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Question Number 91298 by jagoll last updated on 29/Apr/20
$${what}\:{is}\:{the}\:{particular}\:{integral} \\ $$$${of}\:\left({D}^{\mathrm{2}} +\mathrm{6}{D}+\mathrm{5}\right){y}\:=\:{e}^{−\mathrm{5}{x}\:} \: \\ $$
Commented by john santu last updated on 29/Apr/20
Answered by niroj last updated on 29/Apr/20
$$\:\:\mathrm{PI}=\:\frac{\mathrm{e}^{−\mathrm{5x}} }{\mathrm{D}^{\mathrm{2}} +\mathrm{6D}+\mathrm{5}}=\:\frac{\mathrm{x}}{\mathrm{2D}+\mathrm{6}}\mathrm{e}^{−\mathrm{5x}} \\ $$$$\:=\:\frac{\mathrm{x}.\mathrm{e}^{−\mathrm{5x}} }{−\mathrm{10}+\mathrm{6}}=−\:\frac{\mathrm{xe}^{−\mathrm{5x}} }{\mathrm{4}}\://. \\ $$