If we want to apply the forward Euler method to x ″ = x with x ( 0 ) = 0 , x ′ ( 0 ) = 1, we can introduce a new function u := [ x ′ x ] then u ′ = [ 0 1 1 0 ] u and u ( 0 ) = [ 1 0 ] .So then first step of the forward Euler would be u 1 = u ( 0 ) + h u ′ ( 0 ) = [ 1 0 ] + h [ 0 1 1 0 ] [ 1 0 ] = [ 1 h ] . And if we translate it back we get x 1 = h.How can we do a similar thing with the equation x ″ = x 2 (with x ( 0 ) = 0 , x ′ ( 0 ) = 1)? It doesn't look like it can be written as a system, so there should be some other trick.

Question

If we want to apply the forward Euler method to       x    ″    =  x with   x  (  0  )  =  0  ,      x    ′    (  0  )  =  1, we can introduce a new function  u  :=      [                                        x            ′                                                x                      ]  then      u    ′    =      [                            0                          1                                      1                          0                      ]    u        and        u  (  0  )  =      [                            1                                      0                      ]    .So then first step of the forward Euler would be                              u          1                                    =        u        (        0        )        +        h                  u          ′                (        0        )                                          =                  [                                                    1                                                                    0                                              ]                +        h                  [                                                    0                                            1                                                                    1                                            0                                              ]                          [                                                    1                                                                    0                                              ]                                                  =                  [                                                    1                                                                    h                                              ]                .            And if we translate it back we get       x    1    =  h.How can we do a similar thing with the equation       x    ″    =      x    2   (with   x  (  0  )  =  0  ,      x    ′    (  0  )  =  1)? It doesn&#039;t look like it can be written as a system, so there should be some other trick.

Jazmyn Bean · Accepted Answer

Let   u  =      [                            x                                                  x            ′                                ]  . Then,       u    ′    =      [                                        x            ′                                                            x            ″                                ]    =      [                                        x            ′                                                            x            2                                ]   with   u  (  0  )  =      [                            0                                      1                      ]  .So, taking forward Euler steps yields:  u  (  h  )  ≈  u  (  0  )  +  h      u    ′    (  0  )  =      [                            0                                      1                      ]    +  h      [                            1                                                  0            2                                ]    =      [                            h                                      1                      ]    u  (  2  h  )  ≈  u  (  h  )  +  h      u    ′    (  h  )  =      [                            h                                      1                      ]    +  h      [                            1                                                  h            2                                ]    =      [                            2          h                                      1          +                      h            3                                ]    u  (  3  h  )  ≈  u  (  2  h  )  +  h      u    ′    (  2  h  )  =      [                            2          h                                      1          +                      h            3                                ]    +  h      [                            1          +                      h            3                                                (          2          h                      )            2                                ]    =      [                            3          h          +                      h            4                                                1          +          5                      h            3                                ]  and so on. The only difference is that       u    ′   is a non-linear function of   u.

If we want to apply the forward Euler method to x′′=x with x(0)=0,x′(0)=1, we can introduce a new function u:=[x′ x]

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