# what particular solution do i use

what particular solution do i use

i need to solve mz’’ + az’ = -mg where m, a and g are constants
i got the homogeneous solution to be z = A + Be^-(a/m)t

since there is a constant in the homogeneous sol and a (-mg) in the RHS of the original eqn, would it make sense to put the particular solution as z = Ct where C is a constant??(multiplied by t to account for the homogenous sol) or would it be something different because of the B in the homogeneous sol as well??? sorry if i sound dumb hahaha i just feel like im missing something
Original post by yuhyuh726329
what particular solution do i use
i need to solve mz’’ + az’ = -mg where m, a and g are constants
i got the homogeneous solution to be z = A + Be^-(a/m)t
since there is a constant in the homogeneous sol and a (-mg) in the RHS of the original eqn, would it make sense to put the particular solution as z = Ct where C is a constant??(multiplied by t to account for the homogenous sol) or would it be something different because of the B in the homogeneous sol as well??? sorry if i sound dumb hahaha i just feel like im missing something
If there is no z term in the ode, Id solve the (first order) ode for z' so the CF is just an exponential and the PI is constant. Then integrate that as normal to get z. Its also worth noting you could have done seperation of variables rather than CF/PI to get z' then z if you were stuck on the usual ode approach.

The ode is essential a terminal velocity equation where the resistance is proportional to velocity and the weight is the accelerating term. So the PI for the position would be a linear function wrt t as you suggest as the terminal velocity is constant.
(edited 2 months ago)