Question Number 97868 by bemath last updated on 10/Jun/20 $$\mathrm{3y}'\:=\:\mathrm{2x}+\mathrm{y}−\mathrm{1}\: \\ $$ Answered by bobhans last updated on 10/Jun/20 Commented by bemath last updated on…
Question Number 97866 by me2love2math last updated on 10/Jun/20 Commented by me2love2math last updated on 10/Jun/20 $$\mathrm{1}\:\mathrm{2}\:\mathrm{5}\:{and}\:\mathrm{6} \\ $$ Commented by bobhans last updated on…
Question Number 97847 by bemath last updated on 10/Jun/20 $$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}\right)\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:+\:\left(\mathrm{1}−\mathrm{2x}^{\mathrm{2}} \right)\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{1}\right)\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{1} \\ $$ Answered by niroj last updated on 10/Jun/20…
Question Number 97751 by bobhans last updated on 09/Jun/20 $$\left(\mathrm{y}^{\mathrm{2}} −\mathrm{xy}\right)\mathrm{dx}\:+\:\mathrm{2xy}\:\mathrm{dy}\:=\:\mathrm{0}\: \\ $$ Commented by bemath last updated on 09/Jun/20 $$\Leftrightarrow\:\mathrm{2xy}\:\mathrm{dy}\:=\:\left(\mathrm{xy}−\mathrm{y}^{\mathrm{2}} \right)\:\mathrm{dx}\: \\ $$$$\mathrm{set}\:\mathrm{y}\:=\:\mathrm{zx}\:\Leftrightarrow\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{z}\:+\:\mathrm{x}\:\frac{\mathrm{dz}}{\mathrm{dx}} \\…
Question Number 163113 by Lordose last updated on 03/Jan/22 $$\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:+\:\mathrm{ytan}\left(\mathrm{x}\right)\:=\:\mathrm{e}^{\mathrm{x}} \:\:\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{1},\:\mathrm{y}'\left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 97539 by bobhans last updated on 08/Jun/20 $${x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} +\mathrm{2}{xy}\:\frac{{dy}}{{dx}}\:=\:\mathrm{0} \\ $$ Answered by bemath last updated on 08/Jun/20 Commented by bobhans last…
Question Number 97266 by bobhans last updated on 07/Jun/20 $$\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}}…