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anonymous
 5 years ago
can anyone say how to solve a differential equation using method of undetermined coefficients??
anonymous
 5 years ago
can anyone say how to solve a differential equation using method of undetermined coefficients??

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok( D^25D+6)y=e^3x+sinx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just tell me the assumed particular integral.....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0bad diff equation, what is D?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0D is the notation for operator d/dx....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y^{\prime\prime}5y^\prime+6y=\operatorname{e}^{3x}+\sin x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think one of the best ways to solve this equation is by using Laplace transformation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in the question it is specifically mentioned to use method of undetermined coefficients

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The particular solution will be the sum of particular solutions. Try Ae^(3x) + Bsin(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no it will not work as e^3x is a term in complementary function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0particular solution \[y=Ae^{\lambda1 x}+Be^{\lambda2 x}\], lambda are complex numbers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah...i just did it and it's a mess...too much to write out...buggin' out and going to bed...good luck.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I thought the others were going to help you. Since e^(3x) is part of the homogeneous solution, try xAe^(3x) as one part. Since you have a trig. function too, add to this Msin(x) + Ncos(x). So try,\[y_p=Axe^{3x}+M \sin x +N \sin x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think I ended up with A=1, M=N=(1/10). Try it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[C _{1} e^{2x} + C_{2}e^{3x} +0.1(\cos[x] + \sin[x])+xe^{3x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanx i ended up with the same solution guess the answer in the book was wrong
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