MathSofiya
\[y''+2y'+y=xe^{-x}\]
\[r^2+2r+1=0\]
\[r_1=r_2=1\]
\[y_c=c_1e^{-x}+c_2e^{-x}\]
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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lgbasallote
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your y_c is wrong
MathSofiya
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it's -1
MathSofiya
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typo
MathSofiya
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mah bad...
lgbasallote
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it should be \[y_c = c_1 e^{-x} + c_2 x e^{-x}\]
see your mistake?
MathSofiya
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yeah r=-1
lgbasallote
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i was actually referring to the x by c2
lgbasallote
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if you have repeated roots you attach least power of x to the next arbitrary constant, remember?
MathSofiya
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oh I'm sorry I see it now
lgbasallote
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so do you know what y_p should be now?
MathSofiya
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what does that have to do with y_p? isn't y_p based on what the RHS of the original equation is?
lgbasallote
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are you using undetermined coefficients or variation of parameters>
lgbasallote
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based on your question i presumed you were doing variation....which depends on the complementary solution....
MathSofiya
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I don't quite know what either of those mean, but I trying to do it based on \[y_p=Ae^{bx}\]
MathSofiya
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please don't shoot me :S
MathSofiya
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I'm willing to learn =)
lgbasallote
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can you describe to me the method you're familiar with?
does it involve deriving the yp?
MathSofiya
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according to my book I'm doing it based on THE METHOD OF UNDETERMINED COEFFICIENTS
lgbasallote
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oh that
MathSofiya
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yeah lol I guess
THE METHOD OF VARIATION OF PARAMETERS
is the next section....I haven't started that yet :P
lgbasallote
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\[y_p = Ae^{bx}\] only works if you don't have repeated roots. for repeated roots, you use \[y_p = Ae^{bx} + Bxe^{bx}\]
lgbasallote
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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
lgbasallote
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does that make sense?
MathSofiya
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"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
lgbasallote
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yes. that. but in this case, we're doing y_p
MathSofiya
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\[y_p=Ae^{-x}+Bxe^{-x}\]
MathSofiya
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and then I just do the derivative twice to solve for A and B
lgbasallote
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sounds right
MathSofiya
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thank ya!
lgbasallote
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welcome
MathSofiya
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What did I do wrong? I get 0=x
\[y_p=Ae^{-x}+Bxe^{-x}\]
\[y_p'=-Ae^{-x}-Bxe^{-x}(x-1)\]
\[y_p''=Ae^{-x}+Be^{-x}(x-2)\]
Algebraic!
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y ' =-Ae^(-x) -(x-1)Be^(-x)
you have an extra x in there..
MathSofiya
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oh I see
MathSofiya
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ok I'll try it again
MathSofiya
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I still get a zero for x....I'll show my work, give me a second to type it :)
Algebraic!
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Hold up.
Algebraic!
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I think you need a C*x^2*e-x term as part of your solution...
MathSofiya
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e-x?
MathSofiya
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e^(-x)?
MathSofiya
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Why a
\[Cx^2e^{-x}\]
?
MathSofiya
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I think I'm gonna start a new thread. This one is getting too long
Algebraic!
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the problem is that your forcing function is the same as one of the solutions to the homogeneous equation.
Algebraic!
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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'.
Algebraic!
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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 non-homogeneous equation.
Algebraic!
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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.
Algebraic!
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Hope that helps; probably not, but I tried:)
MathSofiya
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Yeah It kinda makes sense.