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Solve using variation of parameters \[y''-2y'+y=e^{2x}\]

Mathematics
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complimentary....
meaning?
\[y_c=c_1e^{-x}+c_2xe^{-x}\]

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Other answers:

right...
now we need the wronskian, do you know about that?
nope
it is the determinant of a matrix
for\[y_c=c_1y_1+c_2y_2\]the Wronskian is\[W(y_1,y_2)=\left|\begin{matrix}y_1&y_2\\y_1'&y_2'\end{matrix}\right|\]
that's a determinant in case you forgot, so find that determinant
\[\left|\begin{matrix}e^{-x}&xe^{-x}\\-e^{-x}&-e^{-x}(x-1)\end{matrix}\right|\] =\[e^{-x}(e^{-x}(x-1))-xe^{-x}e^{-x}\]
completely wrong?
oh it's plus
yep you caught it :)
\[e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}\]
that's it?
you dropped a minus
where?
ohhhh
\[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}\]
right
simplify and be happy :)
does it equal\[ e^{-x}\]? \[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}=e^{-x}\] \[{-x}(e^{-x}(x-1))+xe^{-x}=1\]
\[(e^{-x}(x-1))+xe^{-x}=1\]
typo
no, why did you set it equal to e^-x, we can't set it equal to anything yet...
oh my bad haha
so how should I simplify it then?
\[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}=e^{-x}(e^{-x}(1-x))+xe^{-x}e^{-x}\]now try again while I try to eat ;)
-(x-1)=1-x 1-x=1-x
no no no ...wait a second
now i get 2-x=2-x
nope still wrong
x=1
stop laughing
Let's try that again \[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}=e^{-x}(e^{-x}(1-x))+xe^{-x}e^{-x}\]
they're just equal to each other
\[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}\]
\[-e^{-2x}(x-1)+xe^{-2x}\]
\[e^{-2x}((1-x)+x)\]
\[e^{-2x}\]
yes?
yes, sorry, I was afk
LOL ok
do you know the formula for the particular solution for variation of parameters?
something like that?
almost unidentifiable to me in that form, but correct I'm sure I have the more direct formula memorized http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx (the green box with the formula that says "variation of parameters)
is the first one a y''? \[y''+q(t)y'+r(t)y=g(t)\]
yeah, it looks screwy I know
do we know what q(t) and r(t) are? are they the coefficients of y' and y? Probably not it looks like
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
oh I see, soo... \[y''+q(t)y'+r(t)y=g(t)\] \[y''-2y'+y=e^{2x}\] so now I integrate
right, using the formula... r(t) and r(t) don't even come into the formula for the particular solution in variation of parameters
q(t) and r(t)... they determine the complimentary only, which as you can see *is* part of the formula for the particular
\[Y_p(t)=-e^{-x}\int\frac{xe^{-x}e^{-x}}{e^{-2x}(e^{-x},xe^{-x})}dx+xe^{-x}\int\frac{e^{-x}e^{2x}}{e^{-2x}(e^{-x},xe^{-x})}dx\]
the first integrand is wrong I think
I think you put the wrong g(x)
\[Y_p(t)=-e^{-x}\int\frac{xe^{-x}e^{2x}}{e^{-2x}(e^{-x},xe^{-x})}dx+xe^{-x}\int\frac{e^{-x}e^{2x}}{e^{-2x}(e^{-x},xe^{-x})}dx\]
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
Oh lol so the denominator is just e^{-2x}
right
notice that makes our integrals quite tolerable :)
\[Y_p(t)=-e^{-x}\int\frac{xe^{-x}e^{2x}}{e^{-2x}}dx+xe^{-x}\int\frac{e^{-x}e^{2x}}{e^{-2x}}dx\] That's a lot easier to integrate
oh yeah :) especially after a little simplification
\[-\frac{e^{2x}}{9}(3x-1)+\frac x3e^{3x}\]
so what was the point of doing variation of parameters
well now you have a particular solution
y(x)=yc+yp you got you now so you're good to go
you got yp*
I did not check your integral btw
that's ok I did wolfram
right-o, I'd do the same right now :)
oh I see \[c_1e^{-x}+c_2xe^{-x}-\frac{e^{2x}}{9}(3x-1)+\frac x3e^{3x}\]
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
good, wow, that was a lot THanks Turing!
Always a pleasure Sofiya :D
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
@Zarkon @terenzreignz Is anybody else familiar with ODE's seeing this, or am I missing something?
@zepdrix Anybody?
@Mendicant_Bias For the homogeneous solution? Yes the characteristic equation appears to be giving +1 as a repeated root, as you indicated
So yah, that was a mistake early in the thread :C

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