anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
No I did that wrong, one moment
anonymous
  • anonymous
no it equals \[e^{2x}\]
anonymous
  • anonymous
\[y_p(x)=-e^xx^2+x^2e^x\]

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More answers

TuringTest
  • TuringTest
wrong complimentary
TuringTest
  • TuringTest
\[r^2-r=r(r-1)=0\implies r=\{?,?\}\]
anonymous
  • anonymous
\[r^2-r=0\]
anonymous
  • anonymous
r=0 and r=1
TuringTest
  • TuringTest
so the complimentary is...?
anonymous
  • anonymous
\[y_c=c_1+c_2e^x\]
TuringTest
  • TuringTest
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
anonymous
  • anonymous
sure \[y_c=c_1+c_2xe^x\]
TuringTest
  • TuringTest
\[y_c=c_1x+c_2xe^x\]
anonymous
  • anonymous
why on both c_1 and c_2, how did come to that conclusion?
TuringTest
  • TuringTest
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...
anonymous
  • anonymous
I'm not sure... DunnBros is about to kick me out, they close at 11pm
anonymous
  • anonymous
I'll be back on in like 15 mins when I get home
TuringTest
  • TuringTest
fair enough, I'm about to call it a night as well see ya, I will investigate this :)
anonymous
  • anonymous
Make that 7 mins, ha!
anonymous
  • anonymous
\[W(y_1,y_2)=x^2e^x\]
anonymous
  • anonymous
\[y_1=x\] \[y_2=xe^x\]
anonymous
  • anonymous
\[y_p(x)=-x\int \frac{e^x}xdx+xe^x\int x^{-1}dx\]
wasiqss
  • wasiqss
well ik it
wasiqss
  • wasiqss
my twin :D
anonymous
  • anonymous
hi twin
TuringTest
  • TuringTest
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!!!!!
wasiqss
  • wasiqss
well i can do it
wasiqss
  • wasiqss
tell me equation for which i have to solve\y complimentary
anonymous
  • anonymous
y''-y'=e^x
TuringTest
  • TuringTest
we need variation of parameters @wasiqss
wasiqss
  • wasiqss
yehh just tell me equation i ll take out solution. as y complimentary is solved i ll solve for y particular
TuringTest
  • TuringTest
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
TuringTest
  • TuringTest
ok @wasiqss let's see what you got :)
wasiqss
  • wasiqss
tell me equation man completely cause this seems messy
TuringTest
  • TuringTest
y''-y'=e^x
wasiqss
  • wasiqss
easy one lol
TuringTest
  • TuringTest
then do it, by all mean!
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=integral+of+e^%28x%29%2Fx
wasiqss
  • wasiqss
y complimentary is y= c1 +c2e^-x
TuringTest
  • TuringTest
^yes, answer not important, we need the process...
TuringTest
  • TuringTest
complimentary is y=c1+c2e^x
anonymous
  • anonymous
i disagree with the above answer, shouldn't it be e^x, yeah what turing has
wasiqss
  • wasiqss
yes cause roots are zero and -1
TuringTest
  • TuringTest
no they are not
TuringTest
  • TuringTest
r^2-r=0 r(r-1)=0 r={0,1}
anonymous
  • anonymous
do you think my W is right? x^2 e^x
TuringTest
  • TuringTest
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
TuringTest
  • TuringTest
with W=e^2e^x I don't think that first integral is closed, check the wolf, I have not yet
TuringTest
  • TuringTest
w=x^2e^x *
anonymous
  • anonymous
do we agree that \[y_c=c_1x+c_2xe^x\] or should it be \[y_c=c_1+c_2e^x\]
anonymous
  • anonymous
yeah that's what I had
TuringTest
  • TuringTest
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 :)
TuringTest
  • TuringTest
however if we are to try the first we need to multiply g(x) by x (I'm thinking....)
TuringTest
  • TuringTest
so let's try that
TuringTest
  • TuringTest
otherwise we get W=0 which is a failure
TuringTest
  • TuringTest
@lalaly I know you can help us, please where are you?
anonymous
  • anonymous
\[y_c=c_1x+c_2xe^x\] \[\left|\begin{matrix}x&xe^x\\1&e^x(x-1)\end{matrix}\right|=x^2e^x\]
anonymous
  • anonymous
can you explain the g(x) concept again and why it being e^x would be a problem because that cancels something out?...
