Superposition Principle Undetermined Coefficients

Integrate

Integrate

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If I understand the superposition principle of undetermined coefficients then that means we can take each part on the right hand side and solve for them using the method of undetermined coefficients.


We then can add them back together and be given a particular solution for the entire differential equation.

But

sometimes we don't need to do that?

Screenshot 2024-04-26 074635.png
In this problem we are to find the general form of the particular equation of this non homogeneous linear differential equation.


However the solution doesnt find the general form for sint. only tcost and 10^t


Why?
 
Integrate said:
If I understand the superposition principle of undetermined coefficients then that means we can take each part on the right hand side and solve for them using the method of undetermined coefficients.


We then can add them back together and be given a particular solution for the entire differential equation.

But

sometimes we don't need to do that?

View attachment 37728
In this problem we are to find the general form of the particular equation of this non homogeneous linear differential equation.


However the solution doesnt find the general form for sint. only tcost and 10^t


Why?
Click to expand...
It is telling you that because the auxiliary equation has [imath]r = i[/imath] or [imath]r = -i[/imath], the guess for the particular solution will be [imath]y_p(t) = t(At + B)\cos t + t(Ct + D)\sin t + E10^t[/imath].

In other words, it is telling you that the complementary solution is:

[imath]y_c(t) = c_1\sin t + c_2 \cos t[/imath] and because of that the guess for the particular solution will be [imath]y_p(t) = t(At + B)\cos t + t(Ct + D)\sin t + E10^t[/imath]. And now you will have to find [imath]y''_p(t)[/imath] and plug [imath]y_p(t)[/imath] and [imath]y''_p(t)[/imath] in the original differential equation to find the coefficients [imath]A, B , C, D, \text{and} \ E[/imath], so that the general solution will be:

[imath]y(t) = y_c(t) + y_p(t)[/imath]
 
mario99 said:
It is telling you that because the auxiliary equation has [imath]r = i[/imath] or [imath]r = -i[/imath], the guess for the particular solution will be [imath]y_p(t) = t(At + B)\cos t + t(Ct + D)\sin t + E10^t[/imath].

In other words, it is telling you that the complementary solution is:

[imath]y_c(t) = c_1\sin t + c_2 \cos t[/imath] and because of that the guess for the particular solution will be [imath]y_p(t) = t(At + B)\cos t + t(Ct + D)\sin t + E10^t[/imath]. And now you will have to find [imath]y''_p(t)[/imath] and plug [imath]y_p(t)[/imath] and [imath]y''_p(t)[/imath] in the original differential equation to find the coefficients [imath]A, B , C, D, \text{and} \ E[/imath], so that the general solution will be:

[imath]y(t) = y_c(t) + y_p(t)[/imath]
Click to expand...
So yeah I understand that y = y_c + y_p

But I don't understand how the complementary equation gives us that guess?

I'm curious why in this case we don't solve for all terms on the right side and add them together.

Why does this solution only use tcost and 10^t and doesn't also solve for sint?
 
Integrate said:
So yeah I understand that y = y_c + y_p

But I don't understand how the complementary equation gives us that guess?

I'm curious why in this case we don't solve for all terms on the right side and add them together.

Why does this solution only use tcost and 10^t and doesn't also solve for sint?
Click to expand...
Let us walk step by step to understand why the particular solution depends on the complementary solution.

First look at the right side of the differential equation. We have:

[imath]\sin t[/imath]
[imath]t\cos t[/imath]
[imath]10^t[/imath]

Our initial guess for [imath]\sin t[/imath] is:
[imath]y_{p1} = A\cos t + B\sin t[/imath]
Is this solution also part of the complementary solution? Yes, then we modify our guess by multiplying the solution by [imath]t[/imath].
[imath]y_{p1} = t(A\cos t + B\sin t)[/imath]

Our initial guess for [imath]t\cos t[/imath] is:
[imath]y_{p2} = (Ct + D)\cos t + (Et + F)\sin t[/imath]
Is this solution also part of the complementary solution? Yes, then we modify our guess by multiplying the solution by [imath]t[/imath].
[imath]y_{p2} = t(Ct + D)\cos t + t(Et + F)\sin t[/imath]

Our initial guess for [imath]10^t[/imath] is:
[imath]y_{p3} = G10^t[/imath]
Is this solution also part of the complementary solution? No, then we leave it as it is.

Now the solutions of [imath]y_{p1}[/imath] and [imath]y_{p2}[/imath] are overlapping.

In other words:

[imath]t(A\cos t + B\sin t) = t(D\cos t + F\sin t)[/imath]

Therefore, we don't need [imath]y_{p1}[/imath] and the total particular solution is:

[imath]y_{p} = y_{p2} + y_{p3}[/imath]

Does this answer your doubts or you are still confused?
 
