Inhomogeneous diff EQ, undetermined coefficients

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SUMMARY

The discussion focuses on solving the inhomogeneous differential equation y" + 3y' = 72sin(3t) + 36cos(3t) with initial conditions y(0) = 6 and y'(0) = 9. The user initially finds the homogeneous solution, yielding y = 9 - 3e^(-3t) - 2sin(3t) - 6cos(3t). However, the solution does not satisfy the initial conditions, prompting clarification on when to apply these conditions in the solution process. The correct approach is to determine the constants after finding the particular solution.

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offbeatjumi
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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 I am 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
 
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Hi offbeatjumi! :smile:
offbeatjumi said:
… where y(0) = 6 and y'(0) = 9

using the initial conditions (maybe I am not supposed to use them here?)
I find a = 9, b = -3

Therefore I get the solution:

y = 9 - 3e^(-3t) - 2sin3t - 6cos3t

But y(0) ≠ 6, is it?

So you did need to wait until the end before finding the constants. :wink:
 
thank you! i wasn't sure about where to apply initial conditions
 

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