Why Is a Particular Solution Necessary in Differential Equations?

[LumenPlacidum]

If the purpose of the general form of the solution to a differential equation is to represent a formula with parameters for the solutions to that differential equation, why is it that we typically want to add some particular solution to the general one?

Solution = General Solution + Particular Solution.

I suppose I understand why you'd want it there, but the part that I don't remember from my Diff. Eq. stuff from long ago is why the particular solution is not included in the general solution.

 

[Geofleur]

The particular solution is for the corresponding non-homogeneous equation. In other words, on the right hand side of the differential equation, zero is replaced with some function.

 

[thierrykauf]

A particular solution is a solution that satisfies boundary or initial value conditions. It shows up in inhomogeneous differential equations, as far as I recall.

 

[LumenPlacidum]

So, is this appropriate?

The general solution for a non-homogeneous system of differential equations is analogous to the +C of integration. Because of the superposition principle, any function of the form of the general solution COULD be a part of the solution for the ODE since it would have become exactly zero upon substitution for y.

 

[Mark44]

If the purpose of the general form of the solution to a differential equation is to represent a formula with parameters for the solutions to that differential equation, why is it that we typically want to add some particular solution to the general one?

Solution = General Solution + Particular Solution.

No, this isn't quite right. It is General Solution = Complementary solution + Particular Solution = y_c + y_p. The complementary solution is the solution to the associated homogeneous problem.

I suppose I understand why you'd want it there, but the part that I don't remember from my Diff. Eq. stuff from long ago is why the particular solution is not included in the general solution.

The homogeneous problem is f(t, y, y', ... ,y⁽ⁿ⁾) = 0; for example, y'' + 4y' + 4y = 0.
The nonhomogeneous problem is f(t, y, y', ... ,y⁽ⁿ⁾) = g(t); for example, y'' + 4y' + 4y = t.
The solution to the homogeneous problem is a linear combination of e^-2t and te^-2t. As it turns out for my example, the particular solution is y_p = -5/4 + t/4.

We know that y_c is a solution to the homogeneous problem, which means for my example, y_c'' + 4y_c' + 4y_c = 0. We also know that y_p is a particular solution of the nonhomogeneous problem, so y_p'' + 4y_p' + 4y_p = t.

Then for a general solution y = y_c + y_p, we will have y'' + 4y' + 4y = t, regardless of which linear combination of e^-2t and te^-2t we choose.