Question Number 93510 by Shakhzod last updated on 13/May/20
$${y}^{'} −{y}.\mathrm{tan}\:{x}+{y}^{\mathrm{2}} \mathrm{cos}\:{x}=\mathrm{0} \\ $$
Commented by Shakhzod last updated on 13/May/20
$${Friends}\:{help}\:{me}\:{please}. \\ $$
y-y-tan-x-y-2-cos-x-0-