Hi, i need some help on this second-order ODE equation:

mu'' + f(u') + s(u) = F(t), t > 0, u(0) = U0, u'(0) = V0

How would it look like as a first-order system?

- Nov 25th 2008, 08:31 PM!alex!Formulating a second-order ordinary differential equation as a first-order system?
Hi, i need some help on this second-order ODE equation:

mu'' + f(u') + s(u) = F(t), t > 0, u(0) = U0, u'(0) = V0

How would it look like as a first-order system? - Nov 25th 2008, 09:06 PMDavid24

hey mate,

there are numerous ways to achieve this, but would probably be the simplest way is to let

w(t) = u'(t) --> w'(t) = u''(t)

Thus,

mu''(t)+f(u'(t)) + s(u(t)) = F(t) becomes,

mw'(t) + f(w(t)) + u(t) = F(t) , u(0) = U0 u'(0) = w(0) = V0

here we observe, we have reduced the order of the differential equation, and our system becomes,

mw'(t) + f(w(t)) + u(t) = F(t)

w(t) - u'(t) = 0

Subject to, u(0) = U0 , w(0) = V0

Hope this helps,

let me know if you require any further assistance,

Regards,

David - Nov 25th 2008, 09:29 PM!alex!
Thanks mate, I really appreciate it. I'm going to look better at it later and try to understand it. Right now i just have to use it in some other exercise.