The title itself is self explanatory - I am trying to numerically solve an ODE with an initial value that has an infinite gradient. It seemed problematic to me and I am not certain as to how I should approach this.e.g. d y d x = y x , y ( 0 ) = 1(Obviously this can be solved analytically but I would like to know if there is any numerical method that tackles problems like this)

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The title itself is self explanatory - I am trying to numerically solve an ODE with an initial value that has an infinite gradient. It seemed problematic to me and I am not certain as to how I should approach this.e.g.             d      y              d      x        =      y          x        ,  y  (  0  )  =  1(Obviously this can be solved analytically but I would like to know if there is any numerical method that tackles problems like this)

clietsoriergono · Accepted Answer

Others can probably give a more &quot;conventional&quot; approach but one thing you can do, because       y    ′    (  x  ) is independent of   y  (  x  ), is take the following approach for the first step (at least).For the initial slope think rather angle. You want to proceed with a certain angle but you can&#039;t use       θ    1    =      π    2  ... which is obviously no good.Instead take as angle the mid-angle of       θ    1    =  π      /    2 and the angle of the curve at   x  =  h:      θ    2    :=      tan          −      1        ⁡      (          y      ′        (    h    )    )    =      tan          −      1        ⁡      (          1              h              )    ,and so take as initial angle:      θ    0    =                    θ        1            +              θ        2              2    =                    π        2            +              tan                  −          1                    ⁡              (                  1                      h                          )              2    ,and so an initial slope of:      m    0    :=  tan  ⁡  (      θ    0    )  .From here you can do normal Euler or perhaps keep it going with slope      m    k    =                    y        ′            (              x        k            )      +              y        ′            (              x                  k          +          1                    )        2    .This is an adaptation of Heun&#039;s Method. With a step-size of   h  =  0.1 this gives an error for       y    1    ≈  y  (      x    1    ) of the order of 0.015.

The title itself is self explanatory - I am trying to numerically solve an ODE with an initial value that has an infinite gradient. It seemed problematic to me and I am not certain as to how I should approach this. e.g. dy/dx=y/sqrt x, y(0)=1

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