Diff. Eq. : Undetermined Coefficents

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SUMMARY

The discussion focuses on solving ordinary differential equations (ODEs) using the method of undetermined coefficients. The first equation, y'' + 4y = 3sin(2t), was incorrectly approached with guesses involving terms like A[tsin(2t)] and B[tcos(2t)], leading to incomplete solutions. The second equation, y'' - y' - 2y = (e^t + e^(-t))/2, was also mismanaged with guesses of the form Ae^t + Bte^(-t), resulting in incorrect coefficients. The correct complementary solutions for the first ODE are Asin(2t) + Bcos(2t) - (1/8)sin(2t) - (3/4)tcost(2t, while the second ODE's solution is Ae^(-t) + Be^(-2t) + (1/6)te^(2t) + (1/8)e^(-2t).

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NINHARDCOREFAN
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I have 2 problems that I'm not getting the right answer to:

y"+4y=3sin(2t)

I chose my guess to be: y=A[tsin(2t)]+B[tcos2t]
with this guess I'm getting only part of the answer right

I also tred this guess: y= A[t^2(sin(2t)]+ B[t^2(cos(2t)] + C[tsin(2t)]+D[tcos2t]

I got the same answer as above + some other weird answers


y"-y'-2y=(e^t+e^(-t))/2

I made this guess: Ae^t+Bte^(-t)

I got my answer as -2A=1/2 and -3B=1/2

However these were also wrong. Anything wrong with the guesses?
 
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What were the fundamental (complimentary) set of solutions for each ODE?
 
Answers...

The first one: Asin(2t)+Bcos(2t)-(1/8)sin(2t)-(3/4)tcost(2t)
The second one : Ae^-t+Be^-2t+(1/6)te^(2t)+(1/8)e^(-2t)
 

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