Question Number 90641 by Tony Lin last updated on 25/Apr/20 $${Solve}\:{x}^{\mathrm{2}} {y}''+{xy}'+{x}^{\mathrm{2}} {y}=\mathrm{0} \\ $$ Commented by jagoll last updated on 25/Apr/20 $${Euler}\:−\:{Cauchy}? \\ $$…
Question Number 25094 by ifcrna380w last updated on 03/Dec/17 Answered by mrW1 last updated on 04/Dec/17 $${e}^{−\mathrm{2}{y}} {dy}={e}^{\mathrm{3}{x}} {dx} \\ $$$$\int{e}^{−\mathrm{2}{y}} {dy}=\int{e}^{\mathrm{3}{x}} {dx} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\int{e}^{−\mathrm{2}{y}}…
Question Number 155965 by john_santu last updated on 06/Oct/21 $$\left({x}\:\mathrm{sin}\:\frac{{y}}{{x}}−{y}\:\mathrm{cos}\:\frac{{y}}{{x}}\right){dx}+{x}\:\mathrm{cos}\:\frac{{y}}{{x}}\:{dy}=\mathrm{0} \\ $$ Answered by mindispower last updated on 07/Oct/21 $$\left({tg}\left(\frac{{y}}{{x}}\right)−\frac{{y}}{{x}}\right){dx}+{dy}=\mathrm{0} \\ $$$$\frac{{y}}{{x}}={u} \\ $$$${dy}={xdu}+{udx} \\…
Question Number 90394 by jagoll last updated on 23/Apr/20 $$\left(\mathrm{3x}^{\mathrm{2}} +\mathrm{9xy}+\mathrm{5y}^{\mathrm{2}} \right)\mathrm{dx}\:=\:\left(\mathrm{6x}^{\mathrm{2}} +\mathrm{4xy}\right)\mathrm{dy} \\ $$ Commented by john santu last updated on 23/Apr/20 $$\left(\mathrm{3}+\mathrm{9}\left(\frac{{y}}{{x}}\right)+\mathrm{5}\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} \right)\:=\:\left(\mathrm{6}+\mathrm{4}\left(\frac{{y}}{{x}}\right)\right)\:{dy}…
Question Number 90262 by jagoll last updated on 22/Apr/20 $$\left(\mathrm{y}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\right)\:\mathrm{dx}\:=\:\mathrm{x}\:\mathrm{dy}\: \\ $$ Commented by john santu last updated on 22/Apr/20 $$\left(\frac{{y}}{{x}}+\sqrt{\mathrm{1}+\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} }\right)\:{dx}\:=\:{dy}\: \\…
Question Number 155631 by aunzo last updated on 03/Oct/21 $$\left({x}+\mathrm{3}\left({x}−\mathrm{2}\right)={x}+\mathrm{10}\right. \\ $$ Commented by aunzo last updated on 03/Oct/21 $${plss} \\ $$ Commented by Ar…
Question Number 90090 by john santu last updated on 21/Apr/20 $${xy}\:\frac{{dy}}{{dx}}\:=\:{y}^{\mathrm{2}} \:+\:\left(\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$ Answered by john santu last updated on 21/Apr/20 Terms…
Question Number 89978 by jagoll last updated on 20/Apr/20 $$\left(\mathrm{x}\:\frac{\mathrm{dy}}{\mathrm{dx}}−\mathrm{y}\right)\left(\mathrm{cos}\:\left(\frac{\mathrm{2y}}{\mathrm{x}}\right)\right)\:=\:−\mathrm{3x}^{\mathrm{4}} \\ $$ Commented by john santu last updated on 20/Apr/20 $$\left[\:{let}\:{y}\:=\:{ux}\:\right]\:\Rightarrow\frac{{dy}}{{dx}}\:=\:{u}\:+\:{x}\frac{{du}}{{dx}} \\ $$$$\left\{{ux}\:+{x}^{\mathrm{2}} \:\frac{{du}}{{dx}}\:−{ux}\:\right\}\left(\mathrm{cos}\:\mathrm{2}{u}\right)\:=\:−\mathrm{3}{x}^{\mathrm{4}} \\…
Question Number 89973 by john santu last updated on 20/Apr/20 $${x}\:\frac{{dy}}{{dx}}\:−{y}\:=\:{x}^{\mathrm{2}} \:\mathrm{tan}\:\left(\frac{{y}}{{x}}\right)\: \\ $$ Commented by john santu last updated on 20/Apr/20 $$\left[\:{let}\:\frac{{y}}{{x}}=\:{v}\:\Rightarrow{y}\:=\:{vx}\:\right]\: \\ $$$$\frac{{dy}}{{dx}}\:=\:{v}\:+\:{x}\:\frac{{dv}}{{dx}}…
Question Number 89970 by jagoll last updated on 20/Apr/20 $$\mathrm{xy}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{y}^{\mathrm{2}} \left(\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)\: \\ $$ Commented by john santu last updated on 20/Apr/20 Terms of…