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No I did that wrong, one moment

no it equals \[e^{2x}\]

\[y_p(x)=-e^xx^2+x^2e^x\]

wrong complimentary

\[r^2-r=r(r-1)=0\implies r=\{?,?\}\]

\[r^2-r=0\]

r=0 and r=1

so the complimentary is...?

\[y_c=c_1+c_2e^x\]

sure
\[y_c=c_1+c_2xe^x\]

\[y_c=c_1x+c_2xe^x\]

why on both c_1 and c_2, how did come to that conclusion?

I'm not sure...
DunnBros is about to kick me out, they close at 11pm

I'll be back on in like 15 mins when I get home

fair enough, I'm about to call it a night as well
see ya, I will investigate this :)

Make that 7 mins, ha!

\[W(y_1,y_2)=x^2e^x\]

\[y_1=x\]
\[y_2=xe^x\]

\[y_p(x)=-x\int \frac{e^x}xdx+xe^x\int x^{-1}dx\]

well ik it

my twin :D

hi twin

well i can do it

tell me equation for which i have to solve\y complimentary

y''-y'=e^x

we need variation of parameters @wasiqss

ok @wasiqss let's see what you got :)

tell me equation man completely cause this seems messy

y''-y'=e^x

easy one lol

then do it, by all mean!

http://www.wolframalpha.com/input/?i=integral+of+e^%28x%29%2Fx

y complimentary is y= c1 +c2e^-x

^yes, answer not important, we need the process...

complimentary is
y=c1+c2e^x

i disagree with the above answer, shouldn't it be e^x, yeah what turing has

yes cause roots are zero and -1

no they are not

r^2-r=0
r(r-1)=0
r={0,1}

do you think my W is right? x^2 e^x

with W=e^2e^x I don't think that first integral is closed, check the wolf, I have not yet

w=x^2e^x *

do we agree that \[y_c=c_1x+c_2xe^x\] or should it be \[y_c=c_1+c_2e^x\]

yeah that's what I had

however if we are to try the first we need to multiply g(x) by x (I'm thinking....)

so let's try that

otherwise we get W=0 which is a failure

@lalaly I know you can help us, please where are you?

\[y_c=c_1x+c_2xe^x\]
\[\left|\begin{matrix}x&xe^x\\1&e^x(x-1)\end{matrix}\right|=x^2e^x\]

no I think that is wrong because that leads to the particular having\[\int\frac{e^x}dxx\]

*\[\int\frac{e^x}xdx\]

here is the problem... have you taken linear algebra?

no

now listen

for y particular

is it xe^x

so as the denominator becomes zero

and would you care to explain how/why ?

we multiply by x

cofficeint of e^1x

=1

[xe^]/[(-1)^2-1)]

HENCE we get zero in denominator

now we need to apply binomeal expansion

arghh i wish i was good at making you understand it

we need it!

you cannot have 0 in the denom, that shows you that you have already made a mistake !!!!

well to counter this zero!! we need to multiply x

...as I was saying...

plz no offense but plz go revise DE
s

._.

yes, but I think it is way more work than is necessary
now I just wanna hear @lalaly

kay

oh dear, we seem to have lost her
I am going to try this op paper I guess

hahaha lana dear

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

we have to expand binomeally
[1+(D(D+1)−1)−1

well i have a solution on page i wonder how to give you that

take a picture with your cellphone and attach it

or use \[\LaTeX\]

good idea twin can i do it tomorow because i would need to upload it too?

okay got it :)

latex is something not my type

turing my approach?

do everything normally, then multiply the particular by x at the end
that is all

get used to it, it's a very helpful Latex is a very useful tool

turing binomeal is essential in it!

oh yeah?, just watch...

oh crap I see a mistake

wait you told me w=xe^x!

dude this whole discussion was based on finding what W equaled

ok just ask lana what is the right answer

waittttttttttttt what the....... i was solving by another method not by variation of parameter lol

there we go... stupid algebra tripped me up :/

yeah that's what I get
\[-e^x+e^xx\]

so yc+yp gives what I have, and what wolf has
so we're good :)

good. Hi 5

back atchya!

As always thank you!

As always you're welcome :D