Optimizing Euler Method for Differential Equations with Large Coefficients

[BeeKay]

Homework Statement

Hello,

I have a question about using Eulers Method to approximate a solution to a differential equation. The problem lists forces that would be applied on an object and influences its velocity and therefore its position. I believe I am doing the Euler method correct to approximate what the "new velocity" would be, but based on somewhat wonky results I want to confirm if I am making an error using the Euler Method ( or in the Fortran code, but I feel like that is fine).

Homework Equations

$m \frac{d^ {2}y}{dt^2} + b \frac{dy} {dt} + ky = F_{o} sin(νt), y(0) = y_{o}, \frac{dy} {dt} = ˙y_{o}.$ when t=0

$\frac{dy} {dt} = v, y(0) = y_{o},$

$\frac{dv} {dt} = \frac{− k} {m} y − \frac {b} {m} v + \frac {F_{o}} {m} sin(νt), v(0) = ˙y_{o}$

$\frac {y_{n+1} − y_{n}} {∆t} = v_{n},$

$\frac{v_{n+1} − v_{n}} {∆t} = \frac {− k }{m} y_{n} − \frac {b} {m} v_{n} + \frac {Fo} {m} sin(νt_{n}),$

$t_{n+1} = t_{n + ∆t}$

The Attempt at a Solution

The way I interpreted it was to first find $\frac{v_{n+1} − v_{n}} {∆t}$ by find it at the initial values. If I use 0.02 as the timestep, I solved by substituting in the initial given y value and v value. Some other parameters were given, but they would not be changing. I multiplied this entire solution by the time step to find how much the velocity changed in that time. I then added this change to the original velocity. From there, I am a little confused as to how it applies to the position estimation. If $\frac {y_{n+1} − y_{n}} {∆t} = v_{n},$, then shouldn't I just take the solution to $\frac{v_{n+1} − v_{n}} {∆t} = \frac {− k }{m} y_{n} − \frac {b} {m} v_{n} + \frac {Fo} {m} sin(νt_{n}),$ multiply it by the timestep, and then add it to the previous y value? Thanks in advance and sorry if I messed up LaTeX. If you need more information about the problem or my process, let me know.

 

[BvU]

Hello BK,

You have a second order differential equation and have worked it around to two expressions for $v_{n+1}$ and $y_{n+1}$ in terms of the known $v_n$ and $y_n$. I think that's correct. So "do the step"!

If the results are wonky, perhaps the coefficients are a bit big for the time step ?