y'=(y-1)cot x



1 Answer | Add Yours

tiburtius's profile pic

Posted on (Answer #1)

This is differential equation with separated variables.


`dy/(y-1)=cotx dx`

Now we integrate whole equation.

`int dy/(y-1)=int cotx dx`

`ln(y-1)=ln (sin x)+ln C=ln(C sin x)`


Hence your function (or set of functions to be exact) is

`y(x)=1+Csin x` where `C in RR.` 

We’ve answered 287,753 questions. We can answer yours, too.

Ask a question