
#1
Feb1412, 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 +2sintcost 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. 



#2
Feb1512, 03:47 AM

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hi hurcw! welcome to pf!
(try using the X^{2} button just above the Reply box ) e^{2t}q' +2qe^{2t} = 5e^{2t}sin(t)that's the same as: (qe^{2t})' = 5e^{2t}sin(t)now you integrate both sides: qe^{2t} = ∫ 5e^{2t}sin(t) dtand finally multiply the RHS (after the integration, and don't forget the constant!) by e^{2t}, to give you q 



#3
Feb1512, 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 e^{2t}tso i am left with 5sin(t)+ C.
Or am i being completely retarded which is a major possibility 



#4
Feb1512, 10:33 AM

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Help with Integrating Factord/dt(5e^{2t}sint) = 10e^{2t}sint + 5e^{2t}cost 



#5
Feb1512, 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? 



#7
Feb1512, 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 ?? 



#8
Feb1612, 04:16 AM

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(just got up …)
if you differentiate it, you don't get 5e^{(2t)}sin(t) (also, that "2" on the far left came from nowhere) 



#9
Feb1612, 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 ????? 



#10
Feb1812, 04:02 PM

P: 20

Can anyone offer any more assistance with his please, all the help so far has been greatly appreciated.




#11
Feb1912, 02:26 PM

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hmm … let's rewrite that so that it's readable …
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 shortcuts by missing out lines) 


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