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Why does "y = C.F. + P.I."?

soutioirsim

I'm just doing some FP2 differential equations and I was wondering, why does:

"y = complimentary function + particular integral" When solving 2nd order equations?

Is there a proof or is there some reasoning behind it. The book doesn't explain it :redface:

Reply 1

JamesyB

Original post by soutioirsim

I will assume you know what the complimentary function and particular integral are considering you're doing FP2!

If you take the superposition of the complimentary function and particular integral this will cover every possible solution to the ODE. I'm sure one of the 2nd/3rd year maths grads on here can cover it in more detail though!

EDIT: Should mention that taking the superposition basically means adding them together (or at least it does in this case)

(edited 13 years ago)

Reply 2

Bobifier

The complimentary function is the general part of your differential equation - it is actually the set of functions that will give 0 when put into the differential equation, meaning that they have no bearing on the outcome of it. Given a differential equation and a solution to the differential equation, you can add any function from the complementary function set to the answer you have and it will still be an answer. Basically what I'm saying is that you need a complementary function to have the set of all possible solutions.

The particular integral is necessary to give, if you like, the bit of the equation that isn't 0. For instance, if you have y'' - y' + y = sin(x), clearly the solution must at least in part be a trigonometric function, so you find a solution in the set of trigonometric functions via whatever method you choose. This solution, the particular integral, will account for the specific part of the differential equation.

Thus, you have your complementary function which gives the set of answers to the equation but is 0 when put into the equation so it has no effect on anything on the other side, plus the particular integral which gives you the other side when it is put into the equation. Hopefully I have managed to explain why both are necessary.

Reply 3

Glutamic Acid

The key property of the differential equation is linearity. We may write a general 2^nd order linear differential equation as

\displaystyle \mathcal{L}y \equiv \dfrac{\text{d}^2y}{\text{d}x^2} + p(x) \dfrac{\text{d}y}{\text{d}x} + q(x)y

, and so we wish to solve

\mathcal{L}y = f(x)

for a given function f. Now, what I mean by the differential equation being linear is that

\mathcal{L}

is a linear operator, meaning for any constants a and b, and functions y and z,

\mathcal{L}(ay + bz) = a \mathcal{L}y + b \mathcal{L}z

.

So now we have this property, we may show that the solution of a second order linear differential equation is of your specified form. I will write our general solution y as

y = y_h + y_p

where y_h refers to the homogeneous solution, i.e. that

\mathcal{L}y_h = 0

, and y_p refers to the particular solution, i.e.

\mathcal{L}y_p = f

. First, we must check that y actually solves the differential equation, otherwise that would be somewhat retarded. And

\mathcal{L}y = \mathcal{L}(y_h + y_p) = \mathcal{L}y_h + \mathcal{L}y_p

(by linearity!)

= 0 + f = f

. And now the subtler job of establishing that all solutions of the differential equations are of this form remains. So, suppose we have another solution,

\tilde{y}

(this is a common trick), s.t.

\mathcal{L}\tilde{y} = f

. Then

\mathcal{L}\tilde{y} - \mathcal{L}y = f - f = 0 \implies \mathcal{L}(\tilde{y} - y) = 0

(by linearity!). But hang on, this means that

\tilde{y} - y

is part of the homogeneous solution, so

\tilde{y} - y = y_h \implies \tilde{y} = y + y_h = 2y_h + y_p

. But it is easy to check that

2y_h

is a homogeneous solution, and so all solutions are of this form.

Two things to note. First, that this easily generalizes to solutions of any n^th order linear differential equations. Secondly, we have demonstrated that solutions of this form are indeed solutions, and any solutions must have this form, but we have not shown the existence of such solutions: this is a very different beast.

Edit: homogeneous solution is synonymous with complementary function.

(edited 13 years ago)

Reply 4

soutioirsim