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Question Number 86003 by jagoll last updated on 26/Mar/20 | ||
$$\mathrm{y}\:''\:+\:\mathrm{y}'\:=\:\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{2x} \\ $$ | ||
Commented by mathmax by abdo last updated on 26/Mar/20 | ||
$${let}\:{y}^{'} ={z}\:\:\:\left({e}\right)\Rightarrow{z}^{'} \:+{z}\:={sinx}\:{cos}\mathrm{2}{x} \\ $$$$\left({he}\right)\rightarrow{z}^{'} \:+{z}=\mathrm{0}\:\Rightarrow\frac{{z}^{'} }{{z}}=−\mathrm{1}\:\Rightarrow{ln}\mid{z}\mid=−{x}\:+\lambda\:\Rightarrow{z}\:={K}\:{e}^{−{x}} \: \\ $$$${mvc}\:{method}\:\:{z}^{'} ={K}^{'} \:{e}^{−{x}} \:−{K}\:{e}^{−{x}} \\ $$$$\left({e}\right)\:\Rightarrow{K}^{'} \:{e}^{−{x}} \:−{K}\:{e}^{−{x}} \:\:+{K}\:{e}^{−{x}} \:={sinx}\:{cos}\left(\mathrm{2}{x}\right)\:\Rightarrow \\ $$$${K}^{'} \left({x}\right)\:={e}^{{x}} \:{sin}\left({x}\right){cos}\left(\mathrm{2}{x}\right)\:\Rightarrow{K}\left({x}\right)\:=\int_{.} ^{{x}} \:{e}^{{t}} \:{sint}\:{cos}\left(\mathrm{2}{t}\right){dt} \\ $$$${sint}\:{cos}\left(\mathrm{2}{t}\right)\:={cos}\left(\frac{\pi}{\mathrm{2}}−{t}\right){cos}\left(\mathrm{2}{t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\frac{\pi}{\mathrm{2}}+{t}\right)+{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{3}{t}\right)\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{sint}\:+\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{3}{t}\right)\:\Rightarrow{K}\left({x}\right)\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\int^{{x}} \:{e}^{{t}} {sint}\:{dt}+\frac{\mathrm{1}}{\mathrm{2}}\int^{{x}} \:{e}^{{t}} {sin}\left(\mathrm{3}{t}\right){dt} \\ $$$$\int\:{e}^{{t}} \:{sint}\:{dt}\:={Im}\left(\int\:{e}^{{t}+{it}} \:{dt}\right)\:={Im}\left(\int\:{e}^{\left(\mathrm{1}+{i}\right){t}} \:{dt}\right)\:{and} \\ $$$$\int\:{e}^{\left(\mathrm{1}+{i}\right){t}} \:{dt}\:=\frac{\mathrm{1}}{\mathrm{1}+{i}}{e}^{\left(\mathrm{1}+{i}\right){t}} \:=\frac{\mathrm{1}−{i}}{\mathrm{2}}\:{e}^{{t}} \left\{\:{cost}\:+{isint}\right\} \\ $$$$=\frac{{e}^{{t}} }{\mathrm{2}}\left\{{cost}\:+{isint}\:−{icost}\:+{sint}\right\}\:\Rightarrow\int^{{x}} \:{e}^{{t}} \:{sint}\:{dt}=\frac{{e}^{{t}} }{\mathrm{2}}\left({sint}−{cost}\right)\:+{c}_{\mathrm{0}} \\ $$$$\int\:{e}^{{t}} \:{sin}\left(\mathrm{3}{t}\right){dt}\:={Im}\left(\int\:{e}^{\left(\mathrm{1}+\mathrm{3}{i}\right){t}} \:{dt}\right)\:{and}\: \\ $$$$\int\:{e}^{\left(\mathrm{1}+\mathrm{3}{i}\right){t}} \:{dt}\:=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{3}{i}}{e}^{\left(\mathrm{1}+\mathrm{3}{i}\right){t}} \:+{c}_{\mathrm{1}} =\frac{\mathrm{1}−\mathrm{3}{i}}{\mathrm{10}}{e}^{{t}} \left\{{cos}\left(\mathrm{3}{t}\right)+{isin}\left(\mathrm{3}{t}\right)\right\} \\ $$$$=\frac{{e}^{{t}} }{\mathrm{10}}\left\{\:{cos}\left(\mathrm{3}{t}\right)+{isin}\left(\mathrm{3}{t}\right)−\mathrm{3}{i}\:{cos}\left(\mathrm{3}{t}\right)+\mathrm{3}\:{sin}\left(\mathrm{3}{t}\right)\right\} \\ $$$$\Rightarrow\int^{{x}} \:{e}^{{t}} \:{sin}\left(\mathrm{3}{t}\right){dt}\:=\frac{{e}^{{x}} }{\mathrm{10}}\left\{{sin}\left(\mathrm{3}{x}\right)−\mathrm{3}{cos}\left(\mathrm{3}{x}\right)\right\}\:\Rightarrow \\ $$$${K}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{4}}{e}^{{x}} \left(\:{sinx}−{cosx}\right)\:+\frac{{e}^{{x}} }{\mathrm{20}}\left({sin}\left(\mathrm{3}{x}\right)−\mathrm{3}{cos}\left(\mathrm{3}{x}\right)\right)\:+{C} \\ $$$${z}\left({x}\right)={K}\left({x}\right)\:{e}^{−{x}} \:=−\frac{\mathrm{1}}{\mathrm{4}}\left({sinx}−{cosx}\right)+\frac{\mathrm{1}}{\mathrm{20}}\left({sin}\left(\mathrm{3}{x}\right)−\mathrm{3}{cos}\left(\mathrm{3}{x}\right)\right)+{Ce}^{−{x}} \\ $$$${y}^{'} \:={z}\:\Rightarrow{y}\left({x}\right)\:=\int^{{x}} \:{z}\left({u}\right){du}\:+\lambda \\ $$$$=\int^{{x}} \left\{−\frac{\mathrm{1}}{\mathrm{4}}\left({sinu}−{cosu}\right)+\frac{\mathrm{1}}{\mathrm{20}}\left({sin}\left(\mathrm{3}{u}\right)−\mathrm{3}{cos}\left(\mathrm{3}{u}\right)\right)\:+{C}\:{e}^{−{u}} \right\}{du}\:+\lambda \\ $$$${y}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left({cosx}\:+{sinx}\right)\:+\frac{\mathrm{1}}{\mathrm{20}}\left(−\frac{\mathrm{1}}{\mathrm{3}}\:{sin}\left(\mathrm{3}{x}\right)−{sin}\left(\mathrm{3}{x}\right)\right)−{C}\:{e}^{−{x}} \:+\lambda \\ $$$$ \\ $$ | ||
Answered by john santu last updated on 26/Mar/20 | ||
$${y}_{{h}\:} =\:{C}_{\mathrm{1}} \mathrm{cos}\:{x}+{C}_{\mathrm{2}} \mathrm{sin}\:{x} \\ $$$${y}''\:+\:{y}'\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{3}{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:{x} \\ $$$${y}_{{p}} \:=\:{A}\mathrm{sin}\:\mathrm{3}{x}+{B}\mathrm{cos}\:\mathrm{3}{x}+{Cx}\:\mathrm{sin}\:{x}+\:{Dx}\:\mathrm{cos}\:{x} \\ $$$${y}^{'} \:=\:\mathrm{3}{A}\mathrm{cos}\:\mathrm{3}{x}−\mathrm{3}{B}\mathrm{sin}\:\mathrm{3}{x}+{C}\mathrm{sin}\:{x} \\ $$$$+{Cx}\mathrm{cos}\:{x}+{D}\mathrm{cos}\:{x}−{Dx}\mathrm{sin}\:{x} \\ $$$${y}''\:=\:−\mathrm{9}{A}\mathrm{sin}\:\mathrm{3}{x}−\mathrm{9}{B}\mathrm{cos}\:\mathrm{3}{x}+\mathrm{2}{C}\mathrm{cos}\:{x}\: \\ $$$$−{Cx}\mathrm{sin}\:{x}−\mathrm{2}{D}\mathrm{sin}\:{x}−{Dx}\mathrm{cos}\:{x} \\ $$$$\Rightarrow\:{A}=\:−\frac{\mathrm{1}}{\mathrm{16}}\:,\:{B}\:=\:\mathrm{0}\:,\:{C}\:=\:\mathrm{0} \\ $$$${D}\:=\:\frac{\mathrm{1}}{\mathrm{4}}.\:\:\Rightarrow{y}_{{p}\:} =\:−\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sin}\:\mathrm{3}{x}+\frac{{x}}{\mathrm{4}}\mathrm{sin}\:{x} \\ $$$$ \\ $$ | ||