Methods for solving differential equationsThere are several different ways of solving differential equations, which I'll list in approximate order of popularity. I'll also classify them in a manner that differs from that found in text books. Know it or look it up. Of course! Very many differential equations have already been solved. Some of these you will learn, and others you can look up. This is by far the most common way by which scientists or mathematicians 'solve' differential equations. It is also how some (nonnumerical) computer softwares solve differential equations. Substitution. Often a differential equation can be simplified by a substitution for one or other of the variables. This may turn it into one that is already solved (see above) or that can be solved by one of the other methods. (The software packages do this, too.) This category of solution includes a range of techniques that you will learn in a second year mathematics course. Guess and try. Another very common method of solving differential equations: guess what the solution might be, substitute it and, if it is not a solution, or not a complete solution, modify the guess until one has a complete solution. This is used often – more often than you would guess from reading books and papers, where the process usually appears to be rather elegant. In many cases you know something about the system studied, which gives you a clue. Experience helps, too, of course. However, we'll see below that the guessing is sometimes easy. Modify a simpler solution. If you know a solution to an equation that is a simplified version of the one with which you are faced, then try modifying the solution to the simpler equation to make it into a solution of the more complicated one. Transformation. Some differential equations become easier to solve when transformed mathematically. This is the main use of Laplace transformations. Numerical solution. If all the above fail, then an algorithm, usually implemented on a computer, can solve it explicitly, calculating the derivatives as ratios. This is usually a method of last resort, for two reasons. First, it only gives you the solution for one particular set of boundary conditions and parameters, whereas all the above give you general solutions. Second, it has limited precision: numerical derviatives are inherently noisy. Integration. This technique is elegant but is often difficult (or impossible). Sometimes one can multiply the equation by an integrating factor to make the integration possible. Special types. This vague title is to include special techniques that work for particular types of equations. This, too, is for study in higher year mathematics courses. Analog solution. Some differential equations are easily solved by analog computers. These are extremely fast and so suited to 'real time' control problems. Their disadvantages are limited precision and that analog computers are now rare. Below we show two examples of solution of common equations. They are simple, because they have only constant coefficients, but they are the ones you will encounter in first year physics. These equations could be solved by several of the means above, but we shall illustrate only two techniques. Example 1: Exponential growth and decay
Incidentally, it's worth stopping here to note that differential equations are almost always only approximations. It is impossible to have a system that is described this equation. For instance, the population of any species cannot grow exponentially. To quote just one limit: once the organisms occupy a solid sphere whose radius is increasing at the speed of light, any further growth cannot be exponential. (It is worth remembering this when politicians become obsessed with achieving growth in anything, but especially population.)
Example 2: Simple harmonic motion
In the x direction, Newton's second law tells us that F = ma = m.d^{2}x/dt^{2}, and here the force is − kx. This gives us the differential equation: where x is the displacement from equilibrium of the mass m at time t, and k is the stiffness of the spring to which the mass is attached.
This is a good time to use the Guess and try method of solution. We need a solution that oscillates forever, and that has the property that its second derivative is proportional to itself, but negative. The sine function does all that. Now we can't write x = sin t for dimensional reasons: the argument of the sine function can't have dimensions: it is given in radians (which is a ratio or number). We can write
However, sin (ωt) is a number and we need a length to have the same dimensions as x, so a possible solution is:
However, there is a problem with this proposed solution: it has x = 0 when t = 0. Which would be okay if I gave it a kick to start it from rest, but what if I release the mass from rest at a point away from equilibrium? In the latter case, I'd need x = A cos (ωt). The general solution must allow for these and any other starting condition. So instead we write: x = A sin (ωt + φ)
So let's return to consider φ. If we start the motion (t = 0) with v = 0 at x = A, then φ must be 90°: we have a cos function instead of a sine. Alternatively, if we start with maximum (positive) velocity at x = 0, then we need φ = 0. We give examples of these cases on the background page for oscillations. However, we could start with any combination of initial displacement x = x_{0} and v = v_{0}. So, for the general case (x_{0} ≠ 0, v_{0} ≠ 0), we can substitute to obtain We can solve these in terms of A and φ, first by dividing the two equations, then by squaring them and adding. So, for these given initial conditions, we can find a combination of the constants A and φ, so this is the general solution. How many boundary conditions? In our first example, we only had one constant of integration to find, so only one initial condition (or other boundary condition) to find. The second example was a second order equation, requiring two integrations or two boundary conditions. Here, we might specify two out of the initial displacement, velocity and acceleration, or some other two parameters.
Damped and forced oscillations
Physically, this term corresponds to a force, proportional to the speed. What can we guess about the solution, and how would we go about modifying the solution we had above so that it would satisfy our new differential equation? Again, we can use our knowledge of the physical system: when we a force whose direction is opposite that of the velocity, we slow it down. So we'd expect one of two possible answers: either it should oscillate, with the magnitude of the oscillations gradually decreasing over time, or else (if the damping were large enough) it could slow to a stop without even oscillating. This would suggest to us the possibility of a solution of the form x = A e^{−βt} sin (ωt + φ). We can try this already. But it's not quite a solution. Well, what if the damping force slows down the vibration? Why not try (ω + δω) instead of ω = k/m and see if this gives a solution for a suitable value of δω? Let's add a further complication: let's start shaking the particle, with an extra oscillating force, say F = F_{0} sin Ωt. This gives us a new differential equation:
Partial differential equations: the wave equation
∂y/∂x. Think of this as dy/dx at a given, constant time, t. Imagine taking a photograph (time is constant): in the image at time t, this is the slope of the y(x) shape at the instant of the photograph. ∂y/∂t. Think of this as dy/dt at a given position, x. This is just the velocity in the y direction at a particular point x on the string. (Not the velocity of the wave, by the way). y = A sin(kx − ωt), so
The bottom two graphs are the second derivatives with respect to the same variables:
∂y^{2}/∂t^{2} = − ω^{2}A sin(kx − ωt), which is the acceleration of a point on the string. These have important physical significance: the first one is determines the curvature of the string. If ∂y^{2}/∂x^{2} = 0, then the slope is constant, so it is straight. That means that the tension T acts in opposite directions at opposite ends, giving no nett force. If a segment is curved, however (∂y^{2}/∂x^{2} ≠ 0), it has a force acting on it. For constant curvature over a small length L, the nett force is proportional to L. We know the acceleration so we can apply Newton's second law. The mass of the segment is μL, where μ is the mass per unit length μ. Writing Newton's law as a = F/m gives:
Looking back at our expressions for the two second derivatives, we see that they our original function y = A sin(kx − ωt) is a solution to the wave equation, provided that T/μ = ω^{2}/k^{2}. We also saw, in Waves I, that ω/k is the wave speed, v. Which finally relates the wave speed to the physical properties T and μ of the string:

