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!