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

This Question is Closed

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
No I did that wrong, one moment
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
no it equals \[e^{2x}\]
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[y_p(x)=e^xx^2+x^2e^x\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
wrong complimentary
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[r^2r=r(r1)=0\implies r=\{?,?\}\]
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[r^2r=0\]
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
r=0 and r=1
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
so the complimentary is...?
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[y_c=c_1+c_2e^x\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
sure \[y_c=c_1+c_2xe^x\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[y_c=c_1x+c_2xe^x\]
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
why on both c_1 and c_2, how did come to that conclusion?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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...
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I'm not sure... DunnBros is about to kick me out, they close at 11pm
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I'll be back on in like 15 mins when I get home
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
fair enough, I'm about to call it a night as well see ya, I will investigate this :)
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
Make that 7 mins, ha!
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[W(y_1,y_2)=x^2e^x\]
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[y_1=x\] \[y_2=xe^x\]
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[y_p(x)=x\int \frac{e^x}xdx+xe^x\int x^{1}dx\]
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
well ik it
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
my twin :D
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
hi twin
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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!!!!!
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
well i can do it
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
tell me equation for which i have to solve\y complimentary
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
y''y'=e^x
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
we need variation of parameters @wasiqss
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
yehh just tell me equation i ll take out solution. as y complimentary is solved i ll solve for y particular
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
ok @wasiqss let's see what you got :)
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
tell me equation man completely cause this seems messy
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
y''y'=e^x
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
easy one lol
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
then do it, by all mean!
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
http://www.wolframalpha.com/input/?i=integral+of+e^%28x%29%2Fx
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
y complimentary is y= c1 +c2e^x
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
^yes, answer not important, we need the process...
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
complimentary is y=c1+c2e^x
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
i disagree with the above answer, shouldn't it be e^x, yeah what turing has
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
yes cause roots are zero and 1
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
no they are not
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
r^2r=0 r(r1)=0 r={0,1}
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
do you think my W is right? x^2 e^x
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
with W=e^2e^x I don't think that first integral is closed, check the wolf, I have not yet
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
w=x^2e^x *
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
do we agree that \[y_c=c_1x+c_2xe^x\] or should it be \[y_c=c_1+c_2e^x\]
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
yeah that's what I had
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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 :)
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
however if we are to try the first we need to multiply g(x) by x (I'm thinking....)
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
so let's try that
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
otherwise we get W=0 which is a failure
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
@lalaly I know you can help us, please where are you?
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[y_c=c_1x+c_2xe^x\] \[\left\begin{matrix}x&xe^x\\1&e^x(x1)\end{matrix}\right=x^2e^x\]
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
can you explain the g(x) concept again and why it being e^x would be a problem because that cancels something out?...
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
no I think that is wrong because that leads to the particular having\[\int\frac{e^x}dxx\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
*\[\int\frac{e^x}xdx\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
here is the problem... have you taken linear algebra?
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
@lalaly is here
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
okay. then we have a set of vectors (in this case equations) if their determinant is zero we know they are linearly dependent
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
now listen
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
for y particular
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
is it xe^x
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
so as the denominator becomes zero
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
and would you care to explain how/why ?
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
we multiply by x
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
cofficeint of e^1x
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
[xe^]/[(1)^21)]
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
HENCE we get zero in denominator
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
now we need to apply binomeal expansion
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
@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
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
arghh i wish i was good at making you understand it
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
we need it!
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
you cannot have 0 in the denom, that shows you that you have already made a mistake !!!!
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
well to counter this zero!! we need to multiply x
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
...as I was saying...
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
plz no offense but plz go revise DE s
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yes, but I think it is way more work than is necessary now I just wanna hear @lalaly
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
oh dear, we seem to have lost her I am going to try this op paper I guess
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
hahaha lana dear
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
\[[1+(D(D+1)1)^1[x^2e^x\]
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
we have to expand binomeally [1+(D(D+1)−1)−1
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
well i have a solution on page i wonder how to give you that
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
take a picture with your cellphone and attach it
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
or use \[\LaTeX\]
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
good idea twin can i do it tomorow because i would need to upload it too?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
okay got it :)
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
latex is something not my type
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
turing my approach?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
do everything normally, then multiply the particular by x at the end that is all
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
get used to it, it's a very helpful Latex is a very useful tool
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
turing binomeal is essential in it!
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
oh yeah?, just watch...
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
oh crap I see a mistake
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
wait you told me w=xe^x!
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
dude this whole discussion was based on finding what W equaled
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
ok just ask lana what is the right answer
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
@MathSofiya yes, because the entire problem we were having was keeping our solutions linearly independent, which we check by finding W
 one year ago

wasiqss Group TitleBest ResponseYou've already chosen the best response.0
waittttttttttttt what the....... i was solving by another method not by variation of parameter lol
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[y_c=c_1+c_2e^x\]\[W=e^x\]\[y_p=\int e ^x dx+e^x\int dx=e^x+xe^x\]\[y=y_c+y_p=c_1+c_2e^xe^x+xe^x=c_1+e^x(x1+c_2)\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
there we go... stupid algebra tripped me up :/
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
yeah that's what I get \[e^x+e^xx\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
so yc+yp gives what I have, and what wolf has so we're good :)
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
good. Hi 5
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
back atchya!
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
As always thank you!
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
As always you're welcome :D
 one year ago
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