Beginner Question....

[MSRA]

Well i am just wondering why...when doing variable separble integration...we put Constant(C) on only one side...

so why only on one side...and in which side in particular...any specific rule...?

 

[andre_luis]

Ok...you could put 2 constants; one per each side : C1 and C2.
And, passing the one placed at right position to left place, you will subtract :

C = C1-C2

Did you get ? Constants may assume any value.

 

[MSRA]

but this will change the answer....i guess the sign's....+ve or -ve...

 

[andre_luis]

Ok again...ve is a negative voltage.



+++

 

[shiv_emf]

c1 and c2 may take any real value and more over

C=C1-C2

is thr any prob by using C=C1+C2

Shiv

 

[MSRA]

Ok again...ve is a negative voltage.



+++

sorry but i can't gt it...
what u r saying...?

from where the voltage comes....

 

[A.Anand Srinivasan]

when you integrate the equations you would integrate it on both sides and so you would be getting what you would have had before the differentiation and hence you wouldn't have any idea on what constants were lost during differentiation...
the main idea behind the usage of only one constant is that if you have had constants on both sides you would have subtracted and made a single constant before differentiation....

 

[MNomanA]

it doesn't matter that u put - C1 or + C1..... obviously if u put "-C1" and C1 comes out to be a positive value then putting "+C1" will give a negative value of C1.... is it really difficult !! if u still can't get it then try it by putting bot the signs and see yourself......

your original question why put C on one side..... okay don't put constant on one side put constants(C1 and C2) on both sides... and see if it results in something different or not..... it obviously will result the same..... try!....

 

[wateror]

c can be any constant, so it is the same if you split it to two constant...

 

[aersoy]

gdy = f(x)dx >>form of seperable equation

suppose we add constants both sides

∫ g dy + c1 =∫ f(x) dx + c2 >>this what you expected to be ( not wrong )


∫ g dy =∫ f(x) dx + c1 - c2 >> C = c1 - c2

since C and ( c1 - c2 ) are both constants it does not matter what you write

∫ g dy =∫ f(x) dx + C

 

[cai2]

i agree with aersoy....

 

[pallavir]

Hi ,
You have note a point that the difference between 2 constant is always a constant.

Not only difference but any arithmetic operation between 2 constants is always a constant.

instead of all these we are simply using single constant one side.

 

[Jerechoestolero]

Actually you could still put a constant on the other side. No matter you put it, it is still an arbitrary constant, better put one on C on either side.

exp(3)*exp(t) = t*15
C1 = exp(3)/15
C2 = C1

C2*exp(t) = t