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Question Number 51840 by 33 last updated on 31/Dec/18 $${solve} \\ $$$$\frac{{dy}}{{dx}\:}\:+\:\frac{\mathrm{2}{y}}{\mathrm{3}{x}\:}\:=\:\frac{{x}}{\:\sqrt{{y}}} \\ $$ Answered by ajfour last updated on 31/Dec/18 $$\mathrm{3}{x}\sqrt{{y}}{dy}+\mathrm{2}{y}\sqrt{{y}}{dx}\:=\:\mathrm{3}{x}^{\mathrm{2}} {dx} \\ $$$${d}\left({xy}\sqrt{{y}}\right)=\:\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}}…
Question Number 51709 by sinx last updated on 29/Dec/18 $${y}''=\mathrm{2}{y}\left({y}'\right) \\ $$$$ \\ $$$${solve}\:{the}\:{equation} \\ $$ Answered by mr W last updated on 29/Dec/18 $${y}''=\frac{{d}\left({y}'\right)}{{dx}}=\frac{{d}\left({y}'\right)}{{dy}}×\frac{{dy}}{{dx}}={y}'\frac{{d}\left({y}'\right)}{{dy}}…
Question Number 51710 by sinx last updated on 29/Dec/18 $${solve}\:{the}\:{equation} \\ $$$${y}\:=\:{xy}'+\:\left({y}'\right)^{\mathrm{2}} \\ $$ Answered by Abdulhafeez Abu qatada last updated on 30/Dec/18 $$ \\…
Question Number 117228 by bemath last updated on 10/Oct/20 $$\left(\frac{{x}+{y}}{{y}−\mathrm{1}}\right)\:{dx}\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}+\mathrm{1}}{{y}−\mathrm{1}}\right)^{\mathrm{2}} {dy}\:=\:\mathrm{0} \\ $$ Answered by TANMAY PANACEA last updated on 10/Oct/20 $$\left(\frac{{x}+{y}}{{y}−\mathrm{1}}\right){dx}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}+{y}}{{y}−\mathrm{1}}−\mathrm{1}\right)^{\mathrm{2}} {dy}=\mathrm{0} \\ $$$$\left(\frac{{x}+{y}}{{y}−\mathrm{1}}\right){dx}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}+{y}}{{y}−\mathrm{1}}\right)^{\mathrm{2}}…
Question Number 51494 by Tawa1 last updated on 27/Dec/18 $$\mathrm{Solve}:\:\:\:\:\:\:\:\:\left(\mathrm{t}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:\frac{\mathrm{dp}}{\mathrm{dt}}\:\:=\:\:\mathrm{p}^{\mathrm{t}} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 27/Dec/18 $$\left({t}^{\mathrm{2}} +\mathrm{1}\right)\frac{{d}^{\mathrm{2}} {p}}{{dt}^{\mathrm{2}} }+\frac{{dp}}{{dt}}×\mathrm{2}{t}=\frac{{d}}{{dt}}\left({p}^{{t}} \right)…
Question Number 117029 by bemath last updated on 09/Oct/20 $$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2y}^{\mathrm{2}} \right)\:\mathrm{dx}\:+\:\left(\mathrm{4xy}−\mathrm{y}^{\mathrm{2}} \right)\:\mathrm{dy}\:=\:\mathrm{0} \\ $$ Answered by bemath last updated on 09/Oct/20 $$\mathrm{letting}\:\mathrm{y}\:=\:\varphi\mathrm{x}\:\Rightarrow\mathrm{dy}\:=\:\varphi\:\mathrm{dx}+\mathrm{x}\:\mathrm{d}\varphi \\ $$$$\Rightarrow\left(\mathrm{x}^{\mathrm{2}}…
Question Number 116614 by bemath last updated on 05/Oct/20 $$\:\left(\mathrm{x}^{\mathrm{2}} \:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{y}}\right)−\mathrm{y}\right)\mathrm{dx}\:+\:\mathrm{x}\:\mathrm{dy}\:=\:\mathrm{0} \\ $$$$\mathrm{where}\:\mathrm{y}\left(\mathrm{1}\right)=\:\frac{\pi}{\mathrm{4}} \\ $$ Commented by TANMAY PANACEA last updated on 05/Oct/20 $${i}\:{think}\:{it}\:{should}\:{be}\:{cos}^{\mathrm{2}}…
Question Number 182120 by amin96 last updated on 04/Dec/22 Answered by SEKRET last updated on 04/Dec/22 $$\:\:\:\sqrt[{\mathrm{4}}]{\mathrm{5}+\frac{\mathrm{3}}{\:\sqrt[{\mathrm{6}}]{\mathrm{13}+\frac{\mathrm{8}}{….}}}\:}\:\:\:\:=\:\:\boldsymbol{\mathrm{a}} \\ $$$$\:\:\:\:\:\:\mathrm{5}\:+\frac{\mathrm{3}}{\:\sqrt[{\mathrm{6}}]{\mathrm{13}+\frac{\mathrm{8}}{\boldsymbol{\mathrm{a}}}}}\:\:=\:\boldsymbol{\mathrm{a}}^{\mathrm{4}} \\ $$$$\:\:\:\left(\mathrm{13}+\frac{\mathrm{8}}{\boldsymbol{\mathrm{a}}}\right)=\:\frac{\mathrm{3}^{\mathrm{6}} }{\left(\boldsymbol{\mathrm{a}}^{\mathrm{4}} −\mathrm{1}\right)^{\mathrm{6}} } \\…
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