steady state solution of a differential equation

asi123

1. The problem statement, all variables and given/known data

Hey.


I am taking a matlab class and my instructor is absolutely horrendous. he does not teach us anything, but asks questions on everything.


I'm having a problem with the following question:

http://imageshack.us/photo/my-images/641/55811089.png/


What is a steady state solution of a differential equation? How can i solve the attached problem in matlab?


Thanks a lot.

2. Relevant equations



3. The attempt at a solution

jackmell

Steady-state means the solution is not changing with respect to time, that is, the first derivative is zero. However, when I solve that in Mathematica, I get:

[tex]\left\{\left\{y\to \text{Function}\left[\{t\},e^{\left(-\frac{2}{3}-\frac{\sqrt{19}}{3}\right) t} C[1]+e^{\left(-\frac{2}{3}+\frac{\sqrt{19}}{3}\right) t} C[2]+\frac{36 (17 \text{Cos}[2 t]-8 \text{Sin}[2 t])}{\left(-59+4 \sqrt{19}\right) \left(59+4 \sqrt{19}\right)}\right]\right\}\right\}[/tex]

so I don't see the relevance of that expression Acos(bt+c). Also, looking at the solution I don't see how it will ever reach a steady state.

LCKurtz

Judging from the way the problem is stated, it looks to me like he is using the term "steady state" to refer to a particular solution of the non-homogeneous equation that doesn't include the complementary solution. Doesn't the question specifically say to find A,B, and C?

epenguin

Maybe the question is referring to the way such systems can typically have a transient initial part that dies down and long-term settles down into a regular periodic 'forced' oscillation, that long term part being called the 'steady state'. You almost must have had examples in math and physics already.

Your LHS expressed in operator terms easily factorises into real factors with exponential solutions I think. (Though it alarms me to see one positive and one negative, making me wonder if this has such a steady state; maybe I am mistaken somehow.) Solve the equation and see how it behaves long-term. Oh, I see you have. The solution does not look right to me, but it does seem to have the problem I mentioned.