Y'+ytanx=0 (How about y. y=?)
1 Answer
Sep 18, 2017
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
Now, we say
#y' + ytanx = -sinx + cosx(sinx/cosx) = -sinx + sinx = 0#
As required.
Hopefully this helps!