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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y_c=c_1e^{x}+c_2xe^{x}\]

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3it is the determinant of a matrix

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\left\begin{matrix}e^{x}&xe^{x}\\e^{x}&e^{x}(x1)\end{matrix}\right\] =\[e^{x}(e^{x}(x1))xe^{x}e^{x}\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3yep you caught it :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[e^{x}(e^{x}(x1))+xe^{x}e^{x}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[e^{x}(e^{x}(x1))+xe^{x}e^{x}\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3simplify and be happy :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0does it equal\[ e^{x}\]? \[e^{x}(e^{x}(x1))+xe^{x}e^{x}=e^{x}\] \[{x}(e^{x}(x1))+xe^{x}=1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(e^{x}(x1))+xe^{x}=1\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so how should I simplify it then?

TuringTest
 4 years ago
Best 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 ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no no no ...wait a second

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let's try that again \[e^{x}(e^{x}(x1))+xe^{x}e^{x}=e^{x}(e^{x}(1x))+xe^{x}e^{x}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0they're just equal to each other

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[e^{x}(e^{x}(x1))+xe^{x}e^{x}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[e^{2x}(x1)+xe^{2x}\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3yes, sorry, I was afk

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3almost 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
 4 years ago
Best ResponseYou've already chosen the best response.0is the first one a y''? \[y''+q(t)y'+r(t)y=g(t)\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do we know what q(t) and r(t) are? are they the coefficients of y' and y? Probably not it looks like

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3yes 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
 4 years ago
Best ResponseYou've already chosen the best response.0oh I see, soo... \[y''+q(t)y'+r(t)y=g(t)\] \[y''2y'+y=e^{2x}\] so now I integrate

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3right, using the formula... r(t) and r(t) don't even come into the formula for the particular solution in variation of parameters

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[Y_p(t)=e^{x}\int\frac{xe^{x}e^{x}}{e^{2x}(e^{x},xe^{x})}dx+xe^{x}\int\frac{e^{x}e^{2x}}{e^{2x}(e^{x},xe^{x})}dx\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3the first integrand is wrong I think

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[Y_p(t)=e^{x}\int\frac{xe^{x}e^{2x}}{e^{2x}(e^{x},xe^{x})}dx+xe^{x}\int\frac{e^{x}e^{2x}}{e^{2x}(e^{x},xe^{x})}dx\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3oh 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
 4 years ago
Best ResponseYou've already chosen the best response.0Oh lol so the denominator is just e^{2x}

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[Y_p(t)=e^{x}\int\frac{xe^{x}e^{2x}}{e^{2x}}dx+xe^{x}\int\frac{e^{x}e^{2x}}{e^{2x}}dx\] That's a lot easier to integrate

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{e^{2x}}{9}(3x1)+\frac x3e^{3x}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so what was the point of doing variation of parameters

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's ok I did wolfram

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh I see \[c_1e^{x}+c_2xe^{x}\frac{e^{2x}}{9}(3x1)+\frac x3e^{3x}\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3yep, 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
 4 years ago
Best ResponseYou've already chosen the best response.0good, wow, that was a lot THanks Turing!

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.3Always a pleasure Sofiya :D

Mendicant_Bias
 2 years ago
Best ResponseYou've already chosen the best response.0Looking 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
 2 years ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 @agent0smith

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

Mendicant_Bias
 2 years ago
Best ResponseYou've already chosen the best response.0@zepdrix Anybody?

zepdrix
 2 years ago
Best 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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1So yah, that was a mistake early in the thread :C
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