- #1
dr_mushroom
- 3
- 0
Hello guys!
I have following differential equation mx"(t)+b(x'(t))x'(t)+k(p)x(t)=0. As can be seen, "attenuation term" is dependent of velocity x'(t).
Also stiffness term k(p) is dependent of term p, which is p=k(p)x(t)/A. In this equation A is constant and k(p) means, of course, same term as in our differential equation. And there we have x(t) again... It looks very hard, I know.. :)
This whole mx"(t)+b(x'(t))x'(t)+k(p)x(t)=0, p=k(p)x(t)/A describes pressurizing of liquid in cylinder as function of displacement.
I have no idea can this kind of equation be even solved analytically. I tried to solve it numerically in Matlab and got good results, that's not a problem. I would be however interested to see what kind of analytic solution there might be.
So, my question is that is there some analytic solution?
If there is no analytic solution, would there be analytic solution to even mx"(t)+b(x'(t))x'(t)+kx(t)=0? In this, we have deleted nasty dependence of p term in k(p)-->k
Thanks already for your answers and follow-up questions,
-DRmushroom
I have following differential equation mx"(t)+b(x'(t))x'(t)+k(p)x(t)=0. As can be seen, "attenuation term" is dependent of velocity x'(t).
Also stiffness term k(p) is dependent of term p, which is p=k(p)x(t)/A. In this equation A is constant and k(p) means, of course, same term as in our differential equation. And there we have x(t) again... It looks very hard, I know.. :)
This whole mx"(t)+b(x'(t))x'(t)+k(p)x(t)=0, p=k(p)x(t)/A describes pressurizing of liquid in cylinder as function of displacement.
I have no idea can this kind of equation be even solved analytically. I tried to solve it numerically in Matlab and got good results, that's not a problem. I would be however interested to see what kind of analytic solution there might be.
So, my question is that is there some analytic solution?
If there is no analytic solution, would there be analytic solution to even mx"(t)+b(x'(t))x'(t)+kx(t)=0? In this, we have deleted nasty dependence of p term in k(p)-->k
Thanks already for your answers and follow-up questions,
-DRmushroom