# Math Help - separation of variabes,new constant help needed

1. ## separation of variabes,new constant help needed

I have separated the variables of dy/dx +2ycot2x = 0
By taking -1/2y to the left i ended u with

y = 1/2Asin2x i. 1/2Acosec2x where A in a constant, my question in since the mark scheme gives 1/Acosec2x can i make a new constant B from the 2A which would represent the right answer in the mark scheme?

If not i believe there may be something i'm missing at the stage
-0.5ln2y = 0.5lnsin2x + c which would get me the answer. I have an exam tommorow and really worried if someone couldhe me here this would be great thanks!!

2. Originally Posted by i_zz_y_ill
I have separated the variables of dy/dx +2ycot2x = 0
$y'+2y\cot 2x = 0$ thus $\frac{y'}{y} = - 2\tan 2x$.
Now you can use seperation of variables.

3. Yeah thanks but that didnt realli answer my question, can i vreate a new constant in such a circumstance. Your wright i should have separated it that way as is more convenient but i may not always do it that way. Unless I am wrong in my method please tell me.

4. Originally Posted by i_zz_y_ill
I have separated the variables of dy/dx +2ycot2x = 0
By taking -1/2y to the left i ended u with

y = 1/2Asin2x i. 1/2Acosec2x where A in a constant, my question in since the mark scheme gives 1/Acosec2x can i make a new constant B from the 2A which would represent the right answer in the mark scheme?

If not i believe there may be something i'm missing at the stage
-0.5ln2y = 0.5lnsin2x + c which would get me the answer. I have an exam tommorow and really worried if someone couldhe me here this would be great thanks!!
The entire post is very hard to understand. Could you please reword it, so that we can understand it clearly?

If I understand correctly you are asking if A is a constant then is B = 2A another constant? Then, yes you are right.