Preface
This is a tutorial made solely for the purpose of education and it was designed for students taking Applied Math 0340. It is primarily for students who have some experience using Mathematica. If you have never used Mathematica before and would like to learn more of the basics for this computer algebra system, it is strongly recommended looking at the APMA 0330 tutorial. As a friendly reminder, don't forget to clear variables in use and/or the kernel.
Finally, the commands in this tutorial are all written in bold black font, while Mathematica output is in regular fonts. This means that you can copy and paste all commands into Mathematica, change the parameters and run them. You, as the user, are free to use the scripts to your needs for learning how to use the Mathematica program, and have the right to distribute this tutorial and refer to this tutorial as long as this tutorial is accredited appropriately. The tutorial accompanies the textbook Applied Differential Equations. The Primary Course by Vladimir Dobrushkin, CRC Press, 2015; http://www.crcpress.com/product/isbn/9781439851043
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Introduction to Linear Algebra
Glossary
Variation of Parameters
Suppose that we know a fundamental matrix \( {\bf X} (t) \) of the vector system of homogeneous linear differential equations:
Example:
Variation of Parameters for constant matrices
Consider a non-homogeneous vector differential equstion with constant coefficients:
Example:
1/25 E^(-2 t) (-7 + 4 E^(5 t/2) + 3 E^(5 t)),
2/25 E^(-2 t) (-1 + E^(5 t/2))^2}, {-(6/25)
E^(-2 t) (-1 + E^(5 t/2))^2,
1/25 E^(-2 t) (14 + 2 E^(5 t/2) + 9 E^(5 t)),
2/25 E^(-2 t) (-2 - E^(5 t/2) + 3 E^(5 t))}, {-(6/25)
E^(-2 t) (-2 - E^(5 t/2) + 3 E^(5 t)),
1/25 E^(-2 t) (-28 + E^(5 t/2) + 27 E^(5 t)),
1/25 E^(-2 t) (8 - E^(5 t/2) + 18 E^(5 t))}}
Therefore, the matrix A has three distinct real eigenvalues: 3, -2, and 1/2
We seek a particular solution in the form:
yp = K0 + K1*t +K2*t^2 + K3*E^t ,
where vector columns K0, K1, K2, and K4 should satisfy the following (algebraic) equations
K1 = A K0 + {0,10,0}
2*K2 = A K1
0 = A K2 + {1,0,0}
K3 = A K3 + {1,0,1}
Let us denote Kj = {k1j,k2j,k3j}, then
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