# Particular Integral

1. Jun 9, 2012

### strokebow

Hi Folks,

Can anyone advise me on particular integrals (any relevant websites or vids)?

In 'particular', I am reading a book and the guy talks about 2 particular integrals. Can anyone show me how and why this is achieved:

1.
The particular integral of:

(a-1)cosz

is:

0.5*(1-a)z*sinz

2.
The particular integral of:

v'' + v' = Acos(mz) - Acos(mz)/(m2-1)

Thanks so much!

cheers

2. Jun 10, 2012

### haruspex

I'm afraid your examples don't make much sense out of context.
The expression Particular Integral is (AFAIK) only used in reference to inhomogeneous differential equations. If the equation to be solved for y(t) is
L(y(t)) = f(t) [1]
and the operator L is linear then we can break the problem into two parts.
Because L is linear, if we had any solution h(t) for [1], and a solution g(t) of the homogeneous version:
L(y(t)) = 0 [2]
then g+h would also be a solution of [1].
It is therefore usual to proceed by finding all solutions of [2] and any solution of [1]. Combining these produces all solutions of [1].
The single solution for [1] that was found is called the Particular Integral.
See e.g. http://en.wikipedia.org/wiki/Linear...mogeneous_equation_with_constant_coefficients
If this doesn't explain the equations you're looking at, please provide more background.

3. Jun 11, 2012

### strokebow

Hi,

I get what you are saying. Thanks for replying.

Well, in the book the guy is finding particular periodic solution of a linear, 2nd order equation, with periodic coefficients.

The point at which he does the 1st particular integral, he is equating coefficients of like powers to zero:

c1'' + c1 - cos3z + (a1 - 1)cosz = 0

I've had a quick look on the internet at complementary functions and particular integrals. The only thing is, is that the guy singles out part of that equation (above) and says:

" the particular integral corresponding to (a1 - 1)cosz is the non-periodic function 0.5(1 - a1)*z*sinz "

Any ideas?

4. Jun 11, 2012

### haruspex

Extending what I said before, if there are multiple non-homogeneous terms (summed), you can find the PI for each separately then add them all together. So he's looking for a solution of
f''(z) + f(z) = (1-a)cos(z)
And 0.5(1-a) z sin(z) is indeed a solution (try it).
How he arrived at that is another question.
The PI for the cos 3z term is -cos(3z)/8, right?

5. Jun 12, 2012

### strokebow

Thanks haruspex. That is great help.

I have just tired it and got the answer. How on earth did he manage to get that solution... 0.5(1-a) z sin(z) ??? (Also, do you mean that this is just one of many possible solutions to this PI?)

Also, Why is it that he omits the '-cos(3z)' term in this instance. Why is this particular integral for f''(z) + f(z) = (1-a)cos(z) ...
As opposed to f''(z) + f(z) - cos(3z) = (1-a)cos(z)
Why can that term be omitted. Or is that what you meant by finding the PI for both of them separately and then bringing it all back together?

Furthermore, the PI for the cos3z term you give is correct! How did you that???

Thanks again for your kind help.

6. Jun 12, 2012

### HallsofIvy

Staff Emeritus
It can be shown that solutions to "linear ordinary homogeneous differential equations" must be of the form $e^{ax}$, $cos(bx)$, $sin(bx)$, $e^{ax}cos(bx)$, $e^{ax}sin(bx)$, depending on whether the solutions to the characteristic equation are real (a), imaginary (bi), or complex (a+ bi), and powers of x times those (if a single number is a multiple root of the characteristic equation).

