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

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Answer 1 · 2017-09-18T13:29:12

#y = Acosx#

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!