To be concrete, consider a function of time, $f(t)$ obeying:

${\displaystyle \frac{d^{2}f(t)}{d{t}^2}+m\frac{df(t)}{dt}=-k^{2}f(t)}$

How do we create a discrete equivalent of this DE?

Consider time $t$, taking the discrete values 0, $\Delta$, $2\Delta$, $3\Delta$, ... . $\Delta$ could be a year, day, second or whatever you fancy. Denote, for convenience:

$f_0=f(0)$

$f_1=f(\Delta )$

$f_2=f(2\Delta )$

$f_3=f(3\Delta )$

etc.

Now recall Taylor expansions:

$f(t+x)=f(t)+x.f\text{'}(t)+x^2.f\text{'}\text{'}(t)/2+...$

Substitute $t=k\Delta$, $x=\Delta$ and $x=-\Delta$:

$f(k\Delta +\Delta )=f(k\Delta )+\Delta .f\text{'}(k\Delta )+{\Delta }^{2}f\text{'}\text{'}(k\Delta )/2+...$

$f(k\Delta -\Delta )=f(k\Delta )-\Delta .f\text{'}(k\Delta )+{\Delta }^{2}f\text{'}\text{'}(k\Delta )/2+...$

Using the discrete notation:

$f_{k+1}=f_k+\Delta .f{\text{'}}_k+{\Delta }^{2}f\text{'}{\text{'}}_k/2+...$

$f_{k-1}=f_k-\Delta -f{\text{'}}_k+{\Delta }^{2}f\text{'}{\text{'}}_k/2+...$

Adding and subtracting the last two equations (and ignoring higher terms beyond the third):

$f_{k+1}+f_{k-1}=2.f_k+{\Delta }^{2}f\text{'}{\text{'}}_k$

$f_{k+1}-f_{k-1}=2.\Delta .f{\text{'}}_k$

Rearrange:

$f\text{'}{\text{'}}_k=(f_{k+1}-2.f_k+f_{k-1})/{\Delta }^2$

$f{\text{'}}_k=(f_{k+1}-f_{k-1})/2\Delta$

These are the key equations for 'discretising' a DE.

Substituting in our example:

$(f_{k+1}-2.f_k+f_{k-1})/{\Delta }^2+m.(f_{k+1}-f_{k-1})/2\Delta =-k^2{f}_k$

I don't know whether this is any form of coursework, but if it is, you should declare any help - I would hope that as long as you use your own examples and work through them yourself, you should be OK.

Lets take a (fairly) real life problem: a particle hanging on a spring, with air resistance. We will (i) Use a bit of physics to derive the equation of motion, (ii) Solve the differential equation exactly, (iii) convert the differential equation to a difference equation, (iv) for particular initial conditions, use a spreadsheet to compare the differential/difference equations.

(i) Lets suppose we have a mass $m$ hanging from a spring, the other end being tied to the ceiling. There will be a point at which the mass can hang without motion (i.e. the spring tension equals the force due to gravity). Now suppose the mass is moved a distance, $x$, above this rest point. The force due to gravity is unchanged, but the spring tension changes (reduces) by an amount proportional to $x$. Therefore the net force (downwards) is $kx$. Applying Newton's second law:

$-k.x=m.x\text{'}\text{'}$

Now lets stick in some air resistance. Assume that this is proportional to (and in the opposite direction to) velocity, i.e. the resistance $R=-r.x\text{'}$. Including this in the equation above:

$-k.x-r.x\text{'}=m.x\text{'}\text{'}$

Now we are not interested in actual masses, so let's just drop the $m$, and incorporate it into $k$ and $r$:

$x(t)\text{'}\text{'}+r.x\text{'}(t)+k.x(t)=0$

(ii)This is a fairly easy differential equation to solve. Assume a solution of the form

$x(t)=A.e^{wt}$

Substitute (and a little cancelling):

$w^2+rw+k=0$

So $w=(-r\pm (r^2-4k{)}^{1/2})/2$

Let us assume that $r^2-4k<0$, which amounts to saying that the damping is sufficiently low that the system exhibits oscillations that die away [an underdamped system]. For suitable choices of $A$ we can form solutions

$x(t)=e^{-pt}\sin (qt)$, $x(t)=e^{-pt}\cos (qt)$

where $p=r/2$, $q=(4k-r^2{)}^{1/2}/2$

Lets assume for simplicity that the initial displacement is zero (I'll leave you to do the general case) and the initial velocity, $v_0=x\text{'}(0)$. A bit of thought should convince you that the solution is

$x(t)=v_0/q.e^{-pt}\sin (qt)$

(iii) Now let's convert the equation to a difference equation.

We start with

$x(t)\text{'}\text{'}+r.x\text{'}(t)+k.x(t)=0$ (1)

then use the 'discretising' equations I posted previously to get:

$[(x_{n+1}-2.x_n+x_{n-1})/{\Delta }^2]+r.[(x_{n+1}-x_{n-1})/2\Delta ]+k.x_n=0$

rearranging:

$x_{n+1}=[-k.x_n+r.x_{n-1}/2\Delta +(2x_n-x_{n-1})/{\Delta }^2]/[1/{\Delta }^2+r/2\Delta ]$ (2)

Finally, we know $x_0=0$, and that $x_1=v_0\times \Delta$. This allows us to compute $x_n$ for $n=$2, 3, ...

(iv) Now take a look at the attached spreadsheet. The spring constant and air resistance factor are inputs in E2 and E3 (they are in blue so you can change them). The initial velocity is in F5 (note that with the workings above, initial displacement is assumed to be zero so F4 is not used!). C10 is the value of $\Delta$.

B15 downwards represents time, in accordance with the time spacing, $\Delta$. C15 downwards represents the displacement computed according to the solution of the differential equation (1) above. D15 and down represents the discrete approximation, using equation (2) above.

Finally, I have included a graph, plotting the exact and discrete approximation. All you see is one line - that's because it works well! If you increase delta (to, say, 4) you see some divergence between the two curves.

There's quite a lot in this post, so if you want more explanation, let me know.