Unit problem in differential equation

[sunipa.som]

I have one differential equation like
dN/dt=c*(other terms with no unit)
unit of c is 1/ns. Now if I solve this equation, I will get value of N corresponding to t.
(1) Then what will be the unit of t?
(2) and if I calculate dN/dt1=(other terms with no unit). where t1=t/c.
Then what will be the unit of t1?

 

[hunt_mat]

As I see it you have three possible units [N], [t] and [c], and I nonlimensionalise according to $N=N_{d}\bar{N}$, $t=t_{d}\bar{t}$ and $c=c_{d}\bar{c}$. There I put all the dimensions into the quantities with subscript d. So this turns the differential equation into:
$$\frac{d\bar{N}}{d\bar{t}}=\frac{c_{d}t_{d}}{N_{d}}\bar{c}$$
Where $c_{d}t_{d}/N_{d}$ is a nondimensional quantity. Does this clear things up?

 

[pmsrw3]

t will be in ns. Your RHS has dimensions of time^-1. dN/dt also has dimensions of time^-1. So everything matches.

Are you sure you want to divide t by c? Maybe I'm confused, but I think it would make more sense to multiply t by c, in order to get a dimensionless version of the equation. For instance, if you define $\tau=ct$, then you get $\frac{dN}{d\tau}=\mbox{(other terms)}$ and c has gone away.

 

[sunipa.som]

t will be in ns. Your RHS has dimensions of time^-1. dN/dt also has dimensions of time^-1. So everything matches.

Are you sure you want to divide t by c? Maybe I'm confused, but I think it would make more sense to multiply t by c, in order to get a dimensionless version of the equation. For instance, if you define $\tau=ct$, then you get $\frac{dN}{d\tau}=\mbox{(other terms)}$ and c has gone away.

--------------------------
Thank you. Sorry for mistake. I have to multiply t by c. Then I will get value of N for different times but then time has no unit.

 

[blue_raver22]

(1) The unit of t will depend on the units of the other terms in the differential equation. If all other terms have no units, then the unit of t will be in seconds (s). This is because c has units of 1/ns, meaning it is a rate per second. Therefore, when we solve for N, we are solving for a quantity at a specific time (t), which is measured in seconds.

(2) If we calculate dN/dt1, where t1 = t/c, then the unit of t1 will also be in seconds (s). This is because t1 is simply a scaled version of t, and the units of c (1/ns) will cancel out with the units of 1/t in the differential equation. Therefore, t1 will still be measured in seconds.