Occasionally in this course, we'll come across concepts that you'll have studied before, perhaps a couple of years ago, that you may not have used too often in the intervening time. Everyone has the odd topic that they've gone a bit rusty on, so on this page, I'm going to put links that might help you revise any such concepts that crop up in the lectures.
A common theme in many of the methods we'll use to solve partial differential equations (PDEs) is to reduce them to ordinary differential equations (ODEs), which then need to be solved.
An ODE of the form \[\frac{\mathrm{d}y}{\mathrm{d}x}+P(x)y=Q(x)\] may be solved by multiplying throughout by an integrating factor of \(e^{\int P(x)\mathrm{d}x}\), yielding \[e^{\int P(x)\mathrm{d}x}\frac{\mathrm{d}y}{\mathrm{d}x}+e^{\int P(x)\mathrm{d}x} P(x)y=e^{\int P(x)\mathrm{d}x} Q(x),\] or equivaltently \[\frac{\mathrm{d}}{\mathrm{d}x}\left\{e^{\int P(x)\mathrm{d}x}y\right\}=e^{\int P(x)\mathrm{d}x}Q(x).\]
Integrating both sides with respect to \(x\) and rearranging yields the solution \(y\).
An ODE of the form \[\frac{\mathrm{d}y}{\mathrm{d}x}=f(x)g(y)\] may be solved by "rearranging" to get \[\int\frac{\mathrm{d}y}{g(y)}=\int f(x)\mathrm{d}x.\]
Rearranging yields the solution.
ODEs of the form \[a\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+b\frac{\mathrm{d}y}{\mathrm{d}x}+cy=0\] can be solved by formulating the characteristic solution and determining whether its roots are real, imaginary or repeated. If the right hand side is some function \(f(x,y)\) instead of zero, a particular solution must also be sought.
In the general case of the method of characteristics, we parameterise curves. This essentially means that we write the \((x,y)\) co-ordinates of the points on the curve as \((x(t),y(t))\), where \(x(t)\) and \(y(t)\) are functions of a parameter \(t\). Varying \(t\) varies the point on the curve.
If you want to read further about curve parameterisation, the external resource below is useful.