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MathSofiya
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Solve using variation of parameters
\[y''2y'+y=e^{2x}\]
 2 years ago
 2 years ago
MathSofiya Group Title
Solve using variation of parameters \[y''2y'+y=e^{2x}\]
 2 years ago
 2 years ago

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TuringTest Group TitleBest ResponseYou've already chosen the best response.3
complimentary....
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
meaning?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[y_c=c_1e^{x}+c_2xe^{x}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
right...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
now we need the wronskian, do you know about that?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
it is the determinant of a matrix
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
for\[y_c=c_1y_1+c_2y_2\]the Wronskian is\[W(y_1,y_2)=\left\begin{matrix}y_1&y_2\\y_1'&y_2'\end{matrix}\right\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
that's a determinant in case you forgot, so find that determinant
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[\left\begin{matrix}e^{x}&xe^{x}\\e^{x}&e^{x}(x1)\end{matrix}\right\] =\[e^{x}(e^{x}(x1))xe^{x}e^{x}\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
completely wrong?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
oh it's plus
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
yep you caught it :)
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[e^{x}(e^{x}(x1))+xe^{x}e^{x}\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
that's it?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
you dropped a minus
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[e^{x}(e^{x}(x1))+xe^{x}e^{x}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
simplify and be happy :)
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
does it equal\[ e^{x}\]? \[e^{x}(e^{x}(x1))+xe^{x}e^{x}=e^{x}\] \[{x}(e^{x}(x1))+xe^{x}=1\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[(e^{x}(x1))+xe^{x}=1\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
no, why did you set it equal to e^x, we can't set it equal to anything yet...
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
oh my bad haha
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
so how should I simplify it then?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
\[e^{x}(e^{x}(x1))+xe^{x}e^{x}=e^{x}(e^{x}(1x))+xe^{x}e^{x}\]now try again while I try to eat ;)
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
(x1)=1x 1x=1x
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
no no no ...wait a second
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
now i get 2x=2x
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
nope still wrong
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
stop laughing
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
Let's try that again \[e^{x}(e^{x}(x1))+xe^{x}e^{x}=e^{x}(e^{x}(1x))+xe^{x}e^{x}\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
they're just equal to each other
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[e^{x}(e^{x}(x1))+xe^{x}e^{x}\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[e^{2x}(x1)+xe^{2x}\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[e^{2x}((1x)+x)\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[e^{2x}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
yes, sorry, I was afk
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
do you know the formula for the particular solution for variation of parameters?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
something like that?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
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)
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
is the first one a y''? \[y''+q(t)y'+r(t)y=g(t)\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
yeah, it looks screwy I know
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
do we know what q(t) and r(t) are? are they the coefficients of y' and y? Probably not it looks like
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
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
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
oh I see, soo... \[y''+q(t)y'+r(t)y=g(t)\] \[y''2y'+y=e^{2x}\] so now I integrate
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.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
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
q(t) and r(t)... they determine the complimentary only, which as you can see *is* part of the formula for the particular
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[Y_p(t)=e^{x}\int\frac{xe^{x}e^{x}}{e^{2x}(e^{x},xe^{x})}dx+xe^{x}\int\frac{e^{x}e^{2x}}{e^{2x}(e^{x},xe^{x})}dx\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
the first integrand is wrong I think
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
I think you put the wrong g(x)
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[Y_p(t)=e^{x}\int\frac{xe^{x}e^{2x}}{e^{2x}(e^{x},xe^{x})}dx+xe^{x}\int\frac{e^{x}e^{2x}}{e^{2x}(e^{x},xe^{x})}dx\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
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
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
Oh lol so the denominator is just e^{2x}
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
notice that makes our integrals quite tolerable :)
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[Y_p(t)=e^{x}\int\frac{xe^{x}e^{2x}}{e^{2x}}dx+xe^{x}\int\frac{e^{x}e^{2x}}{e^{2x}}dx\] That's a lot easier to integrate
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
oh yeah :) especially after a little simplification
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{e^{2x}}{9}(3x1)+\frac x3e^{3x}\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
so what was the point of doing variation of parameters
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
well now you have a particular solution
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
y(x)=yc+yp you got you now so you're good to go
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
you got yp*
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
I did not check your integral btw
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
that's ok I did wolfram
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
righto, I'd do the same right now :)
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
oh I see \[c_1e^{x}+c_2xe^{x}\frac{e^{2x}}{9}(3x1)+\frac x3e^{3x}\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
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
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
good, wow, that was a lot THanks Turing!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.3
Always a pleasure Sofiya :D
 2 years ago

Mendicant_Bias Group TitleBest ResponseYou've already chosen the best response.0
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
 one month ago

Mendicant_Bias Group TitleBest ResponseYou've already chosen the best response.0
@jim_thompson5910 @agent0smith
 one month ago

Mendicant_Bias Group TitleBest ResponseYou've already chosen the best response.0
@Zarkon @terenzreignz Is anybody else familiar with ODE's seeing this, or am I missing something?
 one month ago

Mendicant_Bias Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix Anybody?
 one month ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
@Mendicant_Bias For the homogeneous solution? Yes the characteristic equation appears to be giving +1 as a repeated root, as you indicated
 one month ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So yah, that was a mistake early in the thread :C
 one month ago
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