Question Number 97266 by bobhans last updated on 07/Jun/20 $$\mathrm{If}\:\mathrm{x}\:\&\mathrm{y}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{4x}−\mathrm{6y}−\mathrm{1}\:=\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}+\mathrm{y}\:? \\ $$ Commented by bobhans last updated on 07/Jun/20 $$\mathrm{f}\left(\mathrm{x},\mathrm{y},\lambda\right)\:=\:\mathrm{x}+\mathrm{y}+\lambda\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}}…
Question Number 97031 by bobhans last updated on 06/Jun/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 97001 by bemath last updated on 06/Jun/20 $$\mathrm{solve}\:\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{xy}−\mathrm{xy}^{\mathrm{2}} \\ $$ Commented by bobhans last updated on 06/Jun/20 $$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{x}\left(\mathrm{y}−\mathrm{y}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\Leftrightarrow\:\frac{\mathrm{dy}}{\mathrm{y}−\mathrm{y}^{\mathrm{2}} }\:=\:\frac{\mathrm{x}\:\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}…
Question Number 96713 by Tony Lin last updated on 04/Jun/20 $${y}^{\mathrm{2}} \:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{dy}}{{dx}} \\ $$ Answered by mr W last updated on 04/Jun/20 $${y}^{\mathrm{2}}…
Question Number 96669 by john santu last updated on 03/Jun/20 $$\mathrm{find}\:\mathrm{minimum}\:\mathrm{value} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{9}}+\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{30x}+\mathrm{250}} \\ $$ Answered by bemath last updated on 03/Jun/20 $$\mathrm{17}?…
Question Number 96610 by bemath last updated on 03/Jun/20 $$\mathrm{solve}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{x}+\mathrm{y}+\mathrm{4}}{\mathrm{x}−\mathrm{y}−\mathrm{6}} \\ $$ Commented by bobhans last updated on 03/Jun/20 $$\mathrm{set}\:\mathrm{x}+\mathrm{y}+\mathrm{4}\:=\:\mathrm{0}\:\&\:\mathrm{x}−\mathrm{y}−\mathrm{6}=\mathrm{0}\:,\:\mathrm{intercept} \\ $$$$\mathrm{at}\:\mathrm{point}\left(\mathrm{1},−\mathrm{5}\right).\:\mathrm{Let}\:\mathrm{x}=\mathrm{X}+\mathrm{1}\:,\:\mathrm{y}=\mathrm{Y}−\mathrm{5} \\ $$$$\mathrm{dX}=\mathrm{dx}\:,\:\mathrm{dY}=\mathrm{dy}.\:\mathrm{Let}\:\mathrm{Y}=\mathrm{XZ} \\…
Question Number 96596 by bemath last updated on 03/Jun/20 $${x}\left({x}+\mathrm{1}\right)\:\frac{{dy}}{{dx}}−\left(\mathrm{2}{x}+\mathrm{1}\right){y}\:=\:\mathrm{0} \\ $$ Answered by bobhans last updated on 03/Jun/20 $$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}}.{y}\:\Rightarrow\frac{{dy}}{{y}}\:=\:\frac{\left(\mathrm{2}{x}+\mathrm{1}\right){dx}}{{x}^{\mathrm{2}} +{x}} \\ $$$$\mathrm{ln}\mid\mathrm{y}\mid\:=\:\mathrm{ln}\mid\mathrm{C}\left({x}^{\mathrm{2}} +{x}\right)\mid\:\Rightarrow\:\mathrm{y}=\:\mathrm{C}\left({x}^{\mathrm{2}}…
Question Number 96392 by john santu last updated on 01/Jun/20 $$\mathrm{solve}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\left(\mathrm{4x}+\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} \:,\:\mathrm{when}\: \\ $$$$\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{1} \\ $$ Answered by bobhans last updated on 01/Jun/20 $$\mathrm{set}\:\mathrm{s}\:=\:\mathrm{4x}+\mathrm{y}+\mathrm{1}\:\Rightarrow\:\frac{\mathrm{ds}}{\mathrm{dx}}\:=\:\mathrm{4}\:+\frac{\mathrm{dy}}{\mathrm{dx}} \\…
Question Number 96217 by john santu last updated on 30/May/20 $$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\left(\frac{\mathrm{y}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} +\mathrm{y}}{\mathrm{x}}\right)\: \\ $$ Answered by bobhans last updated on 30/May/20 $$\mathrm{set}\:\mathrm{y}\:=\:{xz}\:\Rightarrow\:\frac{{dy}}{{dx}}\:=\:{z}\:+\:{x}\:\frac{{dz}}{{dx}} \\ $$$${z}\:+\:{x}\:\frac{{dz}}{{dx}}\:=\:\frac{{x}^{\mathrm{2}}…
Question Number 96182 by bemath last updated on 30/May/20 $${x}^{\mathrm{2}} {y}''−{xy}'+\mathrm{y}\:=\:\mathrm{0}\: \\ $$ Answered by john santu last updated on 30/May/20 $$\mathrm{let}\:{y}\:=\:{sx}\:\Rightarrow\mathrm{y}'=\:{s}+{xs}^{'} \\ $$$$\mathrm{y}''\:=\:{xs}''+\mathrm{2}{s}' \\…