ODEs: Why do I need Both a General And a Particular Solution?

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I am asking this question as it relates to physics, and in particular how it relates to harmonic oscillation.

Why is the equation not solved when I use only a particular solution? Why is the equation not solved when I use only a general solution?
 
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The obvious thing to say would be that a particular solution is only valid in a specific case, where as a general solution is always valid but is dependent upon intital conditions.
 
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You want to be able to describe all possible solutions. When someone asks you to solve x2 = 1, you give all possible solutions: 1 and -1. So when someone asks you to solve and ODE, you want to give all possible solutions. If you take the general solution (to the related homogenous equation) and add it to a particular solution, you are able to describe all possible solutions. Consider:

y'' - 5y' + 6y = 3

The related homogenous equation is:

y'' - 5y' + 6y = 0

which has general solution

yH(x) = ce2x + de3x

Note the above is really a family of solutions depending on the paramters c and d. A particular solution to the original equation is:

yP(x) = 1/2

So the solution to the original equation is

y(x) = ce2x + de3x + 1/2

Again, this is a family of solutions depending on paramters c and d. This fully describes the set of solutions to the original ODE. By "fully" describes I mean that:

a) if you replace c and d with any two real numbers, you will get a real function that solves the original ODE, i.e. IF y(x) = ce2x + de3x + 1/2 for some c and d THEN y(x) solves the original ODE
b) you can get EVERY solution to the ODE by plugging in numbers for c and d, i.e. IF y(x) solves the original ODE THEN y(x) = ce2x + de3x + 1/2 for some c and d.
 
Why is the equation not solved when I use only a particular solution? Why is the equation not solved when I use only a general solution?

That is not correct. A general solution is as much of a solution as a particular solution. Its just that a general solution does not incorporate initial conditions whereas a particular solution does. So for a physical application it is necessary to place restrictions imposed by physics on a solution obtained by solving the d.e. to restrict its class and to make sure it gives physically meaningful results.
 
The equation is solved if you present a particular solution, simply because you are giving a certain function that agrees with the D.E. Your homework problem might not be solved by only giving 1 particular solution, however to say that the DE is not solved is incorrect.
 
The particular solution often times solves the equation for the forcing function, while the homogeneous solution is the steady-state solution of the equation. if you are dealing with homogenous ode, the particular solution and the homogeneous solution are one in the same.

so everyone is right partially:

the particualar solution is the non-equilibrium solution

please make sure to add the particular and homogeneous solutions together before applying boundary conditions.
 
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