# What's the meaning of the general solution and the particular solution in differential equations?

Can anybody cast some physical insight into this? I've been studying differential equations on my own and don't understand how you can have a whole host of general solutions. It seems like a rather curious situation which we don't come across in other areas of mathematics. Is there anything more to the discussion that I'm missing?

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To study differential equations, you should look at a numerical book, to learn to simulate them. The analytic methods are generally primitive and misleading outside of one and two dimensional phase spaces, because of chaotic behavior. –  Ron Maimon Jul 18 '12 at 17:09

The solutions of a homogeneous linear ordinary differential equation form a vector space, ie $0$ is always a valid solution and arbitrary linear combinations of solutions will yield another valid solution. Finding the general solution of such an equation means finding a basis for that vector space.

In contrast, solutions to an inhomogeneous linear ordinary differential equation form an affine space over the space of homogeneous solutions. This means that $0$ is in general not a solution, and the difference between two solutions of the inhomogeneous equation will be a solution of the homogeneous one.

Thus, you only need to find a single inhomogeneous solution: The general solution of the inhomogeneous equation can be expressed of the sum of that particular solution and the general solution of the homogeneous one.

In case of non-linear differential equations, the integration constants (or equivalently, the choice of initial conditions) still parametrize the space of general solutions, and choosing values for these parameters will get you a particular one. However, the simple relation between the general solution of an inhomogeneous equation and the general solution of the corresponding homogeneous one does not hold for arbitrary differential equations.

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This answer is giving an answer specific to linear ODEs, while the phenomenon is general. But it's a fine answer for the linear case. –  Ron Maimon Jul 19 '12 at 1:55
@Ron: added a small note about the non-linear case –  Christoph Jul 19 '12 at 8:03
I don't agree it is a small note--- the reason I wrote a separate answer is because the linear case is misleading, the real reason for the arbitrary constants is not a vector space argument, it's an initial conditions argument, and it works regardless of the linearity of the equation. Neither are nonlinear equations "more complicated", they are just nonlinear. Sometimes they are simpler. –  Ron Maimon Jul 19 '12 at 16:27
@Ron: I slightly reworded the last part –  Christoph Jul 19 '12 at 16:49
thanks, I appreciate it, +1. –  Ron Maimon Jul 19 '12 at 17:02

The differential equation needs initial conditions to simulate on a computer (which you should do once, if you want to understand what they are). A differential equation is a rule on a time-grid of size $\epsilon$ for $x(t+\epsilon)$, given $x(t)$. For example;

$${dx\over dt}= x + x^2 + x^3$$

means

$$x(t+\epsilon) = x(t) + \epsilon ( x(t) + x(t)^2 + x(t)^3)$$

in the limit $\epsilon\rightarrow 0$.

You can use the equation above to step forward in time, but you need to give the value of $x(0)$. Then from this value, you calculate $x(\epsilon)$, then $x(2\epsilon)$, then $x(3\epsilon)$ and so on into the future. A special solution is one for a given initial value, while the general solution is expressed in terms of $x(0)$ or some equivalent arbitrary constant that can be used to find $x(0)$.

The theorem you want is the uniqueness theorem for the initial value problem of ordinary differential equations. The mathematical proof tends to use a less useful approximation technique than lattice stepping, but people who do computer simulations or computer games always use some sort of lattice for ODE's.

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