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…
Question Number 86657 by john santu last updated on 30/Mar/20 $$\mathrm{solve}\:\left(\mathrm{1}+\mathrm{x}^{\mathrm{3}} \right)\mathrm{dy}\:−\mathrm{x}^{\mathrm{2}} \:\mathrm{y}\:\mathrm{dx}=\mathrm{0} \\ $$$$\mathrm{y}\left(\mathrm{1}\right)\:=\:\mathrm{2} \\ $$ Answered by jagoll last updated on 30/Mar/20 $$\int\:\frac{\mathrm{dy}}{\mathrm{y}}\:=\:\int\:\frac{\mathrm{x}^{\mathrm{2}}…
Question Number 86640 by niroj last updated on 29/Mar/20 $$\:\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{differential}}\:\boldsymbol{\mathrm{equations}}: \\ $$$$\:\:\left(\boldsymbol{\mathrm{i}}\right).\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }\:−\:\boldsymbol{\mathrm{x}}\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}\:+\:\boldsymbol{\mathrm{y}}\:=\:\:\boldsymbol{\mathrm{log}}\:\boldsymbol{\mathrm{x}}. \\ $$$$\:\:\left(\boldsymbol{\mathrm{ii}}\right).\:\left(\boldsymbol{\mathrm{x}}+\mathrm{2}\right)^{\mathrm{2}} \:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }\:−\:\mathrm{4}\left(\boldsymbol{\mathrm{x}}+\mathrm{2}\right)\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}\:+\:\mathrm{6}\boldsymbol{\mathrm{y}}\:=\:\:\boldsymbol{\mathrm{x}}. \\ $$$$\: \\ $$ Answered…
Question Number 86307 by niroj last updated on 28/Mar/20 $$\:\:\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{following}}\:\boldsymbol{\mathrm{equation}}: \\ $$$$\:\:\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }\:−\mathrm{2}\:\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}\:+\boldsymbol{\mathrm{y}}\:=\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} \:\boldsymbol{\mathrm{sin}}\:\boldsymbol{\mathrm{x}}. \\ $$ Answered by TANMAY PANACEA. last updated on 28/Mar/20…
Question Number 86294 by jagoll last updated on 28/Mar/20 $$\mathrm{y}\:=\:\mathrm{2x}\:+\:\left(\mathrm{y}'\right)^{\mathrm{2}} −\mathrm{4y}' \\ $$ Answered by mr W last updated on 28/Mar/20 $${y}'=\mathrm{2}\pm\sqrt{\mathrm{4}−\mathrm{2}{x}+{y}} \\ $$$${let}\:{u}=\mathrm{4}−\mathrm{2}{x}+{y} \\…
Question Number 86142 by jagoll last updated on 27/Mar/20 $$\mathrm{y}'\:.\mathrm{sin}\:\mathrm{t}\:\mathrm{cos}\:\mathrm{t}\:=\:\mathrm{y}\:+\:\mathrm{sin}\:^{\mathrm{3}} \mathrm{t}\: \\ $$$$\mathrm{y}\left(\frac{\pi}{\mathrm{4}}\right)\:=\:\mathrm{0}\: \\ $$ Answered by Kunal12588 last updated on 27/Mar/20 $$\frac{{dy}}{{dt}}=\frac{{y}}{\mathrm{sin}\:{t}\:\mathrm{cos}\:{t}}+\mathrm{sin}\:{t}\:\mathrm{tan}\:{t} \\ $$$$\Rightarrow\frac{{dy}}{{dt}}+\left(−\frac{\mathrm{1}}{\mathrm{sin}\:{t}\:\mathrm{cos}\:{t}}\right){y}=\mathrm{sin}\:{t}\:\mathrm{tan}\:{t}…
Question Number 86120 by jagoll last updated on 27/Mar/20 $$\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\frac{\mathrm{sin}\:\mathrm{2y}}{\mathrm{x}}\:=\:\mathrm{x}^{\mathrm{3}} \:\mathrm{cos}\:^{\mathrm{2}} \:\mathrm{y} \\ $$ Commented by Prithwish Sen 1 last updated on 27/Mar/20 $${sir}\:{it}\:{is}\:{sin}\mathrm{2}{x}\:{not}\:{sin}\mathrm{2}{y}\:. \\…