Question Number 83327 by john santu last updated on 01/Mar/20 $$\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:−\mathrm{2x}\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{p}\left(\mathrm{p}+\mathrm{1}\right)\mathrm{y}\:=\:\mathrm{0}\: \\ $$$$\mathrm{in}\:\mathrm{descending}\:\mathrm{power}\:\mathrm{of}\:\mathrm{x}.\:\mathrm{what}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{solution}? \\ $$ Commented by Joel578 last updated…
Question Number 17735 by tawa tawa last updated on 09/Jul/17 $$\mathrm{Solve}:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{ysec}\left(\mathrm{x}\right)\:=\:\mathrm{tan}\left(\mathrm{x}\right) \\ $$ Answered by alex041103 last updated on 10/Jul/17 $$\mathrm{So}\:\mathrm{let}'\mathrm{s}\:\mathrm{find}\:\mathrm{function}\:\mu\left(\mathrm{x}\right)\:\mathrm{wich}\:\mathrm{satisfies}\:\mathrm{the}\:\mathrm{followinv} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\mu+\mathrm{P}\left(\mathrm{x}\right)\mu\mathrm{y}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{y}\mu\right] \\ $$$$\mathrm{We}\:\mathrm{know}\:\mathrm{that}…
Question Number 17734 by tawa tawa last updated on 09/Jul/17 $$\mathrm{If}\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{3}} \:−\:\mathrm{3xy}\:=\:\mathrm{0}, \\ $$$$\mathrm{Show}\:\mathrm{that},\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=\:\frac{−\:\mathrm{2xy}}{\mathrm{y}^{\mathrm{2}} \:−\:\mathrm{x}^{\mathrm{2}} } \\ $$ Terms of Service Privacy…
Question Number 83159 by jagoll last updated on 28/Feb/20 $$\mathrm{3x}\:\left(\mathrm{xy}−\mathrm{2}\right)\mathrm{dx}\:+\:\left(\mathrm{x}^{\mathrm{3}} +\mathrm{2y}\right)\:\mathrm{dy}\:=\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{solution} \\ $$ Commented by niroj last updated on 28/Feb/20 $$\:\:\:\left(\mathrm{3x}^{\mathrm{2}} \mathrm{y}−\mathrm{6x}\right)\mathrm{dx}+\left(\mathrm{x}^{\mathrm{3}} +\mathrm{2y}\right)\mathrm{dy}=\mathrm{0}…
Question Number 17477 by Sai dadon. last updated on 07/Jul/17 $${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