Question

Y'+ytanx=0 (How about y. y=?)

Answer 1

`y = Acosx`

Explanation:

We have:

`y' = -ytanx`

`(y')/y = -tanx`

`int (y')/(y) = -int tanxdx`

`ln|y| = ln|cosx|+A`

`ln|y| = ln|cosx|+A`

`y = Acos x`

If we check, we know that `d/dx(cosx) = -sinx`.

Now, we say

`y' + ytanx = -sinx + cosx(sinx/cosx) = -sinx + sinx = 0`

As required.

Hopefully this helps!