Can somebody teach me how to do 2nd Order Differential Equations - I'm really curious

unless you're very good at maths you'll need more than a quick brief. have a search around on google, there are loads of explanations out there

Asking about 2nd order ODEs is a very general topic since it can include very horrible ODEs, some of which cannot even be solved by an analytic solution and have to be approximated using numerical methods. I assume you mean second order linear ODEs? If so, these are pretty easy to do.. just look around on the internet as a poster above suggested

I know how to do 1st Order DE, but have never been taught how to do 2nd ODE.

I wish they were included in the A-level (non-FM) specification.

Can somebody please write a quick brief, and then set me 3 questions to solve for practice.

Thanks.

unless you're very good at maths you'll need more than a quick brief. have a search around on google, there are loads of explanations out there

Here's a very (very) quick briefing on 2nd-order linear homogeneous differential equations with constant coefficients.

Say you have $ay'' +by' + cy = 0$. Then when you substitute $y=e^{\lambda x}$, since $y'=\lambda e^{\lambda x}$ and $y''=\lambda^2 e^{\lambda x}$, we get:
$e^{\lambda x}(a\lambda^2 + b\lambda + c) = 0$

This is only equal to zero (as is on the RHS) when $a\lambda^2 + b\lambda + c = 0$ (since $e^{\lambda x}>0$ for all x).

This quadratic is called the characteristic equation of the differential equation. Suppose it has two real roots, $\alpha$ and $\beta$. Then $y=e^{\alpha x}$ and $y=e^{\beta x}$ satisfy the differential equation. But so do multiples of them, and they still do when you add them together, and so the general solution of the differential equation is
$y=Ae^{\alpha x}+Be^{\beta x}$
...for arbitrary constants A and B (analogous to constants of integration). This solution is known as the "general solution".

If $\alpha = \beta$ then instead of the above we get $(Ax+B)e^{\alpha x}$.

If $\alpha,\beta$ are complex, then we can still write it as $y=Ae^{\alpha x}+Be^{\beta x}$, or we can use various identities which allow us instead to express it as $y=e^{px}(C\cos qx + D\sin qx)$, where $\alpha = p+qi$ and $\beta = p-qi$.

If you were then told that $y=0$ when $x=0$ and $y'=0$ when $x=0$ then you could find $A,B$ (or $C,D$) and hence find a particular solution.

The only way to make any sense of what I've just written is to try some for yourself... reading it through probably won't do anything for you.