For harder sketching—and there is some at the far end of the second A-level, depending on your exam board's syllabus—then the expression of the problem is likely to be a set of equations expressed as differentials.
The same basic questions as in the previous page apply: mostly based on looking to see what happens as the various terms tend towards interesting places, such as zero values.
One can draw what is called a tangent field, plotting values of dy/dx for chosen values of x and y, then plotting these and using what occurs to recognise where there might be lines of invariance, or what an economist would call a trend line. They often end up looking like a mathematican's attempt at art. As with the previous page, the essential to grasp here is that A-level is supposed to be allowing you, the student, to demonstrate a grasp of the higher order material. In consequence you are expected to see shortcuts through the evidence and towards a description of what is occurring (in this case, usually a graph).
I copy here the foot of the page on Differential Equations:
Moving on to systems of DEs, where dx/dt is written x‘
then any pair of equations y‘ = px + qy ......1
and x’ = rx + sy ......2
allows from 1: x” = rx’ = sy‘ and from 2: sy = x’-rx so
x” = rx’ + sy’ = rx’ + s(px+qy) = rx’ + spx + sqy = rx’ + spx + q(x’ - rx)
so x” - (q-r) x’ - (ps-qr) x = 0 ......3
Note that ps-qr is the determinant of the square matrix formed by the coefficients.
Similarly, y’ - (q-r)y’ - (ps-qr) y = 0 (which is, oddly, the same).
However, if one was to start with the equations 4 and 5 instead
y‘ = px + qy + a ......4
and x’ = rx + sy + b ......5
then you (will when you have done it for yourself) find that
x” - (q-r)x’ - ps - qr) x = sa - qb ......6
and y” - (q+r) y’ - (ps-qr) y = pb - ra ......7
You ought to write this as a matrix equation to see advantages to using such methods.
There is much mileage to be had in exploring tangent fields for solutions to DEs, particularly for non-linear functions. However, it is not necessary to develop a rigid technique for this: some intelligent use of mathematics can help enormously.
Given x’ and y’, dy/dx is easily written down. An asymptotic version of the general solution occurs when dy/dx is constant, say m.. From equations 4 & 5,
y‘ = px + qy + a gives dy = px + qy + a = m
x’ = rx + sy + b dx rx + sy + b
px + qy + a = m(rx + sy + b) => y = (mr - p ) x + mb - a
q - ms q-ms
the gradient of this (which may not display properly on your browser) is m which implies that
qm - m²s = (mr-p) x,
and hence m²s - (q+r+b)m - p = 0 and m = {q+r ± √((q+r)² + 4sp) } / 2s.
If the discriminant is negative then there are no asymptotes (expect a spiral or concentric rings in the tangent field). If the discriminant is zero, there is one , so .... (you fathom it out)
If the roots are different and real then one can expect one line to be ingress and one egress.
It is worthwhile finding and using the particular lines for which dy/dx is 0, -1, +1 and what happens along the two axes, etc. It is also worthwhile looking to see if there is an equilibrium point, where x’=y’=0. i.e. y = - (px+a) / q = - (rx+b) / s => spx = as = qrx + bq => x = (bq-as) / (ps - qr)
Similarly (you do it) y = (bp-ar) / (ps-qr)
My preferred book is Stewart: Differential Equations. I particularly liked the examples that lay outside Physics; I always wanted more.
