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Help with Integrating Factor

by hurcw
Tags: factor, integrating
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hurcw
#1
Feb14-12, 09:38 PM
P: 20
I am really struggling with proving a ODE by means of using the integrating factor method.
My original problem was a Laplace transform
q'+2q=5sin(t) where q(0)=0
I believe i have got the correct naswer for this as being:- q= e^-2t +2sint-cost
I just need to confirm this i have my integrating factor as e^2x but after that i am not really sure where to go next.
I am surrounded by piles of paper with varying answers on, non of which are the same as the one above.
Please please help someone.
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tiny-tim
#2
Feb15-12, 03:47 AM
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hi hurcw! welcome to pf!

(try using the X2 button just above the Reply box )
Quote Quote by hurcw View Post
q'+2q=5sin(t) where q(0)=0

i have my integrating factor as e^2x but after that i am not really sure where to go next.
yes, you multiply the orignal equation by e2t, giving:
e2tq' +2qe2t = 5e2tsin(t)
that's the same as:
(qe2t)' = 5e2tsin(t)
now you integrate both sides:
qe2t = ∫ 5e2tsin(t) dt
and finally multiply the RHS (after the integration, and don't forget the constant!) by e-2t, to give you q
hurcw
#3
Feb15-12, 08:27 AM
P: 20
Thanks for the reply if i integrate 5e^2t sin(t) do i not just get the same +C, and then if i multiply by e-2t it cancels out the e2ttso i am left with 5sin(t)+ C.
Or am i being completely retarded which is a major possibility

tiny-tim
#4
Feb15-12, 10:33 AM
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Help with Integrating Factor

Quote Quote by hurcw View Post
Thanks for the reply if i integrate 5e^2t sin(t) do i not just get the same+C
nooo
d/dt(5e2tsint) = 10e2tsint + 5e2tcost
hurcw
#5
Feb15-12, 03:56 PM
P: 20
Ahhh i see.....i think,
I then multiply this by e(-2t)
Which cancels out the e(2t) am i right?
i still dont see how i end up at my original answer?
tiny-tim
#6
Feb15-12, 04:31 PM
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Quote Quote by hurcw View Post
i still dont see how i end up at my original answer?
show us what you've done
hurcw
#7
Feb15-12, 06:01 PM
P: 20
Right here it is:-

q'+2q=5sin(t)
IF= e(2t) , Q=5sin(t)
e(2t)q'+2e(2t)q=5e(2t)sin(t)
d/dt2(e(2t)q)=5e(2t)sin(t)

2(e(2t)q)=∫5e(2t)sin(t).dt

= 10e(2t)sin(t)+5e(2t)cos(t)+C

e(2t)q=(5(2e(2t)sin(t)+e(2t)cos(t))+C/2

q=(5(2e(2t)sin(t)+e(2t)cos(t))+C/2e(2t)

I then put in my original boundary condition of q(0)=0 to find C

And end up with C=-5
which kind of tallies up because q = 10e(2t)sin(t)+5e(2t)cos(t)+C

0 = (0x1)+(5x1)-5
Transposed 5 = 5
Hows this look so far ??
tiny-tim
#8
Feb16-12, 04:16 AM
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(just got up )
Quote Quote by hurcw View Post
q'+2q=5sin(t)
IF= e(2t) , Q=5sin(t)
e(2t)q'+2e(2t)q=5e(2t)sin(t)
d/dt2(e(2t)q)=5e(2t)sin(t)

2(e(2t)q)=∫5e(2t)sin(t).dt

= 10e(2t)sin(t)+5e(2t)cos(t)+C
no, that last line is wrong

if you differentiate it, you don't get 5e(2t)sin(t)

(also, that "2" on the far left came from nowhere)
hurcw
#9
Feb16-12, 08:52 AM
P: 20
The '2' is from the original 2q.
I don't get what you mean about my differentiation at the end, i have not differentiated yet.
Where am i going wrong ?
So is it this line that is wrong? q=(5(2e(2t)sin(t)+e(2t)cos(t))+C/2e(2t)
Should it be q=5(2e(2t)sin(t)+e(2t)cos(t))+C/e(2t)
which would cancel down to q=5(e(2t)sin(t)+cos(t))+C wouldn't it ?????
hurcw
#10
Feb18-12, 04:02 PM
P: 20
Can anyone offer any more assistance with his please, all the help so far has been greatly appreciated.
tiny-tim
#11
Feb19-12, 02:26 PM
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hmm let's rewrite that so that it's readable
q=(5(2e2tsin(t)+e2tcos(t))+C/2e2t)
Should it be q=5(2e2tsin(t)+e2tcos(t))+C/e2t
which would cancel down to q=5(e2tsin(t)+cos(t))+C
no your ∫ was wrong (there should be a minus in the first line),

and because you've put your brackets in the wrong place, there are errors in the next two lines also
you must write these proofs out more carefully and in full (ie without taking short-cuts by missing out lines)


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