Consider the following linear system of differential equations: { <mtable columnalign="

William Santiago 2022-05-16 Answered
Consider the following linear system of differential equations:
{ x ˙ = 4 y y ˙ = x
where x ( t ) and y ( t ) are unknown real functions.
One can simply verify that the general solution is
( x ( t ) y ( t ) ) = c 1 ( cos ( 2 t ) 1 2 sin ( 2 t ) ) + c 2 ( 2 sin ( 2 t ) cos ( 2 t ) )
where c 1 and c 2 are real parameters.
Question: Which is the exponential form of this expression? It should by something like
( x ( t ) y ( t ) ) = k 1 e i 2 t ( P 11 P 21 ) + k 2 e i 2 t ( P 12 P 22 )
where k 1 and k 2 should (?) be complex parameters.
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Answers (2)

heilaritikermx
Answered 2022-05-17 Author has 20 answers
I think the key here is to rewrite the linear system as the following matrix equation
( 0 4 1 0 ) ( x y ) = ( x ˙ y ˙ ) A x = x ˙ .
The characteristic equation is λ 2 + 4 = 0 , so you're correct in using cos ( 2 t ) and sin ( 2 t ) . Now, to get at a complex exponential solution, you need to find eigenvectors of A with eigenvalues 2 i , 2 i . These are
( 2 i 1 ) and ( 2 i 1 )
respectively.
Consequently, you can write
( x ( t ) y ( t ) ) = c 1 ( 2 i 1 ) e 2 i t + c 2 ( 2 i 1 ) e 2 i t .
It is worth noticing that the eigenvectors, are attached to their associated eigenfunctions, and that the eigenvalue of the derivative operator is identical to that of the eigenvalue under the action of A ..
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kwisangqaquqw3
Answered 2022-05-18 Author has 3 answers
Well, defining k c 2 i c 1 / 2, you get a real
( x ( t ) y ( t ) ) = k   e i 2 t ( i 1 / 2 ) + k   e i 2 t ( i 1 / 2 ) .
You did not specify anything about the form of the Ps, but you have sufficient freedom to recast these into something of your choice.
Note the problem simplifies to a triviality if you defined z x + i 2 y, so z ˙ = i 2   z... take the c.c. ... do you see the point?
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