I just wanted to see if I went about solving this correctly. The problem didn't provide a step so I'm using .1: x ′ ( t ) = 1 + t sin ⁡ ( t x )Where x ( 0 ) = 0 at t = 1So using the idea that: y 1 = y 0 + h ∙ F ( x 0 , y 0 )I did the following: y 1 = 0 + .1 [ 1 + ( .1 ) sin ⁡ ( 1 ( 0 ) ) ] = .1 y 2 = .1 + .1 [ 1 + ( .1 ) sin ⁡ ( 1 ( .1 ) ) ] = .2I repeated this same formula 6 more times. Did I apply Euler's method correctly?

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I just wanted to see if I went about solving this correctly. The problem didn&#039;t provide a step so I&#039;m using .1:      x    ′    (  t  )  =  1  +  t    sin  ⁡  (  t  x  )Where   x  (  0  )  =  0 at   t  =  1So using the idea that:      y    1    =      y    0    +  h  ∙  F  (      x    0    ,      y    0    )I did the following:      y    1    =  0  +  .1      [    1    +    (    .1    )    sin    ⁡    (    1    (    0    )    )    ]    =  .1      y    2    =  .1  +  .1      [    1    +    (    .1    )    sin    ⁡    (    1    (    .1    )    )    ]    =  .2I repeated this same formula 6 more times. Did I apply Euler&#039;s method correctly?

pradassas66b2d · Accepted Answer

Not quite. A finite difference approximation to       x          ′       gives:                    y                  n          +          1                    −              y                  n                      h    =  1  +  n  h  ⋅  sin  ⁡  (  n  h  ⋅      y          n        )where       y          n        ≈  x  (  n  h  ). Moving some terms around,      y          n      +      1        =      y          n        +      (    1    +    n    h    ⋅    sin    ⁡    (    n    h    ⋅          y              n              )    )    h  .Therefore,                              y                      1                                              =                  y                      0                          +        h                                      y                      2                                              =                  y                      1                          +                  (          1          +          h          ⋅          sin          ⁡          (          h                      y                          1                                )          )                h        .            Using       y          0        =  0 and   h  =  0.1, we get                              y                      1                                              =        0        +        0.1        =        0.1                                      y                      2                                              =        0.1        +                  (          1          +          0.1          ⋅          sin          ⁡          (          0.1          ⋅          0.1          )          )                ⋅        0.1        ≈        0.2.            Though your result is the same, your computation is incorrect (in the first step, you use   t  =  1). Maybe this was a typo?(Also, note that       y    2   is only approximately equal to 0.2; not exactly)

I just wanted to see if I went about solving this correctly. The problem didn't provide a step so I'

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