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anonymous
 4 years ago
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?
anonymous
 4 years ago
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?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No I did that wrong, one moment

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no it equals \[e^{2x}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y_p(x)=e^xx^2+x^2e^x\]

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2so the complimentary is...?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2right, 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
 4 years ago
Best ResponseYou've already chosen the best response.0sure \[y_c=c_1+c_2xe^x\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2\[y_c=c_1x+c_2xe^x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why on both c_1 and c_2, how did come to that conclusion?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2or 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
 4 years ago
Best ResponseYou've already chosen the best response.0I'm not sure... DunnBros is about to kick me out, they close at 11pm

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'll be back on in like 15 mins when I get home

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2fair enough, I'm about to call it a night as well see ya, I will investigate this :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Make that 7 mins, ha!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[W(y_1,y_2)=x^2e^x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y_1=x\] \[y_2=xe^x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y_p(x)=x\int \frac{e^x}xdx+xe^x\int x^{1}dx\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2that 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
 4 years ago
Best ResponseYou've already chosen the best response.0tell me equation for which i have to solve\y complimentary

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

wasiqss
 4 years ago
Best ResponseYou've already chosen the best response.0yehh just tell me equation i ll take out solution. as y complimentary is solved i ll solve for y particular

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=integral+of+e^%28x%29%2Fx

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i disagree with the above answer, shouldn't it be e^x, yeah what turing has

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do you think my W is right? x^2 e^x

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2that 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
 4 years ago
Best ResponseYou've already chosen the best response.2with W=e^2e^x I don't think that first integral is closed, check the wolf, I have not yet

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do we agree that \[y_c=c_1x+c_2xe^x\] or should it be \[y_c=c_1+c_2e^x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah that's what I had

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2that 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
 4 years ago
Best ResponseYou've already chosen the best response.2however if we are to try the first we need to multiply g(x) by x (I'm thinking....)

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y_c=c_1x+c_2xe^x\] \[\left\begin{matrix}x&xe^x\\1&e^x(x1)\end{matrix}\right=x^2e^x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you explain the g(x) concept again and why it being e^x would be a problem because that cancels something out?...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2no I think that is wrong because that leads to the particular having\[\int\frac{e^x}dxx\]

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

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2okay. then we have a set of vectors (in this case equations) if their determinant is zero we know they are linearly dependent

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2that 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
 4 years ago
Best ResponseYou've already chosen the best response.0so as the denominator becomes zero

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

wasiqss
 4 years ago
Best ResponseYou've already chosen the best response.0HENCE we get zero in denominator

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2anyway... 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
 4 years ago
Best ResponseYou've already chosen the best response.0now we need to apply binomeal expansion

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2@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
 4 years ago
Best ResponseYou've already chosen the best response.0arghh i wish i was good at making you understand it

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2you cannot have 0 in the denom, that shows you that you have already made a mistake !!!!

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2...as I was saying...

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

wasiqss
 4 years ago
Best ResponseYou've already chosen the best response.0you 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
 4 years ago
Best ResponseYou've already chosen the best response.2yes, but I think it is way more work than is necessary now I just wanna hear @lalaly

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2oh dear, we seem to have lost her I am going to try this op paper I guess

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

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

wasiqss
 4 years ago
Best ResponseYou've already chosen the best response.0well i have a solution on page i wonder how to give you that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0take a picture with your cellphone and attach it

wasiqss
 4 years ago
Best ResponseYou've already chosen the best response.0good idea twin can i do it tomorow because i would need to upload it too?

wasiqss
 4 years ago
Best ResponseYou've already chosen the best response.0latex is something not my type

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2do everything normally, then multiply the particular by x at the end that is all

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0get used to it, it's a very helpful Latex is a very useful tool

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2oh yeah?, just watch...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2actually 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
 4 years ago
Best ResponseYou've already chosen the best response.2oh crap I see a mistake

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dude this whole discussion was based on finding what W equaled

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2@MathSofiya yes, because the entire problem we were having was keeping our solutions linearly independent, which we check by finding W

wasiqss
 4 years ago
Best ResponseYou've already chosen the best response.0waittttttttttttt what the....... i was solving by another method not by variation of parameter lol

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah that's what I get \[e^x+e^xx\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.2so yc+yp gives what I have, and what wolf has so we're good :)

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

Mendicant_Bias
 2 years ago
Best ResponseYou've already chosen the best response.0I 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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