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

- anonymous

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

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- schrodinger

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- TuringTest

complimentary....

- anonymous

meaning?

- anonymous

\[y_c=c_1e^{-x}+c_2xe^{-x}\]

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## More answers

- TuringTest

right...

- TuringTest

now we need the wronskian, do you know about that?

- anonymous

nope

- TuringTest

it is the determinant of a matrix

- 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

that's a determinant in case you forgot, so find that determinant

- 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

completely wrong?

- anonymous

oh it's plus

- TuringTest

yep you caught it :)

- anonymous

\[e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}\]

- anonymous

that's it?

- TuringTest

you dropped a minus

- anonymous

where?

- anonymous

ohhhh

- anonymous

\[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}\]

- TuringTest

right

- TuringTest

simplify and be happy :)

- 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

\[(e^{-x}(x-1))+xe^{-x}=1\]

- anonymous

typo

- TuringTest

no, why did you set it equal to e^-x, we can't set it equal to anything yet...

- anonymous

oh my bad haha

- anonymous

so how should I simplify it then?

- 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

-(x-1)=1-x
1-x=1-x

- anonymous

no no no ...wait a second

- anonymous

now i get 2-x=2-x

- anonymous

nope still wrong

- anonymous

x=1

- anonymous

stop laughing

- 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

they're just equal to each other

- anonymous

\[-e^{-x}(e^{-x}(x-1))+xe^{-x}e^{-x}\]

- anonymous

\[-e^{-2x}(x-1)+xe^{-2x}\]

- anonymous

\[e^{-2x}((1-x)+x)\]

- anonymous

\[e^{-2x}\]

- anonymous

yes?

- TuringTest

yes, sorry, I was afk

- anonymous

LOL ok

- TuringTest

do you know the formula for the particular solution for variation of parameters?

- anonymous

##### 1 Attachment

- anonymous

something like that?

- 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

is the first one a y''?
\[y''+q(t)y'+r(t)y=g(t)\]

- TuringTest

yeah, it looks screwy I know

- 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

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

oh I see, soo...
\[y''+q(t)y'+r(t)y=g(t)\]
\[y''-2y'+y=e^{2x}\]
so now I integrate

- 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

q(t) and r(t)...
they determine the complimentary only, which as you can see *is* part of the formula for the particular

- 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

the first integrand is wrong I think

- TuringTest

I think you put the wrong g(x)

- 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

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

Oh lol so the denominator is just e^{-2x}

- TuringTest

right

- TuringTest

notice that makes our integrals quite tolerable :)

- 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

oh yeah :) especially after a little simplification

- anonymous

\[-\frac{e^{2x}}{9}(3x-1)+\frac x3e^{3x}\]

- anonymous

so what was the point of doing variation of parameters

- TuringTest

well now you have a particular solution

- TuringTest

y(x)=yc+yp
you got you now so you're good to go

- TuringTest

you got yp*

- TuringTest

I did not check your integral btw

- anonymous

that's ok I did wolfram

- TuringTest

right-o, I'd do the same right now :)

- anonymous

oh I see
\[c_1e^{-x}+c_2xe^{-x}-\frac{e^{2x}}{9}(3x-1)+\frac x3e^{3x}\]

- 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

good, wow, that was a lot
THanks Turing!

- TuringTest

Always a pleasure Sofiya :D

- 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

@jim_thompson5910 @agent0smith

- Mendicant_Bias

@Zarkon @terenzreignz Is anybody else familiar with ODE's seeing this, or am I missing something?

- Mendicant_Bias

@zepdrix Anybody?

- zepdrix

@Mendicant_Bias For the homogeneous solution? Yes the characteristic equation appears to be giving +1 as a repeated root, as you indicated

- zepdrix

So yah, that was a mistake early in the thread :C

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