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