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