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

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lgbasalloteBest ResponseYou've already chosen the best response.3
it should be \[y_c = c_1 e^{x} + c_2 x e^{x}\] see your mistake?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.3
i was actually referring to the x by c2
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.3
if you have repeated roots you attach least power of x to the next arbitrary constant, remember?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
oh I'm sorry I see it now
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.3
so do you know what y_p should be now?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
what does that have to do with y_p? isn't y_p based on what the RHS of the original equation is?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.3
are you using undetermined coefficients or variation of parameters>
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.3
based on your question i presumed you were doing variation....which depends on the complementary solution....
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I don't quite know what either of those mean, but I trying to do it based on \[y_p=Ae^{bx}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
please don't shoot me :S
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I'm willing to learn =)
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.3
can you describe to me the method you're familiar with? does it involve deriving the yp?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
according to my book I'm doing it based on THE METHOD OF UNDETERMINED COEFFICIENTS
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
yeah lol I guess THE METHOD OF VARIATION OF PARAMETERS is the next section....I haven't started that yet :P
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.3
\[y_p = Ae^{bx}\] only works if you don't have repeated roots. for repeated roots, you use \[y_p = Ae^{bx} + Bxe^{bx}\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.3
because 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
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.3
does that make sense?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
"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
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.3
yes. that. but in this case, we're doing y_p
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[y_p=Ae^{x}+Bxe^{x}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
and then I just do the derivative twice to solve for A and B
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
What did I do wrong? I get 0=x \[y_p=Ae^{x}+Bxe^{x}\] \[y_p'=Ae^{x}Bxe^{x}(x1)\] \[y_p''=Ae^{x}+Be^{x}(x2)\]
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
y ' =Ae^(x) (x1)Be^(x) you have an extra x in there..
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
ok I'll try it again
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I still get a zero for x....I'll show my work, give me a second to type it :)
 one year ago

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

MathSofiyaBest ResponseYou've already chosen the best response.1
Why a \[Cx^2e^{x}\] ?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I think I'm gonna start a new thread. This one is getting too long
 one year ago

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

Algebraic!Best ResponseYou've already chosen the best response.0
so 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'.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
There'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.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
Qualitatively 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.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
Hope that helps; probably not, but I tried:)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
Yeah It kinda makes sense.
 one year ago
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