Solving Exam Questions: y^(5) + 3y^(4) - 5y''' - 15'' + 4y' + 12y = 0 & e^-A

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The discussion focuses on solving a fifth-order linear differential equation, specifically y^(5) + 3y^(4) - 5y''' - 15'' + 4y' + 12y = 0, and finding the fundamental matrix for the matrix initial value problem E(t)' = AE(t), E(0) = I, where A is a square matrix. The solutions to the differential equation can be derived using the characteristic equation, which has integer roots. Additionally, the matrix exponential e^-A can be computed using eigenvectors and eigenvalues, with the formula e^(-A) = S (e^D) S^(-1), where D is a diagonal matrix.

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I'm going over old exam questions for the final. I'm not sure what the departament will put on the exams so I'm trying to go over as much as possible, but I having problems figuring certain problems out:

1)
y^(5) + 3y^(4) - 5y''' - 15'' + 4y' + 12y = 0

How do you find the five solutions to the equation and then put it into the 5 dimensional system of first order equations.

2)
Let A be a square matrix, and let Y be a fundamental matrix for the homogeneous linear system x' = Ax.

a) Verify by substituion that e^tA and (YY_0)^-1 are both solutions to the matrix IVP E(t)' = AE(t), E(0) = I, where Y_0 = Y(0).

All I can intepret from this is the e^tA = (YY_0)^-1 so if I can find Y I can slove this...but nothing else...

b) Find e^-A

A = [[1 2],[2,1]]

Any help would be apprecaited, thank you
 
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For (1), the standard trick of assuming y=A*ekt gives you a solution basis fast, as all the roots of the characteristic equation are integers.
 
For 2 b), use eigenvectors and eigenvalues to write -A = S D S^(-1), where D is a diagonal matrix. Note that e^(-A) = e^(S D S^(-1)) = S (e^D) S^(-1).

Regards,
George
 

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