Question

Need a few pointers plz?
Q. Find the first two positive eigenvalues of the following BVE : So lambda>0
y" + (lambda)y = 0
BV(1) : 5y(0) - 2y'(0) = 0
BV(2) : 5y(1) + 2y'(1) =0

Answer 1

Heeeellllpppppp lol

Answer 2

i haven't worked a lot with second order DE's but i think i figured this one out for you.
general solution to y" + (lambda)y = 0 is:
\[y(x) = c1\cos (ux) + c2\sin (ux)\]
where u = sqrt(lambda)
differentiate to get y'
\[y'(x) = -c1u \sin (ux) + c2 u \cos (ux)\]
apply initial conditions :
(i) 5/2*y(0) = y'(0)
(ii) -5/2*y(1) =y'(1)
yielding
\[ \frac{5}{2}c1=uc2 \rightarrow c1 =\frac{2uc2}{5}\]
\[-\frac{5}{2}c1\cos (u)-\frac{5}{2}c2\sin (u) = -uc1\sin (u) + uc2\cos (u)\]
\[\rightarrow \sin (u)(-\frac{5}{2}c2+uc1) = \cos (u)(\frac{5}{2}c1+uc2)\]
\[\rightarrow \frac{\sin (u)}{\cos (u)}=\frac{\frac{5}{2}c1+uc2}{uc1-\frac{5}{2}c2}\]
substitute in value of c1 from (i) for c1 in (ii)
simplifying leaves
\[\tan (u)=\frac{20u}{(2u+5)(2u-5)}\]
solving for u>0 gives approximately 1.86,4.25 ...
lambda = u^2
\[\lambda \approx 3.45, 18.06\]

Answer 3

Excellent, thankyou :-)