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)\: \\ $$ Commented by john santu last updated on 20/Apr/20 Terms of…
Question Number 89925 by john santu last updated on 20/Apr/20 $${x}^{\mathrm{2}} \left({yy}''−{y}^{\mathrm{2}} \right)+{xyy}'\:=\:{y}\sqrt{{x}^{\mathrm{2}} \left({y}'\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 89908 by I want to learn more last updated on 19/Apr/20 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equstion}: \\ $$$$\:\:\:\:\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:\:\:=\:\:\:\frac{\mathrm{y}_{\mathrm{0}} \:\:−\:\:\mathrm{2y}_{−\mathrm{1}} \:\:+\:\:\mathrm{y}_{−\mathrm{2}} }{\mathrm{h}^{\mathrm{2}} } \\ $$ Commented…
Question Number 89835 by john santu last updated on 19/Apr/20 $${x}^{\mathrm{2}} \:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{4}{x}\:\frac{{dy}}{{dx}}\:+\:\mathrm{2}{y}\:=\:{e}^{{x}} \\ $$ Answered by Joel578 last updated on 19/Apr/20 $${Euler}−{Cauchy}\:{eq}.\:\Rightarrow\:{x}^{\mathrm{2}} {y}''\:+\:{axy}'\:+\:{by}\:=\:\mathrm{0}…
Question Number 89809 by jagoll last updated on 19/Apr/20 $$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{xy}−\mathrm{x}^{\mathrm{2}} } \\ $$ Answered by john santu last updated on 19/Apr/20 Answered by mr…