MathSofiya
Solve using variation of parameters
\[y''-2y'+y=e^{2x}\]
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TuringTest
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complimentary....
MathSofiya
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meaning?
MathSofiya
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\[y_c=c_1e^{-x}+c_2xe^{-x}\]
TuringTest
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3
right...
TuringTest
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now we need the wronskian, do you know about that?
MathSofiya
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nope
TuringTest
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it is the determinant of a matrix
TuringTest
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for\[y_c=c_1y_1+c_2y_2\]the Wronskian is\[W(y_1,y_2)=\left|\begin{matrix}y_1&y_2\\y_1'&y_2'\end{matrix}\right|\]
TuringTest
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that's a determinant in case you forgot, so find that determinant
MathSofiya
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\[\left|\begin{matrix}e^{-x}&xe^{-x}\\-e^{-x}&-e^{-x}(x-1)\end{matrix}\right|\]
=\[e^{-x}(e^{-x}(x-1))-xe^{-x}e^{-x}\]
MathSofiya
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completely wrong?
MathSofiya
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oh it's plus
TuringTest
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yep you caught it :)
MathSofiya
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\[e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}\]
MathSofiya
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that's it?
TuringTest
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you dropped a minus
MathSofiya
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where?
MathSofiya
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ohhhh
MathSofiya
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\[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}\]
TuringTest
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right
TuringTest
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simplify and be happy :)
MathSofiya
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does it equal\[ e^{-x}\]?
\[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}=e^{-x}\]
\[{-x}(e^{-x}(x-1))+xe^{-x}=1\]
MathSofiya
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\[(e^{-x}(x-1))+xe^{-x}=1\]
MathSofiya
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typo
TuringTest
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no, why did you set it equal to e^-x, we can't set it equal to anything yet...
MathSofiya
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oh my bad haha
MathSofiya
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so how should I simplify it then?
TuringTest
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\[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}=e^{-x}(e^{-x}(1-x))+xe^{-x}e^{-x}\]now try again while I try to eat ;)
MathSofiya
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-(x-1)=1-x
1-x=1-x
MathSofiya
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no no no ...wait a second
MathSofiya
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now i get 2-x=2-x
MathSofiya
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nope still wrong
MathSofiya
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x=1
MathSofiya
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stop laughing
MathSofiya
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Let's try that again
\[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}=e^{-x}(e^{-x}(1-x))+xe^{-x}e^{-x}\]
MathSofiya
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they're just equal to each other
MathSofiya
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\[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}\]
MathSofiya
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\[-e^{-2x}(x-1)+xe^{-2x}\]
MathSofiya
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\[e^{-2x}((1-x)+x)\]
MathSofiya
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\[e^{-2x}\]
MathSofiya
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yes?
TuringTest
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yes, sorry, I was afk
MathSofiya
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LOL ok
TuringTest
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do you know the formula for the particular solution for variation of parameters?
MathSofiya
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MathSofiya
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something like that?
MathSofiya
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is the first one a y''?
\[y''+q(t)y'+r(t)y=g(t)\]
TuringTest
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yeah, it looks screwy I know
MathSofiya
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do we know what q(t) and r(t) are? are they the coefficients of y' and y? Probably not it looks like
TuringTest
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yes and in this case they are the coefficients, which in this case are constants
the formula is just pointing out that they may not be
MathSofiya
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oh I see, soo...
\[y''+q(t)y'+r(t)y=g(t)\]
\[y''-2y'+y=e^{2x}\]
so now I integrate
TuringTest
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3
right, using the formula...
r(t) and r(t) don't even come into the formula for the particular solution in variation of parameters
TuringTest
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q(t) and r(t)...
they determine the complimentary only, which as you can see *is* part of the formula for the particular
MathSofiya
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\[Y_p(t)=-e^{-x}\int\frac{xe^{-x}e^{-x}}{e^{-2x}(e^{-x},xe^{-x})}dx+xe^{-x}\int\frac{e^{-x}e^{2x}}{e^{-2x}(e^{-x},xe^{-x})}dx\]
TuringTest
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3
the first integrand is wrong I think
TuringTest
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I think you put the wrong g(x)
MathSofiya
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\[Y_p(t)=-e^{-x}\int\frac{xe^{-x}e^{2x}}{e^{-2x}(e^{-x},xe^{-x})}dx+xe^{-x}\int\frac{e^{-x}e^{2x}}{e^{-2x}(e^{-x},xe^{-x})}dx\]
TuringTest
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oh wait, wait, the denominator in each integrand is the wronskian, which we found to be e^(-2x)
I think you are confusing notation with the parentheses
MathSofiya
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Oh lol so the denominator is just e^{-2x}
TuringTest
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right
TuringTest
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notice that makes our integrals quite tolerable :)
MathSofiya
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\[Y_p(t)=-e^{-x}\int\frac{xe^{-x}e^{2x}}{e^{-2x}}dx+xe^{-x}\int\frac{e^{-x}e^{2x}}{e^{-2x}}dx\]
That's a lot easier to integrate
TuringTest
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oh yeah :) especially after a little simplification
MathSofiya
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\[-\frac{e^{2x}}{9}(3x-1)+\frac x3e^{3x}\]
MathSofiya
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so what was the point of doing variation of parameters
TuringTest
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3
well now you have a particular solution
TuringTest
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y(x)=yc+yp
you got you now so you're good to go
TuringTest
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3
you got yp*
TuringTest
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I did not check your integral btw
MathSofiya
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that's ok I did wolfram
TuringTest
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right-o, I'd do the same right now :)
MathSofiya
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oh I see
\[c_1e^{-x}+c_2xe^{-x}-\frac{e^{2x}}{9}(3x-1)+\frac x3e^{3x}\]
TuringTest
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yep, and of course if this is an IVP we now proceed to find c1 and c2
otherwise we're done having found the general solution
MathSofiya
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good, wow, that was a lot
THanks Turing!
TuringTest
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Always a pleasure Sofiya :D
Mendicant_Bias
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Looking back over this, I don't understand how you guys got a negative double root to the characteristic equation. I got m_1 = m_2 = +1. Plugging into the normal formula for repeated roots, getting e^x and xe^x, not e^-x. Am I missing something? @ganeshie8
Mendicant_Bias
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@jim_thompson5910 @agent0smith
Mendicant_Bias
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@Zarkon @terenzreignz Is anybody else familiar with ODE's seeing this, or am I missing something?
Mendicant_Bias
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@zepdrix Anybody?
zepdrix
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@Mendicant_Bias For the homogeneous solution? Yes the characteristic equation appears to be giving +1 as a repeated root, as you indicated
zepdrix
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So yah, that was a mistake early in the thread :C