Solve using variation of parameters \[y''-y'=e^x\] \[r^2-r=0\] \[r_1=r_2=1\] \[y_c=c_1e^x+c_2xe^x\] \[W(y_1,y_2)=0\] Is that possible?

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Solve using variation of parameters \[y''-y'=e^x\] \[r^2-r=0\] \[r_1=r_2=1\] \[y_c=c_1e^x+c_2xe^x\] \[W(y_1,y_2)=0\] Is that possible?

Mathematics
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No I did that wrong, one moment
no it equals \[e^{2x}\]
\[y_p(x)=-e^xx^2+x^2e^x\]

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wrong complimentary
\[r^2-r=r(r-1)=0\implies r=\{?,?\}\]
\[r^2-r=0\]
r=0 and r=1
so the complimentary is...?
\[y_c=c_1+c_2e^x\]
right, now I get a bit dicey.... since you have e^x on the RHS as well I feel you may need to multiply this by x to keep it linearly independent
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?
or do you do it to the particular instead? hm... well normally you do it to the whole RHS particular, but since you need the complimentary for the particular in VP I'm not sure...
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
that first integral is not going to be so nice I don't think.... you know who knows exactly what we're missing @lalaly DE help please!!!!!
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
yehh just tell me equation i ll take out solution. as y complimentary is solved i ll solve for y particular
Our problem is that the solutions to the complimentary are not independent from the particular, which you can tell since we got W=0 on our first try
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
that I doube @MathSofiya because we have e^x on both left and right our solutions are linearly dependet (that means we can't solve it yet) so maybe if we multiply the g(x) by x ? not sure, but worth a try
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
that I am not sure, but based on our attempt to use the second one failing I would now try the first (we are discovering this together right now :)
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?
\[y_c=c_1x+c_2xe^x\] \[\left|\begin{matrix}x&xe^x\\1&e^x(x-1)\end{matrix}\right|=x^2e^x\]
can you explain the g(x) concept again and why it being e^x would be a problem because that cancels something out?...
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
@lalaly is here
okay. then we have a set of vectors (in this case equations) if their determinant is zero we know they are linearly dependent
now listen
for y particular
that is a bad thing in DE's we need all of our solutions to be linearly independent, so the determinant of the solutions, (the wronskian) must not be zero, otherwise our solutions are linearly dependent
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
anyway... to keep the particular linearly independent from the complimentary we often need to multiply by x, so that is what I am thinking must happen somehow @lalaly will confirm if I am making any sense or not now
now we need to apply binomeal expansion
@wasiqss sorry to be so direct, but no we do not need any binomial expansion here please let's just have a listen to @lalaly
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
._.
you know what .. we can approach a DE in differnt ways and this one i will certainly approach by binomeal.. cause we studied this way for this case
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...
actually no need to multiply by x at all\[y_c=c_1+c_2e^x\]\[W=e^x\]\[y_p=-\int e^{-x}+e^x\int dx=e^x+xe^x\]\[y_p=e^x+xe^x\]\[y=y_c+y_p=c_1+c_2e^x+e^x+xe^x=c_1+e^x(1+x+c_2)\]and yes, but is your solution in 5 lines?
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
@MathSofiya yes, because the entire problem we were having was keeping our solutions linearly independent, which we check by finding W
waittttttttttttt what the....... i was solving by another method not by variation of parameter lol
\[y_c=c_1+c_2e^x\]\[W=e^x\]\[y_p=-\int e ^x dx+e^x\int dx=-e^x+xe^x\]\[y=y_c+y_p=c_1+c_2e^x-e^x+xe^x=c_1+e^x(x-1+c_2)\]
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
I don't understand part of the very end of your post, near the solutions. I'll come back and ask about it later, but I do not understand how you obeyed the laws of calculus in factoring out an e^x when integrating with respect to x.

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