A solution for a system of differential equations? <msubsup> x 1 &#x2032; </m

Lena Bell 2022-07-02 Answered
A solution for a system of differential equations?
x 1 = 1 / 5 x 1 + 4 / 5 x 2
x 2 = 4 / 5 x 1 + 1 / 5 x 2
x 1 ( 0 ) = 1, x 2 ( 0 ) = 3
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Answers (2)

SweallySnicles3
Answered 2022-07-03 Author has 21 answers
Do the system as follows:
x = 1 / 5 x + 4 / 5 y ,     y = 4 / 5 x + 1 / 5 y D x = 1 / 5 x + 4 / 5 y ,     D y = 4 / 5 x + 1 / 5 y
where I employ D instead of ' . Now you have:
( D 1 / 5 ) x 4 / 5 y = 0 ,     ( D 1 / 5 ) y 4 / 5 x = 0
Solving the latter system as you always do in pre-calculus, you have
( D 1 / 5 ) 2 y 16 / 25 y = 0 ,     ( D 1 / 5 ) 2 x 16 / 25 x = 0
Both ones are easy ODEs of second orders with constant coefficients.
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Blericker74
Answered 2022-07-04 Author has 5 answers
Your initial value problem (system of ODEs + initial conditions) enjoys global uniqueness, and hence once you find a solution (regardless how you did) and you have verified that this pair of functions satisfies the equations and initial conditions then you are done.Your solution is THE solution of the IVP.
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