-c_1\sqrt\lambda \sin(0)+c_2\sqrt\lambda \cos(0)=0\]\[y_p^\prime(0)=c_2\sqrt\lambda=0;\qquad\qquad c_2=0\]

Answer 21

\[y_p(t)=c_1\cos(\sqrt{\lambda}t)\]\[y_p(L)=c_1\cos(\sqrt{\lambda}L)=0\]
\[\qquad\qquad\sqrt\lambda L=\arccos\left(0\right)\]\[\qquad\qquad\sqrt\lambda L=\frac\pi2+n\pi\]

Answer 22

\[\sqrt\lambda L=\frac\pi2+n\pi=\frac{\pi+2\pi n}{2}\]

Answer 23

\[\sqrt\lambda L=\arccos\left(0\right)=\left(n+\frac1{2}\right)\pi=\left((n+1)-\frac1{2}\right)\pi=\left(n'-\frac1{2}\right)\pi\]
\[\]

Answer 24

your right the only difference is where n starts,
this is pretty arbitrary,
your answer is right

Answer 25

so the roots are the same as the eigen values?