Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

MathSofiyaBest ResponseYou've already chosen the best response.1
\[y_c=c_1e^{x}+c_2xe^{x}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
now we need the wronskian, do you know about that?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
it is the determinant of a matrix
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
for\[y_c=c_1y_1+c_2y_2\]the Wronskian is\[W(y_1,y_2)=\left\begin{matrix}y_1&y_2\\y_1'&y_2'\end{matrix}\right\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
that's a determinant in case you forgot, so find that determinant
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\left\begin{matrix}e^{x}&xe^{x}\\e^{x}&e^{x}(x1)\end{matrix}\right\] =\[e^{x}(e^{x}(x1))xe^{x}e^{x}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yep you caught it :)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[e^{x}(e^{x}(x1))+xe^{x}e^{x}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
you dropped a minus
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[e^{x}(e^{x}(x1))+xe^{x}e^{x}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
simplify and be happy :)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
does it equal\[ e^{x}\]? \[e^{x}(e^{x}(x1))+xe^{x}e^{x}=e^{x}\] \[{x}(e^{x}(x1))+xe^{x}=1\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[(e^{x}(x1))+xe^{x}=1\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
no, why did you set it equal to e^x, we can't set it equal to anything yet...
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
so how should I simplify it then?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[e^{x}(e^{x}(x1))+xe^{x}e^{x}=e^{x}(e^{x}(1x))+xe^{x}e^{x}\]now try again while I try to eat ;)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
no no no ...wait a second
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
Let's try that again \[e^{x}(e^{x}(x1))+xe^{x}e^{x}=e^{x}(e^{x}(1x))+xe^{x}e^{x}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
they're just equal to each other
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[e^{x}(e^{x}(x1))+xe^{x}e^{x}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[e^{2x}(x1)+xe^{2x}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[e^{2x}((1x)+x)\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yes, sorry, I was afk
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
do you know the formula for the particular solution for variation of parameters?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
something like that?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
almost unidentifiable to me in that form, but correct I'm sure I have the more direct formula memorized http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx (the green box with the formula that says "variation of parameters)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
is the first one a y''? \[y''+q(t)y'+r(t)y=g(t)\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yeah, it looks screwy I know
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
do we know what q(t) and r(t) are? are they the coefficients of y' and y? Probably not it looks like
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yes and in this case they are the coefficients, which in this case are constants the formula is just pointing out that they may not be
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
oh I see, soo... \[y''+q(t)y'+r(t)y=g(t)\] \[y''2y'+y=e^{2x}\] so now I integrate
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
right, using the formula... r(t) and r(t) don't even come into the formula for the particular solution in variation of parameters
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
q(t) and r(t)... they determine the complimentary only, which as you can see *is* part of the formula for the particular
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[Y_p(t)=e^{x}\int\frac{xe^{x}e^{x}}{e^{2x}(e^{x},xe^{x})}dx+xe^{x}\int\frac{e^{x}e^{2x}}{e^{2x}(e^{x},xe^{x})}dx\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
the first integrand is wrong I think
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I think you put the wrong g(x)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[Y_p(t)=e^{x}\int\frac{xe^{x}e^{2x}}{e^{2x}(e^{x},xe^{x})}dx+xe^{x}\int\frac{e^{x}e^{2x}}{e^{2x}(e^{x},xe^{x})}dx\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
oh wait, wait, the denominator in each integrand is the wronskian, which we found to be e^(2x) I think you are confusing notation with the parentheses
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
Oh lol so the denominator is just e^{2x}
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
notice that makes our integrals quite tolerable :)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[Y_p(t)=e^{x}\int\frac{xe^{x}e^{2x}}{e^{2x}}dx+xe^{x}\int\frac{e^{x}e^{2x}}{e^{2x}}dx\] That's a lot easier to integrate
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
oh yeah :) especially after a little simplification
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\frac{e^{2x}}{9}(3x1)+\frac x3e^{3x}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
so what was the point of doing variation of parameters
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
well now you have a particular solution
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
y(x)=yc+yp you got you now so you're good to go
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I did not check your integral btw
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
that's ok I did wolfram
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
righto, I'd do the same right now :)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
oh I see \[c_1e^{x}+c_2xe^{x}\frac{e^{2x}}{9}(3x1)+\frac x3e^{3x}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yep, and of course if this is an IVP we now proceed to find c1 and c2 otherwise we're done having found the general solution
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
good, wow, that was a lot THanks Turing!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
Always a pleasure Sofiya :D
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.