Question Number 17477 by Sai dadon. last updated on 07/Jul/17 $${Given}\: \\ $$$$\left(\mathrm{4}{xy}/{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dy}/{dx}=\mathrm{1}, \\ $$$${y}=\mathrm{0},\:{x}=\mathrm{0} \\ $$$${show}\:{that}\:\sqrt{{x}}\left({x}^{\mathrm{2}} −\mathrm{5}{y}^{\mathrm{2}} \right)=\mathrm{1} \\ $$$$ \\ $$$$…
Question Number 148466 by Ar Brandon last updated on 28/Jul/21 $$\mathrm{xdx}+\mathrm{ydy}=\mathrm{xdy}−\mathrm{ydx} \\ $$ Answered by bramlexs22 last updated on 28/Jul/21 $$\left({x}+{y}\right){dx}=\left({x}−{y}\right){dy} \\ $$$$\:\frac{{dy}}{{dx}}=\:\frac{{x}+{y}}{{x}−{y}} \\ $$$$\:{let}\:{y}={ux}\:\Rightarrow\frac{{dy}}{{dx}}\:=\:{u}\:+\:{x}\:\frac{{du}}{{dx}}…
Question Number 82883 by jagoll last updated on 26/Feb/20 $$\left[\frac{\mathrm{e}^{−\mathrm{2}\sqrt{\mathrm{x}}} }{\:\sqrt{\mathrm{x}}}−\frac{\mathrm{y}}{\:\sqrt{\mathrm{x}}}\:\right]\:.\frac{\mathrm{dx}}{\mathrm{dy}}\:=\:\mathrm{1}\:,\:\mathrm{x}\:\neq\:\mathrm{0} \\ $$ Answered by john santu last updated on 26/Feb/20 $$\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{e}^{−\mathrm{2}\sqrt{\mathrm{x}}} }{\:\sqrt{\mathrm{x}}}\:−\:\frac{\mathrm{y}}{\:\sqrt{\mathrm{x}}} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\right)\:\mathrm{y}\:=\:\frac{\mathrm{e}^{−\mathrm{2}\sqrt{\mathrm{x}}}…
Question Number 17323 by tawa tawa last updated on 04/Jul/17 $$\mathrm{Solve}:\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{2cos}\left(\mathrm{2x}\right)}{\mathrm{3}\:−\:\mathrm{2y}}\:\:\:\:\:\:\:\:\:\:\:\mathrm{with}\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)\:=\:−\mathrm{1} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{for}\:\mathrm{what}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:>\:\mathrm{0}\:\mathrm{does}\:\mathrm{the}\:\mathrm{situation}\:\mathrm{exist} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{for}\:\mathrm{what}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{is}\:\mathrm{y}\left(\mathrm{x}\right)\:\mathrm{maximum} \\ $$ Answered by Tinkutara last updated on 04/Jul/17 $$\left(\mathrm{3}\:−\:\mathrm{2}{y}\right)\:{dy}\:=\:\mathrm{2}\:\mathrm{cos}\:\mathrm{2}{x}\:{dx}…
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Question Number 17137 by tawa tawa last updated on 01/Jul/17 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}\: \\ $$$$\mathrm{2x}\left[\mathrm{ye}^{\mathrm{x}} \:−\:\mathrm{1}\right]\mathrm{dx}\:+\:\mathrm{e}^{\mathrm{y}} \:\mathrm{dy}\:=\:\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 16919 by rustylee last updated on 28/Jun/17 $${ax}\mathrm{2}+{bx}+{c}=\mathrm{0} \\ $$ Answered by RasheedSoomro last updated on 28/Jun/17 $${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\Rightarrow \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\…
Question Number 16918 by rustylee last updated on 28/Jun/17 $$\mathrm{2}{x}\mathrm{2}+\mathrm{9}{x}=\mathrm{10} \\ $$ Answered by Joel577 last updated on 28/Jun/17 $$\mathrm{2}{x}^{\mathrm{2}} \:+\:\mathrm{9}{x}\:−\:\mathrm{10}\:=\:\mathrm{0} \\ $$$${x}_{\mathrm{1},\mathrm{2}} \:=\:\frac{−\mathrm{9}\:\pm\:\sqrt{\mathrm{81}\:+\:\mathrm{80}}}{\mathrm{4}} \\…
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