Solve the initial value problem related to H ( x , y ) = x 2 + y 2 − 1 = 0 and ( x 0 , y 0 ) = ( 1 , 0 ), using the Euler method.The problems I've seen on the internet always give the y ′ in terms of y and x so what I did first was tro try to find this y ′ since I don't know it yet.By implicit differentiation I have that y ′ = − x / yNow, If I set my h = 0.1 for example, you can see on the first iteration I'll have an undefined value: x 0 = 1 , y 0 = 0 x 1 = x 0 + h = 1.01 , y 1 = y 0 + y ′ ( x 0 , y 0 ) = 0 + − 1 / 0 → U N D E F I N E DBut I don't think this makes any sense, perhaps I differentiated incorrectly or this is just not the right way to go. I'd like someone to help me understand the very first step which is obtaining an ODE for a given function.

Question

Solve the initial value problem related to   H  (  x  ,  y  )  =      x    2    +      y    2    −  1  =  0 and   (      x    0    ,      y    0    )  =  (  1  ,  0  ), using the Euler method.The problems I&#039;ve seen on the internet always give the       y    ′   in terms of   y and   x so what I did first was tro try to find this       y    ′   since I don&#039;t know it yet.By implicit differentiation I have that       y    ′    =  −  x      /    yNow, If I set my   h  =  0.1 for example, you can see on the first iteration I&#039;ll have an undefined value:      x    0    =  1  ,      y    0    =  0      x    1    =      x    0    +  h  =  1.01  ,      y    1    =      y    0    +      y    ′    (      x    0    ,      y    0    )  =  0  +  −  1      /    0    →  U  N  D  E  F  I  N  E  DBut I don&#039;t think this makes any sense, perhaps I differentiated incorrectly or this is just not the right way to go. I&#039;d like someone to help me understand the very first step which is obtaining an ODE for a given function.

Ezequiel Davidson · Accepted Answer

I suggest another option, namely treat   x and   y as functions of the time   t. From  x  (  t      )    2    +  y  (  t      )    2    =  1we obtain by differentiation with respect to   t the equation  2  x  (  t  )      x    ′    (  t  )  +  2  y  (  t  )      y    ′    (  t  )  =  0.This suggests to take      d          d      t            [                            x          (          t          )                                      y          (          t          )                      ]    =      [                            −          y          (          t          )                                      x          (          t          )                      ]    .This is a system of ordinary differential equations            z        ′    (  t  )  =      f    (  t  ,      z    (  t  )  )  ,where      z    (  t  )  =      [                            x          (          t          )                                      y          (          t          )                      ]    ,        f        (    t    ,          [                                    x                                                y                              ]        )    =      [                            −          y                                      x                      ]    .The initial condition is            z        0    =      z    (  0  )  =      [                            x          (          0          )                                      y          (          0          )                      ]    =      [                                        x            0                                                            y            0                                ]    =      [                            1                                      0                      ]    .Euler&#039;s method takes the form            z              n      +      1        =            z        n    +  h      f    (  t  ,            z        n    )or equivalently      [                                        x                          n              +              1                                                                        y                          n              +              1                                            ]    =      [                                        x            n                                                            y            n                                ]    +  h      [                            −                      y            n                                                            x            n                                ]    =      [                                        x            n                    −          h                      y            n                                                            y            n                    +          h                      x            n                                ]    .

Open question

Answer & Explanation

New Questions in Integral Calculus