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?

- anonymous

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- anonymous

No I did that wrong, one moment

- anonymous

no it equals \[e^{2x}\]

- anonymous

\[y_p(x)=-e^xx^2+x^2e^x\]

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## More answers

- TuringTest

wrong complimentary

- TuringTest

\[r^2-r=r(r-1)=0\implies r=\{?,?\}\]

- anonymous

\[r^2-r=0\]

- anonymous

r=0 and r=1

- TuringTest

so the complimentary is...?

- anonymous

\[y_c=c_1+c_2e^x\]

- 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

sure
\[y_c=c_1+c_2xe^x\]

- TuringTest

\[y_c=c_1x+c_2xe^x\]

- anonymous

why on both c_1 and c_2, how did come to that conclusion?

- 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

I'm not sure...
DunnBros is about to kick me out, they close at 11pm

- anonymous

I'll be back on in like 15 mins when I get home

- TuringTest

fair enough, I'm about to call it a night as well
see ya, I will investigate this :)

- anonymous

Make that 7 mins, ha!

- anonymous

\[W(y_1,y_2)=x^2e^x\]

- anonymous

\[y_1=x\]
\[y_2=xe^x\]

- anonymous

\[y_p(x)=-x\int \frac{e^x}xdx+xe^x\int x^{-1}dx\]

- wasiqss

well ik it

- wasiqss

my twin :D

- anonymous

hi twin

- 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

well i can do it

- wasiqss

tell me equation for which i have to solve\y complimentary

- anonymous

y''-y'=e^x

- TuringTest

we need variation of parameters @wasiqss

- wasiqss

yehh just tell me equation i ll take out solution. as y complimentary is solved i ll solve for y particular

- 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

ok @wasiqss let's see what you got :)

- wasiqss

tell me equation man completely cause this seems messy

- TuringTest

y''-y'=e^x

- wasiqss

easy one lol

- TuringTest

then do it, by all mean!

- anonymous

http://www.wolframalpha.com/input/?i=integral+of+e^%28x%29%2Fx

- wasiqss

y complimentary is y= c1 +c2e^-x

- TuringTest

^yes, answer not important, we need the process...

- TuringTest

complimentary is
y=c1+c2e^x

- anonymous

i disagree with the above answer, shouldn't it be e^x, yeah what turing has

- wasiqss

yes cause roots are zero and -1

- TuringTest

no they are not

- TuringTest

r^2-r=0
r(r-1)=0
r={0,1}

- anonymous

do you think my W is right? x^2 e^x

- 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

with W=e^2e^x I don't think that first integral is closed, check the wolf, I have not yet

- TuringTest

w=x^2e^x *

- anonymous

do we agree that \[y_c=c_1x+c_2xe^x\] or should it be \[y_c=c_1+c_2e^x\]

- anonymous

yeah that's what I had

- 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

however if we are to try the first we need to multiply g(x) by x (I'm thinking....)

- TuringTest

so let's try that

- TuringTest

otherwise we get W=0 which is a failure

- TuringTest

@lalaly I know you can help us, please where are you?

- 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

can you explain the g(x) concept again and why it being e^x would be a problem because that cancels something out?...

- TuringTest

no I think that is wrong because that leads to the particular having\[\int\frac{e^x}dxx\]

- TuringTest

*\[\int\frac{e^x}xdx\]

- TuringTest

here is the problem... have you taken linear algebra?

- anonymous

no

- anonymous

@lalaly is here

- 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

now listen

- wasiqss

for y particular

- 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

is it xe^x

- wasiqss

so as the denominator becomes zero

- TuringTest

and would you care to explain how/why ?

- wasiqss

we multiply by x

- wasiqss

cofficeint of e^1x

- wasiqss

=1

- wasiqss

[xe^]/[(-1)^2-1)]

- wasiqss

HENCE we get zero in denominator

- 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

now we need to apply binomeal expansion

- 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

arghh i wish i was good at making you understand it

- wasiqss

we need it!

- TuringTest

you cannot have 0 in the denom, that shows you that you have already made a mistake !!!!

- wasiqss

well to counter this zero!! we need to multiply x

- TuringTest

...as I was saying...

- wasiqss

plz no offense but plz go revise DE
s

- TuringTest

._.

- 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

yes, but I think it is way more work than is necessary
now I just wanna hear @lalaly

- wasiqss

kay

- TuringTest

oh dear, we seem to have lost her
I am going to try this op paper I guess

- wasiqss

hahaha lana dear

- wasiqss

\[[1+(D(D+1)-1)^-1[x^2e^x\]

- wasiqss

we have to expand binomeally
[1+(D(D+1)−1)−1

- wasiqss

well i have a solution on page i wonder how to give you that

- anonymous

take a picture with your cellphone and attach it

- anonymous

or use \[\LaTeX\]

- wasiqss

good idea twin can i do it tomorow because i would need to upload it too?

- TuringTest

okay got it :)

- wasiqss

latex is something not my type

- wasiqss

turing my approach?

- TuringTest

do everything normally, then multiply the particular by x at the end
that is all

- anonymous

get used to it, it's a very helpful Latex is a very useful tool

- wasiqss

turing binomeal is essential in it!

- TuringTest

oh yeah?, just watch...

- 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

oh crap I see a mistake

- wasiqss

wait you told me w=xe^x!

- anonymous

dude this whole discussion was based on finding what W equaled

- wasiqss

ok just ask lana what is the right answer

- TuringTest

@MathSofiya yes, because the entire problem we were having was keeping our solutions linearly independent, which we check by finding W

- wasiqss

waittttttttttttt what the....... i was solving by another method not by variation of parameter lol

- 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

there we go... stupid algebra tripped me up :/

- anonymous

yeah that's what I get
\[-e^x+e^xx\]

- TuringTest

so yc+yp gives what I have, and what wolf has
so we're good :)

- anonymous

good. Hi 5

- TuringTest

back atchya!

- anonymous

As always thank you!

- TuringTest

As always you're welcome :D

- 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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