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anonymous
 5 years ago
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
anonymous
 5 years ago
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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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 \[\tan (u)=\frac{20u}{(2u+5)(2u5)}\] solving for u>0 gives approximately 1.86,4.25 ... lambda = u^2 \[\lambda \approx 3.45, 18.06\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Excellent, thankyou :)
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