# Help with Integrating Factor

1. Feb 14, 2012

### hurcw

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.

2. Feb 15, 2012

### tiny-tim

welcome to pf!

hi hurcw! welcome to pf!

(try using the X2 button just above the Reply box )
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

3. Feb 15, 2012

### hurcw

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

4. Feb 15, 2012

### tiny-tim

nooo

d/dt(5e2tsint) = 10e2tsint + 5e2tcost

5. Feb 15, 2012

### hurcw

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?

6. Feb 15, 2012

### tiny-tim

show us what you've done

7. Feb 15, 2012

### hurcw

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 ??

Last edited: Feb 15, 2012
8. Feb 16, 2012

### tiny-tim

(just got up :zzz: …)
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)

9. Feb 16, 2012

### hurcw

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 ?????

Last edited: Feb 16, 2012
10. Feb 18, 2012

### hurcw

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

11. Feb 19, 2012

### tiny-tim

hmm … let's rewrite that so that it's readable …
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)​