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

1. Jun 6, 2006

### dimensionless

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?

2. Jun 6, 2006

### Hootenanny

Staff Emeritus
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.

Last edited: Jun 6, 2006
3. Jun 6, 2006

### AKG

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.

4. Jun 6, 2006

### maverick280857

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.

5. Jun 8, 2006

### Wishbone

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.

6. Jun 8, 2006

### lance

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.