# Method of undetermined coefficients

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?

Related Differential Equations News on Phys.org
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).

HallsofIvy
Homework Helper
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.