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Question Number 91417 by jagoll last updated on 30/Apr/20
$${y}'+\mathrm{2}{y}={e}^{−{x}} \\ $$
Commented by john santu last updated on 30/Apr/20
$${IF}\:{u}\left({x}\right)=\:{e}^{\int\:\mathrm{2}\:{dx}} \:=\:{e}^{\mathrm{2}{x}} \\ $$$${solution}\: \\ $$$${y}\:=\:\frac{\int\:{u}\left({x}\right){q}\left({x}\right)\:{dx}\:+{C}}{{u}\left({x}\right)} \\ $$$${y}\:=\:\frac{\int\:{e}^{\mathrm{2}{x}} .{e}^{−{x}} \:{dx}\:+{C}}{{e}^{\mathrm{2}{x}} }\: \\ $$$${y}=\:{Ce}^{−\mathrm{2}{x}} +\:{e}^{−{x}} \: \\ $$