Question

tx''-(t+1)x'+x=0 t>0 f(t)=e^t
need help finding my mistake! Please
In Problems 45 through 48, a differential equation and a non-trivial solution f are given. Find a second linearly independent solution using reduction of order.

Answer 1

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Answer 2

@primeralph think you could help me out on this one?

Answer 3

i was able to do the last problem that was similar to this

Answer 4

would you like to see the last problem i worked?

Answer 5

Hold on.

Answer 6

@KingGeorge

Answer 7

@satellite73

Answer 8

should i close the question? is anyone here?

Answer 9

@dumbcow wanna take a look at this before i close it?

Answer 10

I'm insanely new to this kind of stuff, but I'm kinda just reading it over for the hell of it x_x

Answer 11

lol

Answer 12

I'll have to do it in a month anyway myself, so might as well see what Im in for.

Answer 13

i feel like a zombie all ive been doing all day everyday for the last 3 weeks is differential equations. been neglecting phys 2 which sucks but this class is a prereq for two im taking in the fall so i NEED TO pass it

Answer 14

Yeah, I hear ya. I'm taking it starting at the end of August. I'm reluctant to buy out textbook yet since apparently it's crap. But I have been looking over some of the things that will be in the class and I recognized the reduction of order so I thought Id go look.

Answer 15

im gonna close the question. thanks for looking everyone!

Answer 16

Out of curiosity, did you actually get an answer when trying to solve?

Answer 17

no i couldn't get past the part of substituting and getting a v'' and v' i got two v's for some reason. i did cancel the v though which was good but the two v' throw me off

Answer 18

I see. I got past that part actually. I may nothave a correct answer, but I'm on to something. Almost done and I can give ya an answer, which may be a million miles off, but an answer.

Answer 19

ok

Answer 20

Hmm....Probably a mistake, but somehow I ended up with something where nearly everything cancelled out. No idea if that's an automatic sign of a wrong answer, but the second part of the general solution I got was simply C2(-t-1). After all the substitutions, things just canceled out. Perhaps the linearly independent part that I do not know about throws a monkey wrench into it. But I can maybe show where I got the v'' and v' to work. Although dumbcow may have all your answers, too, haha.

Answer 21

*Had fun trying no matter what* :)

Answer 22

sorry i was looking up some stuff
which method did you use?
here is a good site to use
http://tutorial.math.lamar.edu/Classes/DE/LaplaceTransforms.aspx

also in case you dont have solution:
http://www.wolframalpha.com/input/?i=tx%28t%29%27%27-%28t%2B1%29x%28t%29%27%2Bx%28t%29%3D0

wolfram uses LaPlace transforms

Answer 23

yeah i get excited if i even get close to the ans cause then i know its just a small mistake somewhere but unfortunately this is an even problem so i don't know the answer.

Answer 24

Thats actually where I looked. I tried following along with the reduction of order section. It at least led me to the point where I could even get an answer.

Answer 25

i tried using reduction of order

Answer 26

@rperez36 What is the class this work is from called?

Answer 27

differential equations

Answer 28

hello are you still having trouble?

Answer 29

so where is my mistake? i checked over and over my product rule and distribution but couldn't see why i have two v'

Answer 30

yes still having trouble lol

Answer 31

I had two v' as well if it makes ya feel better, lol.

Answer 32

@rperez36 In the 4th line of the picture, where did you get it from?

Answer 33

i think we are only supposed to have a v'' and v'

Answer 34

i applied red of order using your equations and found that \(v=\int\limits_{}^{}\frac{ e ^{\frac{ -1 }{ t }} }{ t }dt\)

Answer 35

Thats what id be unsure of. I just did reduction of order anyway, even though I had two v'. I don't know the type of problems well enough to know if that's even legal xD

Answer 36

i did v times the known solution e^x which is what i did for the last problem and i was able to work it

Answer 37

I created a variable \(w=v'\) and solved the 1st order, separable equation that resulted

Answer 38

Yeah, that was the step I basically did after I had the two v's. I just did the w' = v'' and w=v'

Answer 39

can we have two v' ?? if so then i shoulve just continued

Answer 40

I continued just for the hell of it to see if I'd get anything that'd look logical.

Answer 41

tx''-(t+1)x'+x=0 t>0 f(t)=e^t
need help finding my mistake! Please
In Problems 45 through 48, a differential equation and a non-trivial solution f are given. Find a second linearly independent solution using reduction of order.

For reduction of order, assume there exists another solution \(x(t)=u(t)e^t\) thus \(x'(t)=u'(t)e^t+u(t)e^t,x''(t)=u''(t)e^t+2u'(t)e^t+u(t)e^t\). Substituting in we get:$$t(u''e^t+2u'e^t+ue^t)-(t+1)(u'e^t+ue^t)+ue^t=0\\tu''e^t+2tu'e^t+tue^t-tu'e^t-tue^t-u'e^t-ue^t+ue^t=0\\tu''e^t+tu'e^t-u'e^t=0\\tu''+(t-1)u'=0$$

Answer 42

after doing this, I got \(\int\limits_{}^{}\frac{ dw }{ w }=\int\limits_{}^{}\frac{ \frac{ 1 }{ t } -1}{ t }dt\)

Answer 43

I got something pretty damn close to what ybarrap got it seems.

Answer 44

If I knew how to understand when math is written that way then maybe I could see my mistake, too.

Answer 45

@oldrin.bataku completed it. Looks good

Answer 46

what happened to the e^t from (t-1)u'

Answer 47

@rperez36 i divided thru by \(e^t\) (this is safe because \(e^t\ne0\))

Answer 48

oh nice! i didn't even realize they were all gone. Nice job!

Answer 49

i had a feeling there was a "trick" involved lol

Answer 50

Since I'm new to this, I apologize for the stupid question, but once you get to v = te^(-t), you don't have to integrate it back into terms of u? Where you were before the reduction oforder I mean.

Answer 51

I made it that far and thought I had to integrate it back.

Answer 52

well you substitute a variable for v' then make variable' equal v'' to make it a first order equation seperate then solve for variable then replace back with v' and integrate from there if that makes any sense. i need to get to bed have class in the am

Answer 53

Night then.

Answer 54

night everyone!

Answer 55

@Psymon you are totally correct I made a very dumb error. sorry @rperez36

Answer 56

Let \(v=u'\) to reduce it to a first-order separable ODE:$$tv'+(t-1)v=0\\tv'=(1-t)v\\\frac1vdv=\frac{1-t}tdt\\\int\frac1vdv=\int\left(\frac1t-1\right)dt\\\log v=\log t-t\\v=e^{\log t}e^{-t}=te^{-t}$$Switching back to \(u\) we have:$$u=\int te^{-t}\,dt=-te^{-t}+\int e^{-t}\,dt=-te^{-t}-e^{-t}=-e^{-t}(t+1)$$... hence our solution is given by \(x=ue^t=-(t+1)\) hence our second solution is merely \(-(t+1)\) (or more simply \(t+1\) as the coefficient does not matter -- our equation is linear)

Answer 57

Hm...I somehow got it right. Ican sleep happy now. Thanks for finishing that problem, I wanted to see the answer as well ^_^