Question Number 214000 by golsendro last updated on 24/Nov/24
$$\:\:\mathrm{Let}\:\mathrm{y}\left(\mathrm{x}\right)\:\mathrm{be}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{diff}\:\mathrm{eq}. \\ $$$$\:\:\mathrm{y}\:'=\:\frac{\mathrm{cos}\:\mathrm{x}+\mathrm{y}}{\mathrm{cos}\:\mathrm{x}}\:,\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\:\:\mathrm{Find}\:\mathrm{y}\left(\frac{\pi}{\mathrm{6}}\right). \\ $$
Commented by mr W last updated on 24/Nov/24
$${do}\:{you}\:{mean}\:\mathrm{cos}\:\left({x}+{y}\right)\:{instead}\:{of} \\ $$$$\mathrm{cos}\:\left({x}\right)+{y}\:? \\ $$
Commented by golsendro last updated on 24/Nov/24
$$\:\mathrm{cos}\:\left(\mathrm{x}\right)\:+\:\mathrm{y}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 24/Nov/24
$${then}\: \\ $$$${y}'−\frac{{y}}{\mathrm{cos}\:{x}}=\mathrm{1} \\ $$$${it}'{s}\:{of}\:{type}\:{y}'+{p}\left({x}\right){y}={q}\left({x}\right) \\ $$