Question Number 153542 by EDWIN88 last updated on 08/Sep/21 $$\:{y}'''+{y}'=\mathrm{sec}\:{x}\: \\ $$ Answered by puissant last updated on 08/Sep/21 $${y}'''+{y}'={secx} \\ $$$$\Rightarrow\:{y}''+{y}={ln}\left({tan}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)\right)+{C}_{\mathrm{1}} \\ $$$$\Rightarrow\:\left({D}^{\mathrm{2}} +{D}\right){y}={ln}\left({tan}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)\right)+{C}_{\mathrm{1}}…
Question Number 87944 by jagoll last updated on 07/Apr/20 $$\mathrm{y}\:'\:=\:\frac{\mathrm{2xy}}{\mathrm{y}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} } \\ $$ Commented by john santu last updated on 07/Apr/20 $$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{2xy}}{\mathrm{y}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }…
Question Number 87905 by john santu last updated on 07/Apr/20 $$\mathrm{y}'\:=\:\mathrm{2}^{\mathrm{y}} \\ $$ Commented by john santu last updated on 07/Apr/20 $$\mathrm{2}^{−\mathrm{y}} .\mathrm{y}'\:=\:\mathrm{1} \\ $$$$\mathrm{e}^{−\mathrm{y}\:\mathrm{ln}\:\mathrm{2}}…
Question Number 87904 by john santu last updated on 07/Apr/20 $$\mathrm{y}\:''\:−\mathrm{3y}'\:+\mathrm{2y}\:=\:\mathrm{10sin}\:\mathrm{x}\:+\:\mathrm{2cos}\:\mathrm{2x} \\ $$ Commented by niroj last updated on 07/Apr/20 $$\:\:\mathrm{y}^{''} −\mathrm{3y}^{'} +\mathrm{2y}=\mathrm{10sin}\:\mathrm{x}+\mathrm{2cos}\:\mathrm{2x} \\ $$$$\:\:\:\left(\mathrm{D}^{\mathrm{2}}…
Question Number 87789 by john santu last updated on 06/Apr/20 $$\left(\mathrm{D}^{\mathrm{3}} −\mathrm{D}^{\mathrm{2}} \right)\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} +\mathrm{1}\:,\mathrm{y}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\mathrm{y}\:'\left(\mathrm{0}\right)=−\mathrm{1}\:,\mathrm{y}\:''\left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$ Commented by mathmax by abdo last updated…
Question Number 87504 by M±th+et£s last updated on 04/Apr/20 $${solve}\: \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}\left(\frac{\mathrm{2}+{y}}{\mathrm{1}+{x}+{y}}\right)^{\mathrm{2}} \\ $$ Answered by TANMAY PANACEA. last updated on 04/Apr/20 $${i}\:{think} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}\left(\frac{{x}+{y}}{\mathrm{1}+{x}+{y}}\right)^{\mathrm{2}}…
Question Number 87059 by jagoll last updated on 02/Apr/20 $$\left(\mathrm{y}\:'\right)^{\mathrm{2}} −\mathrm{xy}'\:+\mathrm{y}\:=\:\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{solution} \\ $$ Answered by mr W last updated on 03/Apr/20 $${y}'=\frac{\mathrm{1}}{\mathrm{2}}\left({x}\pm\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{y}}\right)…
Question Number 86837 by M±th+et£s last updated on 31/Mar/20 $${solve} \\ $$$$\left.\mathrm{1}\right)\sqrt{{xy}}\:\frac{{dy}}{{dx}}=\mathrm{1} \\ $$$$\left.\mathrm{2}\right){e}^{{y}} \:{sec}\left({x}\right){dx}+{cos}\left({x}\right){dy}=\mathrm{0} \\ $$ Answered by TANMAY PANACEA. last updated on 31/Mar/20…
Question Number 21252 by youssoufab last updated on 17/Sep/17 $${prove},\forall{x}_{\mathrm{1}} ,…,{x}_{{n}} {y}_{\mathrm{1}} ,…,{y}_{{n}} \in\mathbb{R}^{+} \\ $$$$\sqrt{{x}_{\mathrm{1}} {x}_{\mathrm{2}} …{x}_{{n}} }+\sqrt{{y}_{\mathrm{1}} {y}_{\mathrm{2}} …{y}_{{n}} }\leqslant\sqrt{\left({x}_{\mathrm{1}} +{y}_{\mathrm{1}} \right)\left({x}_{\mathrm{2}} +{y}_{\mathrm{2}}…
Question Number 86734 by Tony Lin last updated on 30/Mar/20 $${Find}\:{all}\:{functions}\:{that}\:{satisfy}\:{the} \\ $$$${equation} \\ $$$$\left[\int{f}\left({x}\right){dx}\right]\left[\int\frac{\mathrm{1}}{{f}\left({x}\right)}{dx}\right]=−\mathrm{1} \\ $$ Answered by mr W last updated on 30/Mar/20…