Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (fibonaccichick666):

ODE question: Assume that \(\phi "+p(t)\phi ' +q(t) \phi =0\) admits solutions \(\phi=t\) and \(\phi(t)=e^t\). Find p(t) and q(t). Also solvey"+p(t)y'+q(t)y=f(t).

OpenStudy (fibonaccichick666):

I am stuck. I found \(p(t)=c-\frac{1}{t-1}\) but how do I find q(t)?

OpenStudy (fibonaccichick666):

i found p(t) using Abel's theorem. But there is no equivalent I can find for q(t)

OpenStudy (freckles):

Curious did you solve this one since you close it?

OpenStudy (fibonaccichick666):

yea I found my mistake

OpenStudy (fibonaccichick666):

I didn't take the derivative of C first time round... it was stupid

OpenStudy (freckles):

I'm not totally familiar with abel...But I was thinking this and maybe it is not appropriate . I know you have the answer but I kinda wanted to give it a shot too. :) \[ \phi=t \\ \phi'=1 \\ \phi''=0 \\ 0+p(1)+q(t)=0 \\ p+qt=0 \\ \text{ now the other solution } e^t=\phi=\phi'=\phi'' \\ e^t+pe^t +qe^t=0 \\ \\ \text{ now the combination of those solutions } \\ \phi=e^t+t \\ \phi'=e^t +1 \\ \phi''=e^t \\ e^t+p(e^t+1)+q(e^t+t)=0\\ \text{ we sorta have a system to solve now } \\ p=-qt \text{ from first equation } \\ \text{ second equation } e^t+pe^t+qe^t=0 \text{ can be written as } 1+p+q=0 \\ \text{ so } p+q=-1 \\ \text{ so hmmm using first equation with last equation I got we have : } \\ e^t-qt(e^t+1)+q(e^t+t)=0 \\ \text{ so we can solve for } q \text{ I believe } \\ \\ q[-te^t-t+e^t+t]=-e^t \\ \\ q=\frac{-e^t}{e^t(1-t)+0}=\frac{1}{1-t} \\ \text{ then } p+q=-1 \\ p+\frac{1}{1-t}=-1 \\ p(1-t)+1=-1(t-1) , t \neq -1 \\ p(1-t)=1-t-1 \\ p=\frac{-t}{1-t} \\ \text{ now I didn't consider the most general from of the sum solution I think } \]

OpenStudy (fibonaccichick666):

yep that's p(t)

OpenStudy (freckles):

I just plug in the solutions basically and also plug in the sum of the solution into differenitial equation and solved a system of equations

OpenStudy (fibonaccichick666):

abel's is just faster, it's at the bottom of this page http://tutorial.math.lamar.edu/Classes/DE/Wronskian.aspx

OpenStudy (fibonaccichick666):

nice work

OpenStudy (fibonaccichick666):

see if you can get q(t),it's not too far off from p

OpenStudy (freckles):

yah since their sum is suppose to be -1 :) Or that is what I have :)

OpenStudy (freckles):

Anyways cute question

OpenStudy (fibonaccichick666):

hahahaha thanks. It's this "review" sheet for my test. He didn't cover half of it...

OpenStudy (freckles):

I think that is the way that most advance classes go

OpenStudy (freckles):

too much to shove into class lectures

OpenStudy (fibonaccichick666):

no, he's just a spiteful (insert inappropriate word here)

OpenStudy (fibonaccichick666):

he is so awful

OpenStudy (freckles):

lol

OpenStudy (freckles):

Well hopefully you will be done with him soon.

OpenStudy (fibonaccichick666):

check out this question, I still have no answer. No one at my university can do it apparently http://openstudy.com/users/fibonaccichick666#/updates/54e4068ee4b00ad827d6874a

OpenStudy (fibonaccichick666):

only a month and a half left. I cannot wait

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!