Previous page : Using Taylor series to solve differential equations

Next page : Euler’s step method

For simple harmonic motion we have that the force is proportional to the displacement from an equilibrium point, and directed in the opposite direction to the displacement. I.e. the object is pulled towards a point, and the pull in proportional to how far away from the point the object is. I.e.

Where k is a positive constant and *y* is the displacement from the equilibrium point. Since we have that *F*=*ma* we get

or

that can be written as

We previously found that the general solution to this is

Now let us work with this using Taylor series.

**Taylor series**

Let us set *θ* to 0 and let us assume a solution of the form

exists. We then have that

and

So

So

That means that if we know *a*_{0} and *a*_{1} we have the whole series.

So say we have that the solution should go through the origin. We than have *a*_{0}=0. That means that every second term will be 0. Say we choose *a*_{1}=1, that gives us

So

So what does the 1/*ω* stand for? The derivative of the above stand for the velocity, and for *t*=0 we get *v*=*ω* /*ω*=1. We could in principle multiply by an initial velocity *v*_{0}, and that gives us

In this we can recognize

By starting at *a*_{0}=*A* and *a*_{0}=1 we would end up with the series expansion of

Previous page : Using Taylor series to solve differential equations

Next page : Euler’s step methodLast modified: