Help with Integrating Factor

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.

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hi hurcw! welcome to pf!

(try using the X2 button just above the Reply box )
 Quote by hurcw 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

 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

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Help with Integrating Factor

 Quote by hurcw 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

 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?

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 Quote by hurcw … i still dont see how i end up at my original answer?
show us what you've done

 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 ??

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(just got up …)
 Quote by hurcw 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)

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

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