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The solutions of the system of differential equations are [tex]x(t) = c_{1}\cdot e^{t}+c_{2}\cdot t\cdot e^{t}[/tex] and [tex]y(t) = c_{3}\cdot e^{t} + c_{4}\cdot t\cdot e^{t}[/tex].
In this question we must apply systematic elimination to derive the homogeneous solution of the given system of linear differential equations. The basis of this method is base on the consideration that a derivative of a given variable is directly proportional to the same variable.
Now, we can rewrite the system as follows:
[tex]\dot x = 2\cdot x - y[/tex] (1)
[tex]\dot y = x[/tex] (2)
Now,
[tex]D\cdot x = 2\cdot x - y[/tex] (1b)
[tex]D\cdot y = x[/tex] (2b)
And the resulting system is:
[tex](D-2)\cdot x + y = 0[/tex] (1c)
[tex]-x+D\cdot y = 0[/tex] (2c)
We eliminate y by multiplying (1c) by D and add this result to (2c) and we have the solution of x(t):
[tex]x(t) = c_{1}\cdot e^{t}+c_{2}\cdot t\cdot e^{t}[/tex] (3)
And we eliminate x by multiplying (2c) by (D - 2) and we result to (1c) and we have the solution of y(t):
[tex]y(t) = c_{3}\cdot e^{t} + c_{4}\cdot t\cdot e^{t}[/tex] (4)
The solutions of the system of differential equations are [tex]x(t) = c_{1}\cdot e^{t}+c_{2}\cdot t\cdot e^{t}[/tex] and [tex]y(t) = c_{3}\cdot e^{t} + c_{4}\cdot t\cdot e^{t}[/tex]. [tex]\blacksquare[/tex]
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