anonymous
  • anonymous
Solve using variation of parameters \[y''-2y'+y=e^{2x}\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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TuringTest
  • TuringTest
complimentary....
anonymous
  • anonymous
meaning?
anonymous
  • anonymous
\[y_c=c_1e^{-x}+c_2xe^{-x}\]

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

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