Given is the ODE system y ′ = ( 1 1 0 ) + ( 0 0 0 0 k 0 0 − k k ) ywith boundary conditions y ( T ) = 0.Why does for y hold that ∫ t T d d s ( e K ( T − s ) y ( s ) ) d s = ∫ t T e K ( T − s ) ( 1 1 0 ) d sand using the boundary conditions y ( t ) = − e − K ( T − t ) ∫ t T e K ( T − s ) ( 1 1 0 ) d s

Question

Given is the ODE system      y    ′    =      (                            1                                      1                                      0                      )    +      (                            0                          0                          0                                      0                          k                          0                                      0                          −          k                          k                      )    ywith boundary conditions   y  (  T  )  =  0.Why does for   y hold that      ∫    t    T        d          d      s        (      e          K      (      T      −      s      )        y  (  s  )  )  d  s  =      ∫    t    T        e          K      (      T      −      s      )            (                            1                                      1                                      0                      )    d  sand using the boundary conditions  y  (  t  )  =  −      e          −      K      (      T      −      t      )            ∫    t    T        e          K      (      T      −      s      )            (                            1                                      1                                      0                      )    d  s

Ryan Fitzgerald · Accepted Answer

This is like a matrix version of integrating factor: let  K  =      (                            0                          0                          0                                      0                          k                          0                                      0                          −          k                          k                      )    ,  b  =      (                            1                                      1                                      0                      )    ,then       y    ′    =  b  +  K  y  ⇒      e          K      (      T      −      s      )            y    ′    −      e          K      (      T      −      s      )        K  y  =      e          K      (      T      −      s      )        bwhich is        (        e          K      (      T      −      s      )        y  (  s  )                )        ′    =      e          K      (      T      −      s      )        b     .Integrating both side,  y  (  T  )  −      e          K      (      T      −      t      )        y  (  t  )  =      ∫    t    T        e          K      (      T      −      s      )        b  d  s     .But   y  (  T  )  =  0.

Given is the ODE system y ′ </msup> = ( <mtable rowspacin

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