Undetermined Coefficients - Duplication Question

[VitaX]

I'm having some problems recognizing duplication when using the undetermined coefficients method to solve homogeneous type differential equations.

http://college.cengage.com/mathematics/larson/calculus_early/3e/shared/chapter15/clc7eap1504.pdf

Example 2 on Page 1119, could someone explain to me why there is duplication in that problem because I am just not seeing it. And when there is duplication, do you multiply each term in the particular solution by the variable, t, or just a select few? This is my main concern with undetermined coefficients.

To check for duplication, do you look $g(t)$ and $y_c(t)$ and see if there are any alike terms then move on to your initial guess and multiply through by t if there are any like terms? I'm just having some trouble with this concept. If someone could explain it to me in some detail that would be great, as the sites I've found just don't explain it well enough, though the site linked above is the best one I've come across so far. My textbook doesn't even mention duplication sadly.

 

[S_Happens]

Edit- Actually, homogenous is when the right hand side is zero and nonhomogenous is nonzero. You're trying to solve nonhomogenous, which is why you have to worry about duplication.

The duplication is because e^0x (zero is a solution, and e^0x = 1) is a solution of the homogenous equation.

The first term on the right hand side is x*e^0x. This means your INITIAL undetermined coefficient form for that term should be (Ax + B)*e^0x (your book uses switches A and B, but that doesn't matter). But when you distribute, you can see that you get Ax*e^0x + B*e^0x. B*e^0x is a solution to the homogenous equation. This means you multiply ONLY the polynomial (Ax +B) by x, to get (Ax² + Bx)*e^0x.

You leave your other undetermined coefficient term alone (in this case).

 

[VitaX]

Edit- Actually, homogenous is when the right hand side is zero and nonhomogenous is nonzero. You're trying to solve nonhomogenous, which is why you have to worry about duplication.

The duplication is because e^0x (zero is a solution, and e^0x = 1) is a solution of the homogenous equation.

The first term on the right hand side is x*e^0x. This means your INITIAL undetermined coefficient form for that term should be (Ax + B)*e^0x (your book uses switches A and B, but that doesn't matter). But when you distribute, you can see that you get Ax*e^0x + B*e^0x. B*e^0x is a solution to the homogenous equation. This means you multiply ONLY the polynomial (Ax +B) by x, to get (Ax² + Bx)*e^0x.

You leave your other undetermined coefficient term alone (in this case).

The way I look at it is like so: $g(t) = x + 2e^x$ and $y_c(x) = C_1 + C_2 e^x$

So that would mean duplication of the term $e^x$

So to compensate for that, multiply your particular solution by the variable x. But my question is this: Do I multiply every term in the particular solution by x, or just the term containing the duplication? Or is it just safe to say that when duplication occurs, multiply every term by the variable even if its overkill. Just to be safe.

Sorry if it seemed like I was repeating what you just said, I was a little confused by your terminology.

 

[lurflurf]

It is a shame so many books talk about guessing and trial and error, it is easy to know exactly what is needed.
If g has a term s_n(t)e^{r t}cos(a t+b) say
where s_n(t) is a degree n polynomial
and s_m(t)e^{r t}cos(a t+b) is a solution of the homogeneous equation
but t*s_m(t)e^{r t}cos(a t+b) is not a solution of the homogeneous equation
the solution of the nonhomogeneous equation should have
s_m+n+1(t)e^{r t}cos(a t+b)
the solution splits up into
s_m(t)e^{r t}cos(a t+b)+t^m+1s_n(t)e^{r t}cos(a t+b)
where s_m(t) is arbitrary and
s_n(t) is particular

 

[VitaX]

It is a shame so many books talk about guessing and trial and error, it is easy to know exactly what is needed.
If g has a term s_n(t)e^{r t}cos(a t+b) say
where s_n(t) is a degree n polynomial
and s_m(t)e^{r t}cos(a t+b) is a solution of the homogeneous equation
but t*s_m(t)e^{r t}cos(a t+b) is not a solution of the homogeneous equation
the solution of the nonhomogeneous equation should have
s_m+n+1(t)e^{r t}cos(a t+b)
the solution splits up into
s_m(t)e^{r t}cos(a t+b)+t^m+1s_n(t)e^{r t}cos(a t+b)
where s_m(t) is arbitrary and
s_n(t) is particular

I'm having trouble following what you just said there but for a problem like this:

$y'''(t) - 3y''(t) + 2y'(t) = t + e^t$

$y_c(t) = C_1 + C_2 e^{2t} + C_3 e^t$

Initial guess would be: $y_p(t) = At + B + Ce^t$

But since $e^t$ is present in the differential equation and the complimentary solution, should I:

1) Multiply each term by $t$ : $y_p(t) = At^2 + Bt + Cte^t$

2) Multiply polynomial by $t$ : $y_p(t) = At^2 + Bt + Ce^t$

3) Multiply $e^t$ by $t$ : $y_p(t) = At + B +Cte^t$

Which would you recommend to be on the safe side?

 

[lurflurf]

terms in homogeneous problem
{1,e^t,e^2t}
terms generated by the forcing function
{1,t,e^t}
terms duplicated
{1,e^t}
additional terms needed because of duplication
{t²,t e^t}

 

[S_Happens]

The way I look at it is like so: $g(t) = x + 2e^x$ and $y_c(x) = C_1 + C_2 e^x$

So that would mean duplication of the term $e^x$

So to compensate for that, multiply your particular solution by the variable x. But my question is this: Do I multiply every term in the particular solution by x, or just the term containing the duplication? Or is it just safe to say that when duplication occurs, multiply every term by the variable even if its overkill. Just to be safe.

