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FP2: integrating factor

I'm trying to do the question by myself by using the auxiliary equation and trying to make the integrating factor myself in order to improve my understanding, but I'm stuck. I've read through the textbook and watched exam solutions.

I know that m = 0 + (+-)i5 => (p (+-) iq)

when a value for m = 0, we end to multiply the lambda of the integrating factor by x.

the integrating factor a trig function is: lambda(cos(w(x)) + mu(sin(w(x))

therefore, it should be: x.lambda(cos(w(x)) + mu(sin(w(x))?

There is some knowledge/understanding I am missing and is anyone able to fill in the gaps?
(edited 5 years ago)

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Original post by Maths&physics
I'm trying to do the question by myself by using the auxiliary equation and trying to make the integrating factor myself in order to improve my understanding, but I'm stuck. I've read through the textbook and watched exam solutions.

I know that m = 0 + (+-)i5 => (p (+-) iq)


Ok... looks fine..

when a value for m = 0, we end to multiply the lambda of the integrating factor by x.


Not sure what you're saying here.



This should probably help if you're confused about when we need to choose a PI that is being multiplied by xx; https://www.thestudentroom.co.uk/showthread.php?t=5147458#post75629378
If your P.I. is in the differential equation, then a reasonable try is to multiply it by x.

So the term you've written would be correct.

Also you've said integrating factor when you meant particular integral.
Original post by Maths&physics
I'm trying to do the question by myself by using the auxiliary equation and trying to make the integrating factor myself in order to improve my understanding, but I'm stuck. I've read through the textbook and watched exam solutions

And after all that, you still don't know the difference between an integrating factor and a particular integral...
Original post by DFranklin
And after all that, you still don't know the difference between an integrating factor and a particular integral...


:biggrin: well, from my working out, I clearly do and I obviously mixed up the terms. I'm only human.
Original post by RDKGames
Ok... looks fine..



Not sure what you're saying here.



This should probably help if you're confused about when we need to choose a PI that is being multiplied by xx; https://www.thestudentroom.co.uk/showthread.php?t=5147458#post75629378


sorry, I meant P.I, not I.F
Original post by Maths&physics
sorry, I meant P.I, not I.F


Well OK.

From your question, you get m=±5im = \pm 5i therefore yC=Acos(5x)+Bsin(5x)y_C = A\cos(5x) + B\sin(5x).

So when it comes to the P.I., since the RHS of the ODE already appears in the C.F., we cannot use λsin(5x)+μcos(5x)\lambda \sin(5x) + \mu \cos(5x). We need to use

yP=x[λsin(5x)+μcos(5x)]y_P = x[\lambda \sin(5x) + \mu \cos(5x)]

and so if you go through the procedure to figure out λ,μ\lambda,\mu you would find that μ=0\mu = 0.


The question tells you to use yC=λxsin(5x)y_C = \lambda x \sin (5x) because it's already done some of the grunt work for you by imposing the condition μ=0\mu = 0 right from the bat.

That's all there is to it.
Original post by RDKGames
Well OK.

From your question, you get m=±5im = \pm 5i therefore yC=Acos(5x)+Bsin(5x)y_C = A\cos(5x) + B\sin(5x).

So when it comes to the P.I., since the RHS of the ODE already appears in the C.F., we cannot use λsin(5x)+μcos(5x)\lambda \sin(5x) + \mu \cos(5x). We need to use

yP=x[λsin(5x)+μcos(5x)]y_P = x[\lambda \sin(5x) + \mu \cos(5x)]

and so if you go through the procedure to figure out λ,μ\lambda,\mu you would find that μ=0\mu = 0.


The question tells you to use yC=λxsin(5x)y_C = \lambda x \sin (5x) because it's already done some of the grunt work for you by imposing the condition μ=0\mu = 0 right from the bat.

That's all there is to it.


absolutely brilliant, I've seen where ive gone wrong. thank you!
Original post by RDKGames
Well OK.

