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anonymous
 4 years ago
\[y''+2y'+y=xe^{x}\]
\[r^2+2r+1=0\]
\[r_1=r_2=1\]
\[y_c=c_1e^{x}+c_2e^{x}\]
for y_p would I have (before I derive it and solve for A)
\[y_p=Ae^{x}\]
or
\[y_p=Axe^{x}\]
?
anonymous
 4 years ago
\[y''+2y'+y=xe^{x}\] \[r^2+2r+1=0\] \[r_1=r_2=1\] \[y_c=c_1e^{x}+c_2e^{x}\] for y_p would I have (before I derive it and solve for A) \[y_p=Ae^{x}\] or \[y_p=Axe^{x}\] ?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it should be \[y_c = c_1 e^{x} + c_2 x e^{x}\] see your mistake?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i was actually referring to the x by c2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if you have repeated roots you attach least power of x to the next arbitrary constant, remember?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh I'm sorry I see it now

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so do you know what y_p should be now?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what does that have to do with y_p? isn't y_p based on what the RHS of the original equation is?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0are you using undetermined coefficients or variation of parameters>

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0based on your question i presumed you were doing variation....which depends on the complementary solution....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't quite know what either of those mean, but I trying to do it based on \[y_p=Ae^{bx}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0please don't shoot me :S

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm willing to learn =)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you describe to me the method you're familiar with? does it involve deriving the yp?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0according to my book I'm doing it based on THE METHOD OF UNDETERMINED COEFFICIENTS

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah lol I guess THE METHOD OF VARIATION OF PARAMETERS is the next section....I haven't started that yet :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y_p = Ae^{bx}\] only works if you don't have repeated roots. for repeated roots, you use \[y_p = Ae^{bx} + Bxe^{bx}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because like i said before, if you have repeated roots, you attach the least power of x to the succeeding constant. Since you have xe^x on the right side, there must be an e^x before it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0does that make sense?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"least power of x to the succeeding constant" meaning \[y_c = c_1 e^{x} + c_2 x e^{x}\] referring to the x after the c_2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes. that. but in this case, we're doing y_p

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y_p=Ae^{x}+Bxe^{x}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and then I just do the derivative twice to solve for A and B

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What did I do wrong? I get 0=x \[y_p=Ae^{x}+Bxe^{x}\] \[y_p'=Ae^{x}Bxe^{x}(x1)\] \[y_p''=Ae^{x}+Be^{x}(x2)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0y ' =Ae^(x) (x1)Be^(x) you have an extra x in there..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I still get a zero for x....I'll show my work, give me a second to type it :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think you need a C*x^2*ex term as part of your solution...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Why a \[Cx^2e^{x}\] ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think I'm gonna start a new thread. This one is getting too long

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the problem is that your forcing function is the same as one of the solutions to the homogeneous equation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so without a term that doesn't end up being zero, you're not going to be able to ever get your DE to be 'forced'.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There's a quantitative way to justify it, but I can't recall how it goes. The point is xe^x is the forcing term and one of the solutions to the homogeneous. So you need another term in your yp if you're ever going to get an actual solution that satisfies the original nonhomogeneous equation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Qualitatively speaking, you need a faster transient, because the forcing term is resonant with the system, so the whole thing is getting driven to x*e^x harder than any of the terms you're using now can drive it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hope that helps; probably not, but I tried:)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah It kinda makes sense.
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