y ( 0 ) = 1How can we solve this equation with operator splitting methods in matlab by using forward Euler method?

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y  (  0  )  =  1How can we solve this equation with operator splitting methods in matlab by using forward Euler method?

Jaelyn Levine · Accepted Answer

y              ′        +  y  =  3            e              2      t                      y              ′        +  y  =  0              y              ′        =  −  y    ∫                    d        y            y        =  −  ∫      d    t      ln  ⁡      y    =  −  t  +  C    y  =  C            e              −      t          y  =  C      (    t    )              e              −      t                    y              ′        =            C              ′            (    t    )              e              −      t        −  C      (    t    )              e              −      t                    C              ′            (    t    )              e              −      t        −  C      (    t    )              e              −      t        +  C      (    t    )              e              −      t        =  3            e              2      t                    C              ′            (    t    )              e              −      t        =  3            e              2      t                    C              ′            (    t    )    =  3            e              3      t          C      (    t    )    =  ∫      3                  e                    3        t              d  t  =            e              3      t        +            C              1          y  =            e              −      t            (                  e                    3        t              +                  C                    1              )  and consider the condition that   y      (    0    )    =  1  y      (    0    )    =  1  +  C  =  1    C  =  0so the final answer is  y  =            e              2      t

y′+y=3e2t , y(0)=1 How can we solve this equation with operator splitting methods in matlab by using forward Euler method?

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