# y'=(y-1)cot x

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### 1 Answer

This is differential equation with separated variables.

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

`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)`

`y=1+Csinx`

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

`y(x)=1+Csin x` where `C in RR.`