We know that the solution of y ′ = ywith y ( 0 ) = 1 is y = e x As far as I see the Euler's method is explicitly used only to find the numerical approximation of e.g. y ( 3 ). Can we use the Euler's method to solve this differential equation and find the exact solution?Edit:I thought of the following: y ( t 1 + δ t ) = y ( t 0 ) + y ′ ( t 1 ) δ t y ( t 1 + 2 δ t ) = y ( t 1 + δ t ) + y ′ ( t 1 + δ t ) δ t . . .and from that we could find a recursive solution for y ( t 0 ). Then by taking the limit lim δ t → 0 of it, it could be possible to find the general solution of y for every t 0 in the domain.Is it impossible?

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We know that the solution of      y    ′    =  ywith   y  (  0  )  =  1 is  y  =      e          x      As far as I see the Euler&#039;s method is explicitly used only to find the numerical approximation of e.g.   y  (  3  ). Can we use the Euler&#039;s method to solve this differential equation and find the exact solution?Edit:I thought of the following:  y  (      t    1    +  δ  t  )  =  y  (      t    0    )  +      y    ′    (      t    1    )  δ  t  y  (      t    1    +  2  δ  t  )  =  y  (      t    1    +  δ  t  )  +      y    ′    (      t    1    +  δ  t  )  δ  t  .  .  .and from that we could find a recursive solution for   y  (      t    0    ). Then by taking the limit       lim          δ      t      →      0       of it, it could be possible to find the general solution of   y for every       t    0   in the domain.Is it impossible?

Monserrat Cole · Accepted Answer

I think that the process you&#039;re envisioning, if you did it rigorously, becomes equivalent to Picard&#039;s iterative process: Given an initial value problem       y    ′    (  t  )  =  f  (  t  ,  y  ),   y  (  0  )  =      y    0  , the sequence of functions defined by                              ϕ          0                (        t        )                            =                  y          0                                              ϕ                      k            +            1                          (        t        )                            =                  y          0                +                  ∫          0          t                f        (        s        ,                  ϕ          k                (        s        )        )                d        s            converges to a solution   ϕ  (  t  ).(You can think of that integral as summing   f  (  s  ,      ϕ    k    (  s  )  ) at many discrete points between   s  =  0 and   s  =  t, multiplied by a small amount   Δ  s, and taking the limit as   Δ  s  →  0.)In your problem, you have   f  (  t  ,  y  )  =  y and       y    0    =  1. Picard&#039;s process gives:                              ϕ          0                (        t        )                            =        1                                      ϕ          1                (        t        )                            =        1        +                  ∫          0          t                          ϕ          0                (        s        )                d        s        =        1        +                  ∫          0          t                1                d        s        =        1        +        t                                      ϕ          2                (        t        )                            =        1        +                  ∫          0          t                (        1        +        s        )                d        s        =        1        +        t        +                              1            2                                    t          2                                              ϕ          3                (        t        )                            =        1        +                  ∫          0          t                          (          1          +          s          +                                    1              2                                            s            2                    )                        d        s        =        1        +        t        +                              1            2                                    t          2                +                              1            6                                    t          3                    You can see that       ϕ    k    (  t  ) is the k-th degree Taylor polynomial for   ϕ  (  t  )  =      e    t  , and that is the solution to the IVP.

EnvivyEvoxys6 · Accepted Answer

No. Euler&#039;s method is only an approximation. To determine the exact value of   y at time   t  +  δ  t (regardless of whether the ODE has an exact solution), you would need to keep all terms of the Taylor expansion for the solution. Euler&#039;s method gives  y  (  t  +  δ  t  )  ≈  y  (  t  )  +      y    ′    (  t  )  δ  t  ,whereas the exact solution at   t  +  δ  t is  y  (  t  +  δ  t  )  =  y  (  t  )  +      y    ′    (  t  )  δ  t  +      1    2        y    ″    (  t  )  δ      t    2    +  ⋯  .In particular, Euler&#039;s method will only be exact if the solution is affine (of the form   y  =  a  x  +  b) so that all derivatives beyond the first derivative are zero.

We know that the solution of y ′ </msup> = y with y (

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