Method of undetermined coefficients

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Discussion Overview

The discussion revolves around solving the differential equation y'' + 9y = sin(3t) using the method of undetermined coefficients. Participants explore the challenges in finding the particular solution, particularly when the assumed form leads to a trivial solution.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant identifies the need to find a particular solution for the equation and notes that their initial guess leads to a trivial result.
  • Another participant points out that the assumed form of the particular solution must include functions of t (A(t) and B(t)) rather than constants, as the left side reduces to zero.
  • A suggestion is made that A(t) and B(t) can be simplified to A*t and B*t, where A and B are constants, although this assumes knowledge of the form of the function.
  • There is a discussion about the complexity of the problem if the form of A(t) and B(t) is not known, indicating that more work would be required in that case.
  • A participant expresses gratitude for the responses and confirms that using A*T instead of just "A" resolved their issue.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach initially, as there are differing views on how to handle the coefficients in the particular solution. However, there is agreement that the original assumption was incorrect and that a modification is necessary.

Contextual Notes

Participants discuss the implications of assuming constant coefficients versus functions of t, highlighting the need for careful consideration of the form of the particular solution in the context of the differential equation.

Who May Find This Useful

This discussion may be useful for students and practitioners dealing with differential equations, particularly those learning about the method of undetermined coefficients and the nuances involved in finding particular solutions.

phantomcow2
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y''+9y=sin(3t)

I need to solve the above using the method of undetermined coefficients.
I have already found the solution to y''+9y=0 is c1(Cos3t)+c2(sin3t).

The problem is finding the particular solution. From class I am aware that the general form of the solution is ASin(3t)+BCos(3t).

Taking the second derivative, I get -9ASin(3t) - 9BCos(3t). And substituting it in I get

-9ASin(3t) - 9BCos(3t) + 9(ASin(3t)+BCos(3t))=Sin(3t)

But entire left side reduces to zero! How, then, can I solve for the correct A and B? I understand I must have made a mistake somewhere, but I'm having difficulty spotting it. Could someone shed some light?
 
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Taking the second derivative, I get -9ASin(3t) - 9BCos(3t). And substituting it in I get
-9ASin(3t) - 9BCos(3t) + 9(ASin(3t)+BCos(3t))=Sin(3t)
But entire left side reduces to zero! How, then, can I solve for the correct A and B? I understand I must have made a mistake somewhere, but I'm having difficulty spotting it.
No wonder "entire left side reduces to zero" since y=-9ASin(3t)-9BCos(3t) is solution of y''+9y=0 . It isn't solution of y''+9y=Sin(3t)
To be solution of y''+9y=Sin(3t), the unknown A and B should be not constant, but function of t :
y=-9A(t)Sin(3t)-9B(t)Cos(3t)
In the dérivatives of y, don't forget the derivatives of A(t) and B(t).
 
Slightly simpler: A(t) and B(t) can be written as A times t and B times t where A and B are now constants.
 
Slightly simpler: A(t) and B(t) can be written as A times t and B times t where A and B are now constants.
Not "slightly" simpler, but "very much" simpler ! Of course, but this supposes that the form of the function is already known y=A*t*sin(t)+B*t*cos(t)
If one doesn't know it, he will have to let A(t) and B(t). And unfortunately, he will have to work a lot more...
 
Awesome. Thanks for the helpful responses -- I solved the differential equation and plugging in A*T instead of just "A" worked. I recall my professor mentioning that once during class.

I realize why my original guess wasn't a solution, too.
 

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