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?

Question

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 · Answer

No I did that wrong, one moment

anonymous · Answer

no it equals $$e^{2x}$$

anonymous · Answer

$$y_p(x)=-e^xx^2+x^2e^x$$

turingtest · Answer

wrong complimentary

turingtest · Answer

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

anonymous · Answer

$$r^2-r=0$$

anonymous · Answer

r=0 and r=1

turingtest · Answer

so the complimentary is...?

anonymous · Answer

$$y_c=c_1+c_2e^x$$

turingtest · Answer

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 · Answer

sure
$$y_c=c_1+c_2xe^x$$

turingtest · Answer

$$y_c=c_1x+c_2xe^x$$

anonymous · Answer

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

turingtest · Answer

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 · Answer

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

anonymous · Answer

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

turingtest · Answer

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

anonymous · Answer

Make that 7 mins, ha!

anonymous · Answer

$$W(y_1,y_2)=x^2e^x$$

anonymous · Answer

$$y_1=x$$
$$y_2=xe^x$$

anonymous · Answer

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

wasiqss · Answer

well ik it

wasiqss · Answer

my twin :D

anonymous · Answer

hi twin

turingtest · Answer

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 · Answer

well i can do it

wasiqss · Answer

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

anonymous · Answer

y''-y'=e^x

turingtest · Answer

we need variation of parameters @wasiqss

wasiqss · Answer

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

turingtest · Answer

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 · Answer

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

wasiqss · Answer

tell me equation man completely cause this seems messy

turingtest · Answer

y''-y'=e^x

wasiqss · Answer

easy one lol

turingtest · Answer

then do it, by all mean!

anonymous · Answer

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

wasiqss · Answer

y complimentary is y= c1 +c2e^-x

turingtest · Answer

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

turingtest · Answer

complimentary is
y=c1+c2e^x

anonymous · Answer

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

wasiqss · Answer

yes cause roots are zero and -1

turingtest · Answer

no they are not

turingtest · Answer

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

anonymous · Answer

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

turingtest · Answer

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 · Answer

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

turingtest · Answer

w=x^2e^x *

anonymous · Answer

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

anonymous · Answer

yeah that's what I had

turingtest · Answer

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 · Answer

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

turingtest · Answer

so let's try that

turingtest · Answer

otherwise we get W=0 which is a failure

turingtest · Answer

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

anonymous · Answer

$$y_c=c_1x+c_2xe^x$$
$$\left|\begin{matrix}x&xe^x\1&e^x(x-1)\end{matrix}\right|=x^2e^x$$

anonymous · Answer

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

turingtest · Answer

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

turingtest · Answer

*$$\int\frac{e^x}xdx$$

turingtest · Answer

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

anonymous · Answer

no

anonymous · Answer

@lalaly is here

turingtest · Answer

okay. then we have a set of vectors (in this case equations) if their determinant is zero we know they are linearly dependent

wasiqss · Answer

now listen

wasiqss · Answer

for y particular

turingtest · Answer

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 · Answer

is it xe^x

wasiqss · Answer

so as the denominator becomes zero

turingtest · Answer

and would you care to explain how/why ?

wasiqss · Answer

we multiply by x

wasiqss · Answer

cofficeint of e^1x

wasiqss · Answer

=1

wasiqss · Answer

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

wasiqss · Answer

HENCE we get zero in denominator

turingtest · Answer

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 · Answer

now we need to apply binomeal expansion

turingtest · Answer

@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 · Answer

arghh i wish i was good at making you understand it

wasiqss · Answer

we need it!

turingtest · Answer

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

wasiqss · Answer

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

turingtest · Answer

...as I was saying...

wasiqss · Answer

plz no offense but plz go revise DE
s

turingtest · Answer

._.

wasiqss · Answer

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 · Answer

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

wasiqss · Answer

kay

turingtest · Answer

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

wasiqss · Answer

hahaha lana dear

wasiqss · Answer

$$[1+(D(D+1)-1)^-1[x^2e^x$$

wasiqss · Answer

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

wasiqss · Answer

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

anonymous · Answer

take a picture with your cellphone and attach it

anonymous · Answer

or use $$\LaTeX$$

wasiqss · Answer

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

turingtest · Answer

okay got it :)

wasiqss · Answer

latex is something not my type