How can I show that the Euler method fails to approximate the exact solution y ( x ) = ( 2 x / 3 ) 3 / 2 to the IVP y ′ = y 1 3 y ( 0 ) = 0Here we have f ( t , y ) = y 1 3 , y 0 = 0 and so f ( t 0 , y 0 ) = f ( 0 , 0 ) = 0 and y n + 1 = y n + h f ( t n , y n )Thus y 1 = 0 y 2 = 0 ⋮ y n = 0So, I can't understand why it fails. Could you help me?

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How can I show that the Euler method fails to approximate the exact solution  y  (  x  )  =  (  2  x      /    3      )          3              /            2      to the IVP      y    ′    =      y          1      3        y  (  0  )  =  0Here we have   f  (  t  ,  y  )  =      y          1      3      ,       y    0    =  0 and so   f  (      t    0    ,      y    0    )  =  f  (  0  ,  0  )  =  0 and      y          n      +      1        =      y    n    +  h  f  (      t    n    ,      y    n    )Thus      y    1    =  0        y    2    =  0    ⋮        y    n    =  0So, I can&#039;t understand why it fails. Could you help me?

Irene Simon · Accepted Answer

Note that the general solution,  y  (  x  )  =            (                        2          x          +          C                3            )              3              /            2      Can be achieved by integration:  ∫            d      y              y              1                  /                3              =  ∫  d  xThis assumes that   y  ≠  0, which is not necessarily true. You therefore have two solutions at   x  =  0, and you can&#039;t force the Euler method to &quot;choose&quot; the right one.The reason you don&#039;t have a unique solution at   x  =  0 is that   d  (      y          1              /            3        )      /    d  x does not exist on any open interval containing   x  =  0. Choosing an IV point with   x  &amp;gt;  0 will guarantee local uniqueness.

How can I show that the Euler method fails to approximate the exact solution y(x)=(2x/3)^3/2 to the IVP y′=y^1/3 y(0)=0

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