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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?
 2 years ago
 2 years 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?
 2 years ago
 2 years ago

This Question is Closed

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

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

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

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

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

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

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

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

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

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[y_c=c_1x+c_2xe^x\]
 2 years 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?
 2 years 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...
 2 years 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
 2 years 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
 2 years 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 :)
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
w=x^2e^x *
 2 years 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 \[y_c=c_1+c_2e^x\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
yeah that's what I had
 2 years 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 :)
 2 years 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....)
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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