# Particular integral question with 2nd order diff eq'ns

1. Sep 29, 2007

### rock.freak667

If there is a differential equation to solve of the form
$$a\frac{d^2y}{dx^2} +b\frac{dy}{dx} + cy = tan(x)$$

you would put the LHS=0 and get the complementary function. But what would the the particular integral of tan(x) ?

2. Sep 30, 2007

### HallsofIvy

You are talking about using the "method of undermined coefficients"? That only works when the right hand side is on of the types of functions that you can get as solutions of linear constant coefficients equations: exponentials, sine or cosine, polynomials, or products of those. tan(x) is not of that type so "undetermined coefficients" will not work. Try "variation of parameters".

3. Sep 30, 2007

### genneth

A quick run with Mathematica shows that the particular integral is... unpleasant. Where has this problem come up? Not as a homework problem, I hope?

4. Sep 30, 2007

### rock.freak667

Ah ok I shall have to do some more reading now..thanks

Not as a homework problem...On Friday my teacher officially taught 2nd order diff. equations with constant coefficients. In one of the notes if the RHS was a sine or cosine you'd use Acos(bx)+Bsin(bx) or if "b" was a root(or a repeated root) of the auxiliary equation you'd use x{Acos(bx)+Bsin(bx)} as the P.I.

so out of curiosity I wanted to know what would happen if the RHS was tan(x)

5. Sep 30, 2007

### genneth

Aha. Then in this case, I would not keep trying with this example. The particular integral is *extremely* complex. It is worth keep learning about ways to solve differential equations -- the more the better -- but to solve this one will take a *very* long time to learn.

6. Sep 30, 2007

### real10

let $$y1$$ and $$y2$$ be the homogenous solutions and $$Yp$$ be the particular solution}[/tex]
$$Yp = V1(x) y1 + V2(x)y2$$
$$V1(x) = \int{\frac{-g(x)y2(x)}{W(y1,y2)(x)}dx}$$
$$V2(x) = \int{\frac{g(x)y1(x)}{W(y1,y2)(x)}dx}$$

this will give u the particular solution for 2nd order ode (also known as variation of parameters as "HallsofIvy" suggested)

where $$W(y1,y2)(x)$$ is the wronskian.
http://en.wikipedia.org/wiki/Wronskian

Last edited: Sep 30, 2007
7. Jun 12, 2008

### bsunder

Particular Integral

Actually its called "Method of Undetermined Coefficients" for finding solutions in terms of Particular Integral and Complementary Function for Non-homogeneous Linear Equations with constant coefficients for Second Order Differential Equations.

There are rules for the same in case we find the "forcing functions"(the terms on the right hand side) such as these.

Rules are based on forcing functions

RULE 1
If form of forcing function is in the form of A.exp(kx)
then form of PI will be
C.exp(kx), when k is not a root.
If k is a single root, then C.x.exp(kx)
If k is a double root, then C.x(square).exp(kx)
=====================================================================

RULE 2
If form of forcing function is in the form of Ploynomial then
if k=0 is not a root, PI will be C0+C1.x(raise to 1)+C2.x(square)+...
if k=0 is a single root, PI will be x(C0+C1.x+...)
if k=0 is a double root, PI will be x(square)(C0+C1.x+...)
=====================================================================

RULE 3
If form of focring function is A coskx and
if roots are of nature such as ik and its not a root, then PI will be C coskx + D sinkx

If form of corcing function is A sinkx and
if roots are of nature such as ik and its a single root, then PI will be x(C cos kx + D sinkx)

I hope that gives you a light, instead of the solution directly. :-)

8. Jun 12, 2008

### bsunder

Particular Integral... ways to find

Actually its called "Method of Undetermined Coefficients" for finding solutions in terms of Particular Integral and Complementary Function for Non-homogeneous Linear Equations with constant coefficients for Second Order Differential Equations.

There are rules for the same in case we find the "forcing functions"(the terms on the right hand side) such as these.

Rules are based on forcing functions

RULE 1
If form of forcing function is in the form of A.exp(kx)
then form of PI will be
C.exp(kx), when k is not a root.
If k is a single root, then C.x.exp(kx)
If k is a double root, then C.x(square).exp(kx)
=====================================================================

RULE 2
If form of forcing function is in the form of Ploynomial then
if k=0 is not a root, PI will be C0+C1.x(raise to 1)+C2.x(square)+...
if k=0 is a single root, PI will be x(C0+C1.x+...)
if k=0 is a double root, PI will be x(square)(C0+C1.x+...)
=====================================================================

RULE 3
If form of focring function is A coskx and
if roots are of nature such as ik and its not a root, then PI will be C coskx + D sinkx

If form of corcing function is A sinkx and
if roots are of nature such as ik and its a single root, then PI will be x(C cos kx + D sinkx)

9. Jun 12, 2008

### HallsofIvy

As I said above, the "method of undetermined coefficients" only works when the right hand side is on of the types of functions that you can get as solutions of linear constant coefficients equations: exponentials, sine or cosine, polynomials, or products of those.
Look up "variation of parameters" in your text book. It gives a reasonably simple method of reducing the problem to a pair of integrals. In general, however, those integrals do not have any anti-derivative in terms of elementary functions.