R help - Apr 2018 - ODE

Hello All,


I'm struggling to solve this ODE using R,


vdpol <- function (h, v, t) (
          list(c (
            -0.1*v/(pi*(2*10*h-h^2)),
              (v = (-0.1*v/(pi*(2*10*h-h^2))^2) + 2*9.81*h))

                  ))
          library(deSolve)
          yini <- (c(h = 20, v=0))
          nonstiff <- ode(y = yini, func = vdpol,
                          times= seq(0, 30, by = 0.01),
                                     parms = 1)

It says that


 The number of derivatives returned by func() (4) must equal the length of the
initial conditions vector (2)


I'm not sure how when i only have two derivates in the equation. I have
little to no knowledge of R and I'm following guidance of the internet so
any help is appreciated.


Kind Regards,


H



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You cross-posted this to StackOverflow (where there is already a response.)
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> On Apr 12, 2018, at 1:36 PM, hudeyfa jama <Hjama_96 at hotmail.com>
wrote:
> 
> Hello All,
> 
> 
> I'm struggling to solve this ODE using R,
> 
> 
> vdpol <- function (h, v, t) (
>          list(c (
>            -0.1*v/(pi*(2*10*h-h^2)),
>              (v = (-0.1*v/(pi*(2*10*h-h^2))^2) + 2*9.81*h))
> 
>                  ))
>          library(deSolve)
>          yini <- (c(h = 20, v=0))
>          nonstiff <- ode(y = yini, func = vdpol,
>                          times= seq(0, 30, by = 0.01),
>                                     parms = 1)
> 
> It says that
> 
> 
> The number of derivatives returned by func() (4) must equal the length of
the initial conditions vector (2)
> 
> 	[[alternative HTML version deleted]]
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