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Question Number 118574 by bramlexs22 last updated on 18/Oct/20
$${find}\:{particular}\:{solution}\:{of}\: \\ $$$$\left({D}^{\mathrm{3}} +\mathrm{4}{D}\right)\:{y}\:=\:\mathrm{sin}\:\mathrm{2}{x}\:{by}\:{using} \\ $$$${inverse}\:{operation}\:? \\ $$
Answered by TANMAY PANACEA last updated on 18/Oct/20
$${y}=\frac{{sin}\mathrm{2}{x}}{{D}\left({D}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$${y}=\frac{{e}^{{i}\mathrm{2}{x}} }{{D}\left({D}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$$=\frac{{e}^{{i}\mathrm{2}{x}} }{\left({i}\mathrm{2}\right)\left\{\left({D}+{i}\mathrm{2}\right)^{\mathrm{2}} +\mathrm{4}\right\}} \\ $$$$=\frac{{e}^{{i}\mathrm{2}{x}} }{\left({i}\mathrm{2}\right)\left({D}^{\mathrm{2}} +\mathrm{4}{iD}\right)} \\ $$$$=\frac{{e}^{{i}\mathrm{2}{x}} }{\left({i}\mathrm{2}\right)}×\frac{\mathrm{1}}{{D}×\mathrm{4}{i}×\left(\mathrm{1}+\frac{{D}}{\mathrm{4}{i}}\right)} \\ $$$$=\frac{\mathrm{1}}{−\mathrm{8}}×\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{{i}\mathrm{2}}{\mathrm{4}{i}}\right)}×\frac{{e}^{{i}\mathrm{2}{x}} }{\mathrm{1}}×{x} \\ $$$$=\frac{\mathrm{1}}{−\mathrm{8}}×\frac{\mathrm{2}}{\mathrm{3}}×\frac{{cos}\mathrm{2}{x}+{isin}\mathrm{2}{x}}{\mathrm{1}}×{x} \\ $$$$=\frac{{x}}{−\mathrm{12}}\left({cos}\mathrm{2}{x}\right)+{i}×\frac{{x}}{−\mathrm{12}}\left({sin}\mathrm{2}{x}\right) \\ $$$$\boldsymbol{{P}}.{I}=\frac{{x}}{−\mathrm{12}}×{sin}\mathrm{2}{x}\rightarrow{y}=\frac{{xsin}\mathrm{2}{x}}{−\mathrm{12}} \\ $$$${pls}\:{check}\:{my}\:{answer} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$