Sorry if it seemed like I was repeating what you just said, I was a little confused by your terminology.

Actually this second example is a bad one as well, because both terms will include duplications. The general form for the first term will be (Ax +B), where B is a solution to the homogenous equation. That means you end up multiplying everything by x, which defeats the purpose of your question.

The answer to the question that you are trying to ask is that NO, you do not multiply the entire right side by x. You only multiply duplicate solutions by x (or x²).

Ignoring lurflurf's post that alludes to a simple pattern (hidden somewhere in that post that is extremely difficult to look at), here is the way I do it.

Write down the general form. Find solutions (or use given solutions) to the homogenous equation. Multiply any (and ONLY) duplicated terms by x until they are no longer duplicate solutions to the homogenous equation. This means you could multiply part by x, part by x², and/or part by nothing at all.


In your second example number 1 is the correct choice, but AGAIN it's because you chose terms that are ALL duplicates. B is a duplicate and so is e^t. Why not put something in that is certainly not a solution like cos(3x), then you could see that it does not get multiplied by x (or t, whatever you want to use).

 

[VitaX]

Actually this second example is a bad one as well, because both terms will include duplications. The general form for the first term will be (Ax +B), where B is a solution to the homogenous equation. That means you end up multiplying everything by x, which defeats the purpose of your question.

The answer to the question that you are trying to ask is that NO, you do not multiply the entire right side by x. You only multiply duplicate solutions by x (or x²).

Ignoring lurflurf's post that alludes to a simple pattern (hidden somewhere in that post that is extremely difficult to look at), here is the way I do it.

Write down the general form. Find solutions (or use given solutions) to the homogenous equation. Multiply any (and ONLY) duplicated terms by x until they are no longer duplicate solutions to the homogenous equation. This means you could multiply part by x, part by x², and/or part by nothing at all.


In your second example number 1 is the correct choice, but AGAIN it's because you chose terms that are ALL duplicates. B is a duplicate and so is e^t. Why not put something in that is certainly not a solution like cos(3x), then you could see that it does not get multiplied by x (or t, whatever you want to use).

Ok that makes it a little clearer now. I used the above O.D.E. because it was on my homework and I was wondering which Particular Solution to go with. I can obviously see the duplication of $e^t$. But, what I'm having trouble understanding is what is duplicated to cause the polynomial $At + B$ to be multiplied by $t$? I mean when you look for duplication, you are checking $y_c(t)$ and $g(t)$ for any like terms right? So what is $t$ from $g(t)$ duplicated with in $y_c(t)$? This is probably the main thing I'm not understanding, if you can explain this that would be very helpful. Or perhaps I'm just looking at the wrong thing, I'm not sure.

 

[lurflurf]

The 1 is duplicated. A*t+B is a degree 1 polynomial and C₁ is a degree 0 polynomial so the general solution will have a degree 1+0+1=2 degree polynomial.

 

[S_Happens]

The 1 is duplicated. A*t+B is a degree 1 polynomial and C₁ is a degree 0 polynomial so the general solution will have a degree 1+0+1=2 degree polynomial.

Interesting. I've never come across anything that was difficult enough that I couldn't simply see what was duplicated and needed to be multiplied by x.

To the OP: The above answer is quite clear, but I had already said it a few times now as well, so I'll try again.

The way you're typing it out, it should be pretty obvious that B is a duplicate of C₁. Maybe you're looking at the polynomial like it's a single entity? If you're going to do that then you should look at it as

$$(Ax+B)e^{0x}$$

Then distribute the e^0x and look at the terms separately. You should now see that B is a duplicate of C₁. With a polynomial, it's always going to be the last term or two that is a duplicate. If any of the terms in the polynomial produce duplicates, then multiply the ENTIRE polynomial by x (and repeat if necessary).

 

[VitaX]

Interesting. I've never come across anything that was difficult enough that I couldn't simply see what was duplicated and needed to be multiplied by x.

To the OP: The above answer is quite clear, but I had already said it a few times now as well, so I'll try again.

The way you're typing it out, it should be pretty obvious that B is a duplicate of C₁. Maybe you're looking at the polynomial like it's a single entity? If you're going to do that then you should look at it as

$$(Ax+B)e^{0x}$$

Then distribute the e^0x and look at the terms separately. You should now see that B is a duplicate of C₁. With a polynomial, it's always going to be the last term or two that is a duplicate. If any of the terms in the polynomial produce duplicates, then multiply the ENTIRE polynomial by x (and repeat if necessary).

Yeah I understand how its duplicated from what you said but I had thought that dupilcates only arise from looking at the terms that are present in $y_c(t)$ and $g(t)$ so that's why I'm confused because you are telling me to look at a term in the particular solution, $y_p(t)$, and see if it has duplicates with $y_c(t)$ now. That's why I was a bit confused. But, when in doubt, I can always multiply through by t, that shouldn't change my final answer for the particular solution right? Just gives me more algebra to work with in the end.

Edit: Think I know what you mean now. My line of thinking was correct, but I overlooked something. If a term in the particular solution is a solution to the homogeneous equation, multiply that term by a variable to get rid of duplication. Think that's what you were saying.

 

[S_Happens]

OK. I see where your confusion was. I missed where you explicitly asked that in post 8. You start with g(t), but that's not where you look for the duplication.

Your edit is correct, the particular solution is where you are looking for duplication.