As long as the "right hand side" of a non-homogeneous equation is one of those, you can use the method of "undetermined coefficients".
In this particular case, v'' + v' = Acos(mz) - Acos(mz)/(m2-1), the associated homogeneous equation, v''+ v'= 0 has characteristic equation $r^2+ r= r(r+1)= 0$ which has roots 0 and -1 so the general solution to the homogeneous equation is of the form $C_1e^{0z}+ C_2e^{-1(z)}= C_1+ C_2e^{-z}$. Since the "right hand side", $A(1- 1/(m^2-1)cos(mz)$ is not of either of those forms, we would look for a solution of the form $v(z)= Bcos(mz)+ Dsin(mz)[/tex]. Then [itex]v'= -mBsin(mz)+ mDcos(mz)$ and $v''= -m^2Bcos(mz)- m^2Dsin(mz)$ so the differential equation becomes
$$v''+ v= -m^2Bcos(mz)- m^2Dsin(mz)-mBsin(mz)+ mDcos(mz)= (mD- m^2B)cos(mz)+ (-m^2D- mB)sin(mz)= A(1- 1/(m^2-1)cos(mz)$$
so we have the two equations $mD- m^2B= A(1- 1/(m^2-1)$ and $-m^2D- mB= 0$ to solve for B and D.

7. Jun 12, 2012

### haruspex

Given a linear DE of the form
L(y) = f1(x) + f2(x) + ..
we can break it into:
L(y0) = 0
L(y1) = f1(x)
L(y2) = f2(x)

:
Given any solution of y0, y1 etc., y = y0+y1+y2... is a solution of the original equation.
That's because L, being linear, satisfies L(Ʃyi) = ƩL(yi).
So if y1 is a solution of L(y1) = f1(x) then so is y0+y1, and there may be many possible y0s. But just one y1 will do for the PI.

Essentially what HallsofIvy said. From experience, I expect solutions to be like cos(nx). Given the cos(3x) on the RHS, I tried Acos(3x). Then it was just a matter of finding the value of A which balanced the equation.
The other term is a bit more interesting. Putting y = Acos(x) produces A = 0. As HoI explained, the trick then is to introduce an affine factor, (A+Bx)cos(x). (This is connected with the special case of repeated roots in various contexts, e.g. partial fractions.)

8. Jun 13, 2012

### strokebow

Thanks alot for your inputs; it is very helpful! Really appreciate it.

@HallsofIvy
Thanks for your help. Please allow me to express my understanding to you and please correct me where I fall short:

Complete Solution = complementary function + particular integral
Is it correct to say that in your post you are solving for the complementary function?

Following what you said, In this case we have:
v'' + v - cos3z + (a1 - 1)cosz = 0

The associated homogeneous eqn is: v'' + v = 0
So then as you say:
This homogeneous equation has the characteristic eqn: r2+r=r(r+1)=0 which has roots 0 and -1 so the general solution to the homogeneous equation is of the form C1e0z+C2e−1(z)=C1+C2e−z

Why do you say that:
"In this particular case, v'' + v' = Acos(mz) - Acos(mz)/(m2-1)"??? Not sure how this was come by? How did you get this?

At the end you have assumed a solution for v(z) and then equated it to the Acos(mz) - Acos(mz)/(m2-1) term.
And the post finally ends up with 2 equations and 4 unknowns (A, D, B, m). How can we solve this?

@haruspex

I see what you've done and it makes sense. however, how come you have used v(z) = Acosx
and HoI has used v(z) = Acos(mz) - Acos(mz)/(m2-1)

Interestingly, but also confusingly, for the PI of cos(3z):
Your solution of -1/8cos(3z) is the same as:
the second part of the equation given by HoI: - Acos(mz)/(m2-1)
Whereby A = 1 and m = 3 in this case and you get the same answer. Probably I have not understood something.

Im a tad confused in trying to match everything up.

Thanks again the help. Sorry if I am a bit slow in taking things up

cheers

Last edited: Jun 13, 2012
9. Jun 13, 2012

### haruspex

I've not been able to figure out where HoI's factor (1-1/(m2-1)) came from, but it just means our two A constants differ by that factor. It makes no difference to the answer; you still get the same factor of cos(mz).