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complimentary....

meaning?

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

right...

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

nope

it is the determinant of a matrix

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

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 :)

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

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?

-(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?

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

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

the first integrand is wrong I think

I think you put the wrong g(x)

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

right

notice that makes our integrals quite tolerable :)

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

good, wow, that was a lot
THanks Turing!

Always a pleasure Sofiya :D

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

@zepdrix Anybody?

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