# Integration by Separating variables

• January 6th 2010, 08:19 AM
matlabnoob
Integration by Separating variables
ok so i have the differential equation:

dx/dt = ((x-1)^2)*(t+1)

this is separable and separation of variables gives

integral ( dx/((x-1)^2)) = integral (t+1) dt

im told the correct answer is..

-1 / (x-1) = ((t+1)^2)/2 + c

but am wondering, why cant i go back to the differential equation and integrate the RHS as:

(t^2)/2 + t + c

why is (t+1) integrated together... why not integrate t first... and then 1... ? surely thats more simple?? (Thinking)
• January 6th 2010, 08:45 AM
Shanks
They make no difference, you will get the same result.
• January 6th 2010, 08:51 AM
matlabnoob
maybe im the only one not seeing it,when i integrate the t and 1 separately i dont get an extra 1/2 .... which i get when i integrate (t+1) together...

im sure this is v.basic but im not seeing it!
• January 6th 2010, 08:54 AM
akane
Quote:

Originally Posted by matlabnoob
ok so i have the differential equation:

dx/dt = ((x-1)^2)*(t+1)

this is separable and separation of variables gives

integral ( dx/((x-1)^2)) = integral (t+1) dt

im told the correct answer is..

-1 / (x-1) = ((t+1)^2)/2 + c

but am wondering, why cant i go back to the differential equation and integrate the RHS as:

(t^2)/2 + t + c

why is (t+1) integrated together... why not integrate t first... and then 1... ? surely thats more simple?? (Thinking)

$\int (t+1) dt = \frac{t^2}{2} + t + C$ ( I would also say that this way is more simple)

but also

$\frac{(t+1)^2}{2} = \frac {t^2 + 2t +1}{2} + C = \frac{t^2}{2} + t + \frac{1}{2} + C = \frac{t^2}{2} + t + C$

1/2 is a constant
• January 6th 2010, 09:11 AM
matlabnoob
thanks!!
1/2 is a constant...of course..!

believe it or not i am doing a maths degree (Itwasntme) how embarrassing