When to ask the homogeneous question

  • Thread starter Thread starter TG3
  • Start date Start date
  • Tags Tags
    Homogeneous
TG3
Messages
66
Reaction score
0
When determining a particular solution to a differential equation, one of the necessary steps is to ask the "homogeneous question" aka Does any term in yp solve the homogeneous equation for this problem. When it does, I know that it is necessary to multiple by t.

My question is, do I multiply yp by T and then derive, or do I derive and then multiply by T?

For example, A cos(t) solves the homogeneous equation for a problem I'm working on. Do I derive it first:
y(p) = Acos(T)
Y'(p) = -Asin(T)
Y''(p) = -Acos(T)
And then multiply each of those by T, or do I multiply by T first and then derive? It makes a big difference!
 
Physics news on Phys.org
TG3 said:
When determining a particular solution to a differential equation, one of the necessary steps is to ask the "homogeneous question" aka Does any term in yp solve the homogeneous equation for this problem. When it does, I know that it is necessary to multiple by t.

My question is, do I multiply yp by T and then derive, or do I derive and then multiply by T?
Try to keep your variables straight. t and T usually have different meanings.
TG3 said:
For example, A cos(t) solves the homogeneous equation for a problem I'm working on. Do I derive it first:
y(p) = Acos(T)
Y'(p) = -Asin(T)
Y''(p) = -Acos(T)
And then multiply each of those by T, or do I multiply by T first and then derive? It makes a big difference!

The short answer is you multiply by t first.

If cos(t) is a solution to the homogeneous equation, then so is sin(t), so your particular solution can't be a linear combination of cos(t) and sin(t). In this case, your particular solution will include Atcos(t) + Btsin(t), unless of course, tcos(t) and tsin(t) happen to be solutions of the homogeneous equation. If that happens, then your particular solution will include t2cos(t) and t2sin(t).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top