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anonymous
 3 years ago
linear D.E of nth order with constant coefficient...help guys
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anonymous
 3 years ago
linear D.E of nth order with constant coefficient...help guys .

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1381226444353:dw

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1This WILL be messy... are you up for it? :D

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1For these kinds of differential equations, the general form of the solution is given by \[\Large f^{(h)}(x) + f^{(p)}(x)\] where the f^h is the general solution to the homogeneous equation while f^p is any particular solution of this equation. Can you find the general solution to the homogeneous equation?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1What does that mean? D:

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Oh wait... there's no y' term?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0get the higher order first D.E

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1So basically, it's \[\Large y''  11y = xe^x\]?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then m= square root of 11

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then get the vlue of y

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y=C{1}^{+\sqrt{11}x}+C{2}^{\sqrt{11}x}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Im right? what is the next get the other y which from Xe^x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0take derivative until second derivative

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ang get the value of A I stocked here

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0help guys if Im wrong :(

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Well, if it doesn't work with only one undetermined coefficient, did you try 2? Specifically this: \[\Large y = \color{blue}Axe^x + \color{red}Be^x\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0subst the value of Y" ,Y' and Y to the y′′+ y 12y

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0where did you get this y=Axex+Bex

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1are you sure the second term in the question isn't a derivative?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Oh by the way, I didn't actually "get" that Axe^x + Be^x, for all intents and purposes, it IS just a guess... see if it gets you the coefficients you want, since Axe^x alone doesn't seem to do it.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1if not (yes strange), then i think you (@melmel ) got the complementary homogenous solution \[\large f^h(x)=y_c(x)=C_{1}e^{+\sqrt{11}x}+C_{2}e^{\sqrt{11}x}\] {forgot to type the e's in the bases }

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1Which method to calculate the particular solution \[f^h(x)=y_p(x)\] are you going to use; undetermined coefficients, operator D , or variation of parameters, @melmel ?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1apparently offline :( Oh well, onwards and upwards XD
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