Integrate said:
So yeah I understand that y = y_c + y_p

But I don't understand how the complementary equation gives us that guess?

I'm curious why in this case we don't solve for all terms on the right side and add them together.

Why does this solution only use tcost and 10^t and doesn't also solve for sint?
Click to expand...
Because the guess of the particular solution of [imath]\sin t[/imath] is already included in the guess of the particular solution of [imath]t \cos t[/imath]. You don't see it at first, but if you look carefully and read post #4, you will see it.

If we have the guess:

[imath]t(D\cos t + F\sin t)[/imath]

we don't need the guess:

[imath]t(A\cos t + B\sin t)[/imath]

Why? Simply because they are the same guess!

You will say, but they have different constants? It doesn't matter because at the end you will combine these constants, and they will equal to another constant.

If you include [imath]t(A\cos t + B\sin t)[/imath] to the particular solution, it is not wrong, but you are increasing the messy calculations on yourself.

Look at it in this way:

The guess of [imath]10^t[/imath] is [imath]A10^t[/imath]. Is it necessary to add another guess [imath]B10^t[/imath]?

No because at the end you will combine them as:

[imath]A10^t + B10^t = (A + B)10^t = C10^t[/imath]

In simple words, if you have two guesses that have the same structure, get rid of one of them because its inclusion among the particular solutions is not necessary.
 
One final step to show you what happens when we add [imath]t(A\cos t + B\sin t)[/imath] among the other particular solutions.

[imath]y_{p1} = t(A\cos t + B\sin t)[/imath]

[imath]y_{p2} = t(Ct + D)\cos t + t(Et + F)\sin t[/imath]

[imath]y_{p3} = G10^t[/imath]

[imath]y_{p} = y_{p1} + y_{p2} + y_{p3} = t(A\cos t + B\sin t) + t(Ct + D)\cos t + t(Et + F)\sin t + G10^t[/imath]


[imath]= At\cos t + Bt\sin t + Ct^2\cos t + Dt\cos t + Et^2\sin t + Ft\sin t + G10^t[/imath]


[imath]= (A + D)t\cos t + (B + F)t\sin t + Ct^2\cos t + Et^2\sin t + G10^t[/imath]


[imath]= Ht\cos t + Kt\sin t + Ct^2\cos t + Et^2\sin t + G10^t[/imath]


[imath]= t(Ct + H)\cos t + t(Et + K)\sin t + G10^t[/imath]

which is the same as if we write directly:

[imath]y_{p} = t(Ct + D)\cos t + t(Et + F)\sin t + G10^t[/imath]

because we don't care if the constants are [imath]H,K[/imath] or [imath]D,F[/imath] or whatever.
 
mario99 said:
Let us walk step by step to understand why the particular solution depends on the complementary solution.

First look at the right side of the differential equation. We have:

[imath]\sin t[/imath]
[imath]t\cos t[/imath]
[imath]10^t[/imath]

Our initial guess for [imath]\sin t[/imath] is:
[imath]y_{p1} = A\cos t + B\sin t[/imath]
Is this solution also part of the complementary solution? Yes, then we modify our guess by multiplying the solution by [imath]t[/imath].
[imath]y_{p1} = t(A\cos t + B\sin t)[/imath]

Our initial guess for [imath]t\cos t[/imath] is:
[imath]y_{p2} = (Ct + D)\cos t + (Et + F)\sin t[/imath]
Is this solution also part of the complementary solution? Yes, then we modify our guess by multiplying the solution by [imath]t[/imath].
[imath]y_{p2} = t(Ct + D)\cos t + t(Et + F)\sin t[/imath]

Our initial guess for [imath]10^t[/imath] is:
[imath]y_{p3} = G10^t[/imath]
Is this solution also part of the complementary solution? No, then we leave it as it is.

Now the solutions of [imath]y_{p1}[/imath] and [imath]y_{p2}[/imath] are overlapping.

In other words:

[imath]t(A\cos t + B\sin t) = t(D\cos t + F\sin t)[/imath]

Therefore, we don't need [imath]y_{p1}[/imath] and the total particular solution is:

[imath]y_{p} = y_{p2} + y_{p3}[/imath]

Does this answer your doubts or you are still confused?
Click to expand...
This is fantastic.


Thank you so much.

No need to repeat because it's already in the previous guess.



Would you know why its called the superposition principle however?
 
Integrate said:
This is fantastic.


Thank you so much.

No need to repeat because it's already in the previous guess.



Would you know why its called the superposition principle however?
Click to expand...
Your question is ambiguous.

If you literally mean why did the superposition principle inherit the name "the superposition principle", then I don't know. If you mean what does the superposition principle mean in differential equations, I can answer this question.

A second order differential equation can have two or more solutions. The superposition principle is the combination of these solutions.
 
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