Matrix methods for systems of differential equations
Uploader Comments (DrChrisTisdell)
All Comments (14)
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@DrChrisTisdell Well the problems we were given go up to 4x4 =P 1st I find the solution to the homogeneous system, then I go ahead to the whole transformation business to find particular solutions. The question is: what to do when one of the transformed solutions is general, because one of the transformed equations is homogeneous. When you use those particular solutions to transform the system back, the one to the homogeneous equation is the general solution with an arbitrarily chosen K?
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I'm so glad you put these on youtube!
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Many thanks!
Keep up the great work!
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Thank you Dr Tisdell.
I know the following case is outside the scope of the presentation, but it can often be found in spring-mass-damper model. For example, I tried to diagonalize some real matrix with MATLAB and it gave me complex eigenvalues and eigenvectors. After I applied the multiplication PDP^(-1), I got the original (real) matrix back. What matrix can have such "complex" diagonalization and what are the properties of its eigenvalues and eigenmatrix?
Thank you very much! One question: if you have one of the y-equations as not being inhomogeneous, is the particular solution 0, or is it the homogeneous solution with 1 as a constant? Or can you set the constant as 0 and say 0 is a particular solution?
MarlosZappa 2 months ago
@MarlosZappa Sorry, I'm finding it hard to fully understand your question. Are you meaning the case when all the h_i functions are zero at 04:45? If so then then you have the system y' = Dy which will give rise to exponential solutions.
DrChrisTisdell 1 month ago
@DrChrisTisdell Yeah, that's it, but I meant actually when you're solving the transformed system P^(-1)*A*Py + P^(-1)g. You might have one of the lines of g as a 0. Since you're looking for particular solutions, and you get exponential solutions with general constants in that case, how do you get a valid particular solution useful for this method? Can you use 0, or do you have to use the exponential solution with, say, 1 as a constant, as a particular solution?
MarlosZappa 1 month ago
@MarlosZappa OK, so let's say we have a 2x2 system and h_1 is zero (but h_2 isn't). Our system is then y_1' = \lambda_1 y_1 and y_2' = \lambda_2 y_2 + h_2(t). These can be solved individually, there is no need to worry about particular solutions.
DrChrisTisdell 1 month ago
which playlist is this video on in your channel? or which playlist include videos of the course in which this video was teaching to?
lecolis 3 months ago
@lecolis Engineering Mathematics 2E playlist.
DrChrisTisdell 3 months ago