TuringTest
  • TuringTest
no I think that is wrong because that leads to the particular having\[\int\frac{e^x}dxx\]
TuringTest
  • TuringTest
*\[\int\frac{e^x}xdx\]
TuringTest
  • TuringTest
here is the problem... have you taken linear algebra?
anonymous
  • anonymous
no
anonymous
  • anonymous
@lalaly is here
TuringTest
  • TuringTest
okay. then we have a set of vectors (in this case equations) if their determinant is zero we know they are linearly dependent
wasiqss
  • wasiqss
now listen
wasiqss
  • wasiqss
for y particular
TuringTest
  • TuringTest
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
wasiqss
  • wasiqss
is it xe^x
wasiqss
  • wasiqss
so as the denominator becomes zero
TuringTest
  • TuringTest
and would you care to explain how/why ?
wasiqss
  • wasiqss
we multiply by x
wasiqss
  • wasiqss
cofficeint of e^1x
wasiqss
  • wasiqss
=1
wasiqss
  • wasiqss
[xe^]/[(-1)^2-1)]
wasiqss
  • wasiqss
HENCE we get zero in denominator
TuringTest
  • TuringTest
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
wasiqss
  • wasiqss
now we need to apply binomeal expansion
TuringTest
  • TuringTest
@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
wasiqss
  • wasiqss
arghh i wish i was good at making you understand it
wasiqss
  • wasiqss
we need it!
TuringTest
  • TuringTest
you cannot have 0 in the denom, that shows you that you have already made a mistake !!!!
wasiqss
  • wasiqss
well to counter this zero!! we need to multiply x
TuringTest
  • TuringTest
...as I was saying...
wasiqss
  • wasiqss
plz no offense but plz go revise DE s
TuringTest
  • TuringTest
._.
wasiqss
  • wasiqss
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
TuringTest
  • TuringTest
yes, but I think it is way more work than is necessary now I just wanna hear @lalaly
wasiqss
  • wasiqss
kay
TuringTest
  • TuringTest
oh dear, we seem to have lost her I am going to try this op paper I guess
wasiqss
  • wasiqss
hahaha lana dear
wasiqss
  • wasiqss
\[[1+(D(D+1)-1)^-1[x^2e^x\]
wasiqss
  • wasiqss
we have to expand binomeally [1+(D(D+1)−1)−1
wasiqss
  • wasiqss
well i have a solution on page i wonder how to give you that
anonymous
  • anonymous
take a picture with your cellphone and attach it
anonymous
  • anonymous
or use \[\LaTeX\]
wasiqss
  • wasiqss
good idea twin can i do it tomorow because i would need to upload it too?
TuringTest
  • TuringTest
okay got it :)
wasiqss
  • wasiqss
latex is something not my type
wasiqss
  • wasiqss
turing my approach?
TuringTest
  • TuringTest
do everything normally, then multiply the particular by x at the end that is all
anonymous
  • anonymous
get used to it, it's a very helpful Latex is a very useful tool
wasiqss
  • wasiqss
turing binomeal is essential in it!
TuringTest
  • TuringTest
oh yeah?, just watch...
TuringTest
  • TuringTest
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?
TuringTest
  • TuringTest
oh crap I see a mistake
wasiqss
  • wasiqss
wait you told me w=xe^x!
anonymous
  • anonymous
dude this whole discussion was based on finding what W equaled
wasiqss
  • wasiqss
ok just ask lana what is the right answer
TuringTest
  • TuringTest
@MathSofiya yes, because the entire problem we were having was keeping our solutions linearly independent, which we check by finding W
wasiqss
  • wasiqss
waittttttttttttt what the....... i was solving by another method not by variation of parameter lol
TuringTest
  • TuringTest
\[y_c=c_1+c_2e^x\]\[W=e^x\]\[y_p=-\int e ^x dx+e^x\int dx=-e^x+xe^x\]\[y=y_c+y_p=c_1+c_2e^x-e^x+xe^x=c_1+e^x(x-1+c_2)\]
TuringTest
  • TuringTest
there we go... stupid algebra tripped me up :/
anonymous
  • anonymous
yeah that's what I get \[-e^x+e^xx\]
TuringTest
  • TuringTest
so yc+yp gives what I have, and what wolf has so we're good :)
anonymous
  • anonymous
good. Hi 5
TuringTest
  • TuringTest
back atchya!
anonymous
  • anonymous
As always thank you!
TuringTest
  • TuringTest
As always you're welcome :D
Mendicant_Bias
  • Mendicant_Bias
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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