don't use the initial conditions here, use it at the endusing the initial conditions (maybe Im not supposed to use them here?)
I find a = 9, b = -3
so
so and
and
so and
then finally you end up with
Find the solution of:
y" + 3y' = 72sin(3t) + 36cos(3t)
where y(0) = 6 and y'(0) = 9
I first found the solution to the homogeneous eq:
the roots (R^2 + 3R = 0) are R = 0, -3
which gives the family of solutions:
y = a(1) + be^(-3t)
and y' = -3be^(-3t)
using the initial conditions (maybe Im not supposed to use them here?)
I find a = 9, b = -3
For the inhomogeneous eq:
I try (guess)
y = Asin(3t) + Bsin(3t)
y' = 3Acos(3t) - 3Bsin(3t)
y" = -9Asin(3t) - 9Bcos(3t)
substitute those values into the original equation (left hand side) I find
sin3t(-9A-9B) + cos3t(-9B-9A) = 72sin3t + 36cos3t
therefore
-9A - 9B = 72
-9B - 9A = 36
giving B = -6, A = -2
Therefore I get the solution:
y = 9 - 3e^(-3t) - 2sin3t - 6cos3t
What did I do wrong (this answer is incorrect)
Thanks