anonymous
  • anonymous
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
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Heeeellllpppppp lol
dumbcow
  • dumbcow
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 \[\lambda \approx 3.45, 18.06\]
anonymous
  • anonymous
Excellent, thankyou :-)

Looking for something else?

Not the answer you are looking for? Search for more explanations.