- #1
phantomcow2
- 52
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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?
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?