Question Number 104342 by bemath last updated on 21/Jul/20
$${solve}\:{x}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−\frac{{dy}}{{dx}}−\mathrm{4}{x}^{\mathrm{3}} {y}\:=\:\mathrm{8}{x}^{\mathrm{3}} \mathrm{sin}\left({x}^{\mathrm{2}} \right) \\ $$
Answered by bramlex last updated on 21/Jul/20
$${multiplying}\:{both}\:{sides}\:{by}\:{x} \\ $$$$\Rightarrow{x}^{\mathrm{2}} .\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−{x}\frac{{dy}}{{dx}}−\mathrm{4}{x}^{\mathrm{4}} {y}=\mathrm{8}{x}^{\mathrm{4}} \mathrm{sin}\:\left({x}^{\mathrm{2}} \right)…\left(\mathrm{1}\right) \\ $$$${set}\:{x}^{\mathrm{2}} \:=\:{z}\:\rightarrow\:\frac{{dz}}{{dx}}\:=\:\mathrm{2}{x}\: \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{dz}}.\frac{{dz}}{{dx}}\:=\:\mathrm{2}{x}\:\frac{{dy}}{{dz}}\rightarrow\:\frac{{dy}}{{dx}}\:=\:\mathrm{2}{x}.\frac{{dy}}{{dz}}…\left(\mathrm{2}\right) \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:\frac{{d}}{{dx}}\left(\mathrm{2}{x}\:\frac{{dy}}{{dz}}\right)\:=\:\mathrm{2}\frac{{dy}}{{dz}}+\mathrm{2}{x}\:\frac{{d}}{{dx}}\left(\frac{{dy}}{{dx}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\frac{{dy}}{{dz}}\:+\mathrm{2}{x}\:\frac{{d}}{{dz}}\left(\frac{{dy}}{{dz}}\right).\frac{{dz}}{{dx}}…\left(\mathrm{3}\right) \\ $$$${using}\:\left(\mathrm{2}\right)\:\&\left(\mathrm{3}\right)\:{in}\:\left(\mathrm{1}\right) \\ $$$$\left(\rightarrow\right){x}^{\mathrm{2}} \left[\mathrm{2}\frac{{dy}}{{dx}}+\mathrm{4}{x}^{\mathrm{2}} \:\frac{{d}^{\mathrm{2}} {y}}{{dz}^{\mathrm{2}} }\:\right]−{x}\left(\mathrm{2}{x}\frac{{dy}}{{dz}}\right) \\ $$$$−\mathrm{4}{x}^{\mathrm{4}} {y}\:=\:\mathrm{8}{x}^{\mathrm{4}} \:\mathrm{sin}\:\left({x}^{\mathrm{2}} \right) \\ $$$$\left(\rightarrow\right)\:\mathrm{4}{x}^{\mathrm{4}} \:\left(\frac{{d}^{\mathrm{2}} {y}}{{dz}^{\mathrm{2}} }\:−{y}\right)\:=\:\mathrm{8}{x}^{\mathrm{4}} \:\mathrm{sin}\:\left({z}\right) \\ $$$$\left(\rightarrow\right)\:\frac{{d}^{\mathrm{2}} {y}}{{dz}^{\mathrm{2}} }\:−{y}\:=\:\mathrm{2sin}\:\left({z}\right)\: \\ $$$$\left(\rightarrow\right)\:\left({D}^{\mathrm{2}} −\mathrm{1}\right){y}\:=\:\mathrm{2sin}\:\left({z}\right) \\ $$$${Homogenous}\:{part} \\ $$$${y}_{{c}} \:=\:{C}_{\mathrm{1}} {e}^{−{x}} +{C}_{\mathrm{2}} {e}^{{x}} \\ $$$${particular}\:{integral} \\ $$$${y}_{{p}} =\:\frac{\mathrm{1}}{{D}^{\mathrm{2}} −\mathrm{1}}\left(\mathrm{2sin}\:\left({z}\right)\right)=\:−\mathrm{sin}\:\left({z}\right)=−\mathrm{sin}\:\left({x}^{\mathrm{2}} \right) \\ $$$$\mathcal{G}{eneral}\:{solution} \\ $$$$\therefore\:{y}\:=\:{C}_{\mathrm{1}} {e}^{−{x}} +{C}_{\mathrm{2}} {e}^{{x}} −\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)\:\bigstar \\ $$
Commented by bemath last updated on 21/Jul/20
$${great}…{nice}\: \\ $$