differential equation

dN/dt = kNCos(0.02t) , where k is a constant and N is taken to be a continuous variable. It is given N =125 when t = 0

Solve the differential equation and obtain a relationship in term of N , k and t

∫1/ N dN =∫ k.Cos(0.02t) dt
ln A| N| = (k.Sin(0.02t))/0.02

N= A.e^[(k/0.02).Sin(0.02t)]
when t = 0 , N = 125
So A = 125.
N= 125.e^[(k/0.02).Sin(0.02t)]

How do i find k? if i try to rearrange and make k subject of the equation, it becomes zero which im pretty sure is wrong

Reply 1

hugototh

3

Original post by Carlos Nim

dN/dt = kNCos(0.02t) , where k is a constant and N is taken to be a continuous variable. It is given N =125 when t = 0

Solve the differential equation and obtain a relationship in term of N , k and t

∫1/ N dN =∫ k.Cos(0.02t) dt
ln A| N| = (k.Sin(0.02t))/0.02

N= A.e^[(k/0.02).Sin(0.02t)]
when t = 0 , N = 125
So A = 125.
N= 125.e^[(k/0.02).Sin(0.02t)]

How do i find k? if i try to rearrange and make k subject of the equation, it becomes zero which im pretty sure is wrong

You have answered the question, you have a solution in terms of

N, k

and

t

. Unless you are given more boundary conditions there aren't any more constants you can solve for.

Reply 2

Carlos Nim

OP

9

Original post by hugototh

You have answered the question, you have a solution in terms of

N, k

and

t

. Unless you are given more boundary conditions there aren't any more constants you can solve for.

oh ok i thought it was solvable since 2 unknown was given.

Reply 3

thotproduct

19

You're done and dusted, you've obtained it in the general form that they wanted, you don't know enough prior info or initial conditions to resolve k, and the answer wanted in terms of those variables, so you're good to go.