What is the Correct Way to Use the Integrating Factor Method for Proving an ODE?

In summary, the conversation was about solving a ODE using the integrating factor method. The problem given was q'+2q=5sin(t) where q(0)=0, and the integrating factor was found to be e^2t. The final answer was q= e^-2t +2sint-cost. However, there was some confusion on the steps to get to the final answer and the correct integration method was discussed.
  • #1
hurcw
23
0
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.
 
Physics news on Phys.org
  • #2
welcome to pf!

hi hurcw! welcome to pf! :smile:

(try using the X2 button just above the Reply box :wink:)
hurcw said:
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 :smile:
 
  • #3
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
hurcw said:
Thanks for the reply if i integrate 5e^2t sin(t) do i not just get the same+C …

nooo :redface:

d/dt(5e2tsint) = 10e2tsint + 5e2tcost :wink:
 
  • #5
Ahhh i see...i think,
I then multiply this by e(-2t)
Which cancels out the e(2t) am i right?
i still don't see how i end up at my original answer?
 
  • #6
hurcw said:
… i still don't see how i end up at my original answer?

show us what you've done :smile:
 
  • #7
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:
  • #8
(just got up :zzz: …)
hurcw said:
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) :redface:

(also, that "2" on the far left came from nowhere)
 
  • #9
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:
  • #10
Can anyone offer any more assistance with his please, all the help so far has been greatly appreciated.
 
  • #11
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 :redface:

you must write these proofs out more carefully and in full (ie without taking short-cuts by missing out lines)​
 

1. What is an integrating factor?

An integrating factor is a function used in solving differential equations. It is multiplied by both sides of the equation in order to make it easier to solve.

2. When should I use an integrating factor?

An integrating factor is typically used when solving first-order linear differential equations. It can also be used to solve other types of equations, such as Bernoulli equations.

3. How do I find the integrating factor?

The integrating factor can be found by multiplying the entire equation by the appropriate function. This function is found by taking the coefficient of the highest order derivative and finding the exponential of its integral.

4. Can an integrating factor be used for all types of differential equations?

No, an integrating factor is only used for first-order linear differential equations or equations that can be transformed into this form.

5. What is the purpose of using an integrating factor?

The purpose of using an integrating factor is to make the equation easier to solve by simplifying the differential equation into an exact differential equation.

Similar threads

  • Differential Equations
Replies
8
Views
3K
  • Differential Equations
Replies
2
Views
1K
Replies
2
Views
2K
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Differential Equations
Replies
4
Views
2K
  • Poll
  • Differential Equations
Replies
2
Views
2K
  • Differential Equations
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
876
  • Differential Equations
Replies
4
Views
2K
Back
Top