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anonymous
 5 years ago
Having trouble with another boundary condition question:
y"+4y'+13y=0
Boundary condition:
y'(0) = 0 and y'(pi) = 1
I get the general solution:
y = e^(2x) (Acos3x + Bsin3x)
I think I may be doing this one wrong cause the last boundary condition doesn't work???
anonymous
 5 years ago
Having trouble with another boundary condition question: y"+4y'+13y=0 Boundary condition: y'(0) = 0 and y'(pi) = 1 I get the general solution: y = e^(2x) (Acos3x + Bsin3x) I think I may be doing this one wrong cause the last boundary condition doesn't work???

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Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0did you find the Lagrange multiplier?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1i will scan something and see if it helps you k? just a sec

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1do you see the two equations i got to solve for the constants

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1is that what you got?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1i will try to do it after i finish eating

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeh I got that, it's the y'(pi) bit i dont get, cheers enjoy your food...

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1I couldn't find the constants with those boundary conditions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So what would the solution look like?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1i think the solution that satisfies both boundary conditions and the different equation is y=0 but wait 1 does not equal 0 so let me think about this some more it can't simply be there is no solution or can it...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not sure, not come across a question where the boundary condition fails??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmmm, I may have to investigate further, don't really know enough about these types of questions yet??

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1no solution it is possible

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you think I should leave it as no solution then??

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1http://tutorial.math.lamar.edu/Classes/DE/BoundaryValueProblem.aspx it happened here

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1yes i would say no solution but show your work leading to that conclusion

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.1i would say there is based on the boundary conditions there is no solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Excellent, you've been ace, thanks...
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