Differential Equations

[Warrior4Christ]

This is probably a stupid question, because it's probably really obvious, but here goes...

What is the solution form of a second order constant coefficient DE, with a constant forcing function?

ie. y''(t) + ay'(t) + by(t) = c

I know you first need the homogenious solution:

y''(t) + ay'(t) + by(t) = 0
(solution in the form y(t) = e^(st))

But what is the particular solution? Is it just in the form y = K, which just solves to be K=c, or is there something more?

[Technomancer]

It's not that trivial of a question. Like you said, the actual solution is the sum of the homogenous solution and a particular solution of the non-homogeneous equation.

y=yh+yp

Once you have a particular solution (which will generally be some function), you need to sub it in for y. For example,

y''-4y'+3y=10exp(-2x) y(0)=1, y'(0)=-3

solving for the homogeneous part yields:

yh=c1*exp(x)+c2*exp(3x)

A particular solution which may be dervived from observation may be:
yp=C*exp(-2x)

substituting into the original equation:
4C*exp(-2x)-4*(-2C*exp(-2x))+3C*exp(-2x)=10*exp(-2x)

some algebra will show C to be C=2/3. Further algebra will allow you to solve for c1 and c2.

So far, I've shown you how to use the particular solution to solve the nonhomogeneous DE. However, it is not always possible to find yp by observation, so further mathematics is typically required. These are fairly involved, and you would be better served to pick up a book on DE's for guidance here (Kreyzig is fairly good). In particular, you should read up on the method of undetermined coefficients and the method of variation of parameters.

[Warrior4Christ]

I've already done a differential equations subject last semester, and this problem appeared in a subject this semester, but I can't remember what form the particular solution takes. We did variation of parameters and undetermined coefficients methods, and I looked back on the table of undetermined coeffient solution forms, but it doesn't have the solution form of a constant, eg:

y''-4y'+3y=10exp(-2x) (take your example, but change the 10exp(-2x))

y''-4y'+3y=20 <- constant "forcing function"

I can't remember the form of yp when you use method of undetermined coefficients.
Also, the initial conditions are substituted into the total solution, right? (The total solution being the y = yh +yp)

[Technomancer]

In this case, I'd actually recommend using Laplace transforms, which will solves the problem quite handily. As far as your second question goes, the boundary values are incorporated into the total solution when you solve for the constants.

[Warrior4Christ]

So yp is not needed then? It's all in the yh and y(0) and y'(0)?

[Technomancer]

You don't need to explictly solve for yp, although I believe that you will find the previous formulation to ultimately be equivalent to what you will get by applying the Laplace transform. In your example, for a 2nd order nonhomogeneous linear ODE:

a*y''+b*y'+c*y=f(x)

a*(s^2)*Y(s)+b*s*Y(s)+c*Y(s)=F(s)+s*f(0)+f'(0)

solve for Y and then take the inverse transform, to obtain y(x). The trick of course is to be able to invert the transform. The other methods I mentioned will still work for equations with constant coefficients though, although they may involve more work (depending on what you need to invert).

[Warrior4Christ]

Thanks. Yes, in the subject I mentioned earlier we use the Laplace domain/s domain quite a bit, so I'm familiar with finding the inverse transform.