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

Question

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

SweallySnicles3 · Accepted Answer

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  ywhere I employ   D instead of &#039; . Now you have:  (  D  −  1  /  5  )  x  −  4  /  5  y  =  0  ,        (  D  −  1  /  5  )  y  −  4  /  5  x  =  0Solving 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  =  0Both ones are easy ODEs of second orders with constant coefficients.

Blericker74 · Accepted Answer

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.

A solution for a system of differential equations? <msubsup> x 1 ′ </m

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