Second order differential equation

freeski
Messages
4
Reaction score
0

Homework Statement


if given the general solution to a differential equation
y(t)=a cos t + b sin t + ((cos wt)/1-w^2),
Find out how long it takes the solution to approach the steady state solution.
original second order differential equation:
y'' + y = cos wt such that 0 < w < 10

Homework Equations





The Attempt at a Solution


I am not sure where to begin.
 
Physics news on Phys.org
So for y'' + y = cos wt

What is your homogeneous solution?

EDIT: Remember that your general solution will be y=yh+yss.
 
Last edited:
freeski said:

Homework Statement


if given the general solution to a differential equation
y(t)=a cos t + b sin t + ((cos wt)/1-w^2),
The good news is that you are being very diligent in your use of parentheses, but the bad news is that some of them aren't in the right place. You should write the last expression above as cos(wt)/(1 - w^2). As you have it, this expression would be read as cos(wt)/1 - (w^2), or cos(wt) - w^2, and I'm pretty sure that's not what you intended.
freeski said:
Find out how long it takes the solution to approach the steady state solution.
original second order differential equation:
y'' + y = cos wt such that 0 < w < 10

Homework Equations





The Attempt at a Solution


I am not sure where to begin.
A good start would be to find out what the term "steady state solution" means.
 
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