From your question, you get m=±5im = \pm 5i therefore yC=Acos(5x)+Bsin(5x)y_C = A\cos(5x) + B\sin(5x).

So when it comes to the P.I., since the RHS of the ODE already appears in the C.F., we cannot use λsin(5x)+μcos(5x)\lambda \sin(5x) + \mu \cos(5x). We need to use

yP=x[λsin(5x)+μcos(5x)]y_P = x[\lambda \sin(5x) + \mu \cos(5x)]

and so if you go through the procedure to figure out λ,μ\lambda,\mu you would find that μ=0\mu = 0.


The question tells you to use yC=λxsin(5x)y_C = \lambda x \sin (5x) because it's already done some of the grunt work for you by imposing the condition μ=0\mu = 0 right from the bat.

That's all there is to it.


I spoke too soon. where have I gone wrong?
Original post by Maths&physics
I spoke too soon. where have I gone wrong?

Your 2nd derivative is wrong.
(edited 5 years ago)
Original post by DFranklin
Your 2nd derivative is wrong.


I've just had my lunch and tried again but I'm still stuck. what am I doing wrong?

I know the consequent steps after getting the PI, but I can't derive this PI.
(edited 5 years ago)
Original post by Maths&physics
I've just had my lunch and tried again but I'm still stuck. what am I doing wrong?

I know the consequent steps after getting the PI, but I can't derive this PI.

You choose lambda and mu so the LHS equals the RHS.

Again, I have to ask how you can possibly claim to have covered the material and not know this? [Although I can guess; you're "going through the material" without doing any of the exercises/examples and it's basically going in one ear and out the other].
Original post by DFranklin
You choose lambda and mu so the LHS equals the RHS.

Again, I have to ask how you can possibly claim to have covered the material and not know this? [Although I can guess; you're "going through the material" without doing any of the exercises/examples and it's basically going in one ear and out the other].


I'll do a few examples and come back to this.
Original post by DFranklin
You choose lambda and mu so the LHS equals the RHS.

Again, I have to ask how you can possibly claim to have covered the material and not know this? [Although I can guess; you're "going through the material" without doing any of the exercises/examples and it's basically going in one ear and out the other].


im trying to derive the P.I equation myself (y = λx sin 5x), I'm not struggling to solve it.
Original post by Maths&physics
im trying to derive the P.I equation myself (y = λx sin 5x), I'm not struggling to solve it.

Since this isn't really "the PI equation", what you're saying doesn't make a lot of sense.

The reason I (and others) are getting so frustrated is that 95% of your problems are caused by jumping into exam questions without a proper understanding of the material. If you actually understood the material, you would hopefully understand the difference between what the question is asking, and what you are actually doing.
Original post by DFranklin
Since this isn't really "the PI equation", what you're saying doesn't make a lot of sense.

The reason I (and others) are getting so frustrated is that 95% of your problems are caused by jumping into exam questions without a proper understanding of the material. If you actually understood the material, you would hopefully understand the difference between what the question is asking, and what you are actually doing.


I understand that and I genuinely spent a couple of days going over this topic. the question gives us the PI but I was trying derive it myself and I can't based on the textbook and exam solutions.
Original post by Maths&physics
I understand that and I genuinely spent a couple of days going over this topic. the question gives us the PI but I was trying derive it myself and I can't based on the textbook and exam solutions.

Most people are going to need to spend a lot more than two days on this material. You are clearly not an exception.
Original post by DFranklin
Most people are going to need to spend a lot more than two days on this material. You are clearly not an exception.


Should I be able to find that particular integral? Well, I can do the actual question now.
Original post by Maths&physics
Should I be able to find that particular integral? Well, I can do the actual question now.


Yes. If they didn't give you a P.I. to put in, you should be able to recognise the appropriate form and determine the coefficients λ,μ\lambda,\mu. This process is just differentiating and subbing things in correctly... which shouldn't be an issue at this level.

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