How to determine the particular solution to the DE y'' +8y' +20y = sinx?

- anonymous

How to determine the particular solution to the DE y'' +8y' +20y = sinx?

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- schrodinger

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- anonymous

\[y_c = e^{-4x} (Acos(2x) + Bsin(2x))\]

- TuringTest

wqith either variation of parameters or undetermined coefficients. here I'd recommend undetermined coefficients.
http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

- anonymous

Oh.. undetermined coefficient.. not variation of parameters. :\

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- anonymous

I don't understand how to guess the particular solution. Usually, when we see sinx / cosx on the right, we would guess Csinx + Dcosx
But when sinx/cosx appears on the left, then we should guess yp = Cxsinx +Dxcosx
Probably I'm wrong here...

- TuringTest

since you have sin(2x) in the left, and sinx on the right, you are okay in terms of linear independence and so you can make the standard guess Asinx+Bcosx
if you had sin/cos(2x) on the right then you'd need to multiply by x in your guess

- anonymous

But when we expand sin2x, we get 2sinxcosx..

- anonymous

functions f1 f2 are linearly independent if there exist constants c1, c2, not both zero such that c1 f1(x) + c2 f2(x) = 0 for all x in your domain.
You can expand that definition to check for more than two functions...

- anonymous

So, your particular solution should look something like a * sin x + b * cos x.
(I haven't done this in years. hope this is right...)
So, for them to be dependent (what you hope is not true), then you need to have c1, c2, c3, not all zero such that:
c1 * yc(x) + c2 a* sin x + c3 b * cos x = 0 for every x.

- anonymous

I think that should be impossible.
(I've forgotten exactly why the independence thing applies though... so I might have made some interpretational error above. But the definition of independence is right...)

- anonymous

The particular solution is indeed asinx + bcosx, but I don't quite understand why x isn't there. Probably the most fundamental thing I don't understand is how linearly independence is related to solving the DE using undetermined coefficients. My teacher didn't tell us much about linearly independence except when the Wronskian=0, it's linearly dependent...

- anonymous

undetermined coefficients is basically just a method where you guess what the particular solution looks like, and then use that guess to determine what the coefficients in your guess look like.
I think there's a solution that says if the non-homogeneous part (the right hand side part... whatever it's called) is a polynomial, a sin or cos function, and... maybe also exponential... then the undetermined method will find a particular solution.
There's no x in your guess because you're trying to get to a a function that is sin x. Basically you guess something of the right form that, after taking all derivatives, will take you to the right form of the right hand side.

- anonymous

That book I mentioned before explains all of this stuff about as clearly as is possible. (it's a classic) Your university library might have it. It'll explain what all the linear independence is for, if no one else here can... http://www.amazon.com/Ordinary-Differential-Equations-Dover-Mathematics/dp/0486649407/ref=sr_1_1?ie=UTF8&qid=1354197995&sr=8-1&keywords=ordinary+differential+equations

- anonymous

|dw:1354198038401:dw|

- anonymous

Oh yeah, so you can use it on e^x.

- anonymous

Okay I'm looking through my book, and it's reminding me that an ordinary linear differential equation of order n has n linearly independent solutions.

- anonymous

Yes!

- anonymous

So the reason that you want to make sure that your particular solutions etc are linearly independent is because you want to form a combination of all the possible solutions. And then there's another theorem somewhere that says that combination is then the general solution. i.e., it encapsulates all the solutions...

- anonymous

Am I answering your question, or just rambling about stuff you already know?

- anonymous

Hmm... My question is basically why we don't need the x for yp.. :\

- anonymous

Well x is implicitly in a cos x + b sin x. Not sure if that's what you mean.

- anonymous

But also, you're just guessing what a particular solution looks like. If you thought that yp = a cos x + b sin x + c * x, then you'd then go take the derivatives, then equate coefficients, and you'd expect that c to be not zero, if you think it needs an x in there. But it just doesn't need to be there... because it's all about the trigonometric functions.

- anonymous

Oh are you saying why didn't you need an x because you think the functions are linearly dependent? Since if they're linearly independent, usually you might just throw an x in front to fix it?

- anonymous

I don't really understand what you mean.

- anonymous

Would you mind explaining why it is linearly independent...?

- anonymous

For all those functions, for them to be linearly dependent, you need to find constants to place before them so that the sum of them all will be zero, for every x. So at every x, you need to be able to combine them in the same way so that their sum is zero. If you look at their graph, against each other, you can sort of see that their values, relative to each other, don't really line up... so the only option is that all the constants are zero, which means they're linearly independent.
|dw:1354199000025:dw|

- anonymous

I'm a bit hazy on the details of the differential equation thing. Are you trying to show that, for starters, just sin x and sin (2x) are linearly independent? Or you also want to show (to convince yourself) that sin x and cos x are linearly independent?

- anonymous

Sorry, I'm probably not much help. Maybe someone else can better explain... I haven't done this in years.

- anonymous

It's okay, at least now I know it has something to do with linear independence, which I didn't know in the past!
But why are you showing that sin2x and sinx are not linearly independent?

- anonymous

That was just me doing a proof by contradiction sort of thinking. Just posing "what if they were dependent, then..." and then you can prove it's impossible, which means they're independent... That was how I was thinking, and so phrased it like that.

- anonymous

But actually it occurs to me that you don't care if the particular solution is independent from the general solution.
It's second order, so there will only be two linearly independent solutions. And that's why, in your final form, you'll have two undetermined constants, in front of your sin (2x), cos (2x) things.
For the particular solution, that's really separate from the general solution. It's just some fixed value that's added on, in order to make sure that the final, general form, satisfies the right hand side.

- anonymous

You should only have two variable coefficients in your final general solution. Because it's a second order linear differential equation. And you only care about linear independence when you have these variables, because you want to be sure that they are all distinct, that you can't find some coefficients that make it so you had less of those functions associated to the coefficients than you thought...

- anonymous

Hmm... I guess I have to try more question to see if I get your point.. Thanks for your help!!

- anonymous

*questions

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