# Thread: where does he get this term from?

1. ## where does he get this term from?

I have the differential equation

dT/ds = C + BT

which is derived from

dT = Fds, F = umg + (1/2)CrAv^2, v^2 = 2T/m

and where

T = Kinetic energy
s = distance
C and B are constants

I have been shown a solution

(1/B)*ln(C+BT) = s + D

I understand that he has done the integral of;

dT/(C+BT) = ds

but how does he get the above result?

integral of 1/x = ln(x)

Where does the 1/B come from.

The solution he provoded does not match with my numerical solution.

Any help would be greatly appreciated

2. Originally Posted by revolution2000
I have the differential equation

dT/ds = C + BT

which is derived from

dT = Fds, F = umg + (1/2)CrAv^2, v^2 = 2T/m

and where

T = Kinetic energy
s = distance
C and B are constants

I have been shown a solution

(1/B)*ln(C+BT) = s + D

I understand that he has done the integral of;

dT/(C+BT) = ds

but how does he get the above result?

integral of 1/x = ln(x)

Where does the 1/B come from.

The solution he provoded does not match with my numerical solution.

Any help would be greatly appreciated
To integrate $\int \frac{dT}{C + BT}$ if you use the substitution $x = C+ BT$ then $dx = B dT$ and your integral becomes $\frac{1}{B} \int \frac{dx}{x}$