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# Find the general solution?

Find the general solution of x'=(2, 1, -5, -2)x+(-cos(t), sin(t)).

(this is 2x2 matrix with 2 and 1 on the left, -5 and -2 on the right. And -cos(t) on top, sin(t) on bottom. I've found the answer for the 2x2 matrix but have trouble finding it for the other part. I want to solve it using undetermined coefficients but I don't know what should the form be.

You found the homogeneous solution to be a combination of terms involving sin(t) and cos(t). Since your inhomogeneous term (-cos(t), sin(t)) is of the same functional form, the inhomogeneous solution can contain terms of the form

x= (A1, A2) cos(t) + (B1, B2) sin(t) + (C1, C2) t cos(t) + (D1, D2) t sin(t)

To determine the 8 unknown coefficients A1 through D2 you would substitute this x into your original equation and get 8 linear equations.

An easier method in this case is given by variation of parameters.

How to do the variation of parameter?

There is a shortcut for this method and it goes like this: Find a fundamental matrix X for the homogeneous system and find its inverse X-1. Then if b is the inhomogeneous term, the inhomogeneous part of the solution is given by

x = X ∫ X-1b dt.
What original equation do I plug this x into it to find 8 different values? And how do you get the 8 linear equations?
You plug x into x'=(2, 1, -5, -2)x+(-cos(t), sin(t)). On the left side, you need to take the derivative. On the right side, you need to do matrix multiplication. It will look messy.
Since this is a vector equation, you really have 2 equations, one for the first component and one for the second component. Write them out. From each of these two equations, you extract 4 equations by comparing the terms with a cos(t), with a sin(t), with a t cos(t), and with a t sin(t) on the left and right sides. You get a total of 8 equations.