Solve 3rd order ode using variation of parameters

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The discussion centers on solving the third-order ordinary differential equation y''' - 2y'' - y' + 2y = exp(4t) using the method of variation of parameters. The homogeneous solutions identified are 1, -1, and 2, leading to a general solution form that includes a function g(t). The user expresses frustration with their professor's explanation of variation of parameters for higher-order equations and seeks guidance on calculating the determinant and particular solution. There is also a note about a potential sign error in the calculation of the Wronskian. The conversation highlights the complexity of using variation of parameters for such equations, questioning its practicality for finding particular solutions.
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Homework Statement



Solve using variation of parameters
y''' - 2y'' - y' + 2y = exp(4t)


Homework Equations


Solve using variation of parameters


The Attempt at a Solution



I got the homogenous solutions to be 1, -1, and 2.

So, y = Aexp(t) + Bexp(-t) + Cexp(2t) + g(t)

I got W{exp(t), exp(-t), exp(2t)} = 6exp(2t)


My professor did quite an unsatisfactory job explaining variation of parameter for high order equations (spent an hour and 30 minutes trying to do 1 problem, and still didn't finish it correctly). So I just need to be pointed in the right direction.

Thanks
 
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Actually, I got -6 exp(2t), but maybe I made a sign error in my calculation.

So the general idea is now to replace the ith column in the matrix by (0, 0, exp(4t)) and calculate the determinant |Wi| of that, for i = 1, 2, 3.
Then if you calculate
c_i(t) = \int \frac{|W_i|}{|W|} \, dt
the general solution is
y(t) = c_1(t) e^{t} + c_2(t) e^{-t} + c_3(t) e^{-4t}
 
abstracted6 said:

Homework Statement



Solve using variation of parameters
y''' - 2y'' - y' + 2y = exp(4t)


Homework Equations


Solve using variation of parameters

As a learning exercise, fine. But I assume you know nobody in their right mind would use variation of parameters to find a particular solution, eh?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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