Question Number 89776 by john santu last updated on 19/Apr/20 $${what}\:{is}\:{the}\:{stability}\:{of}\:{the}\: \\ $$$${following}\:{function}\: \\ $$$$\frac{{dy}}{{dt}}\:=\:{y}^{\mathrm{3}} \:−\mathrm{2}{y}^{\mathrm{2}} \:+\:{y}\: \\ $$ Commented by jagoll last updated on…
Question Number 24233 by Nayon.Sm last updated on 14/Nov/17 $${if}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }={ksiny}\:{then}\:{y}=? \\ $$ Answered by ajfour last updated on 15/Nov/17 $${let}\:{y}\:\rightarrow{x}\:\:\:{and}\:{x}\rightarrow{t} \\ $$$$\Rightarrow\:{then}\:\:\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}}…
Question Number 89748 by jagoll last updated on 19/Apr/20 $$\mathrm{dx}\:=\:\left(\mathrm{1}+\mathrm{2xtan}\:\mathrm{y}\right)\:\mathrm{dy}\: \\ $$ Commented by mr W last updated on 19/Apr/20 $$\frac{{dx}}{{dy}}−\left(\mathrm{2}\:\mathrm{tan}\:{y}\right)\:{x}=\mathrm{1} \\ $$$$−\int\mathrm{2}\:\mathrm{tan}\:{y}\:{dy}=\mathrm{2}\int\frac{{d}\:\left(\mathrm{cos}\:{y}\right)}{\mathrm{cos}\:{y}}=\mathrm{2ln}\:\left(\mathrm{cos}\:{y}\right)=\mathrm{ln}\:\mathrm{cos}^{\mathrm{2}} \:{y} \\…
Question Number 24151 by Nayon.Sm last updated on 13/Nov/17 $${y}={f}\left({t}\right)\:{and}\:{y}''={ksiny}\:{y}=? \\ $$ Commented by Nayon.Sm last updated on 14/Nov/17 $${mrw}\mathrm{1}\:{ans}\overset{} {\:} \\ $$ Terms of…
Question Number 89632 by jagoll last updated on 18/Apr/20 $$\mathrm{y}\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:\mathrm{dx}\:+\:\mathrm{x}\sqrt{\mathrm{y}^{\mathrm{2}} −\mathrm{1}}\:\mathrm{dy}\:=\mathrm{0} \\ $$ Answered by john santu last updated on 18/Apr/20 Terms of Service…
Question Number 89624 by M±th+et£s last updated on 18/Apr/20 $$\left.{Q}\mathrm{1}\right){find}\:{tow}\:{power}\:{series}\:{solutions}\:{of}\:{the}\: \\ $$$${given}\:{D}.{E}\:{about}\:{x}=\mathrm{0} \\ $$$${y}^{''} −\mathrm{2}{xy}^{'} +{y}=\mathrm{0} \\ $$$$ \\ $$$$\left.{Q}\mathrm{2}\right){use}\:{the}\:{power}\:{series}\:{method}\:\:{to}\:{solve}\:{the} \\ $$$${given}\:{intial}\:{value}\:{problem} \\ $$$${y}^{''} −\mathrm{2}{xy}^{'}…
Question Number 89600 by jagoll last updated on 18/Apr/20 $$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{y}+\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}{\mathrm{x}}\: \\ $$ Commented by mr W last updated on 18/Apr/20 $$\frac{{dy}}{{dx}}=\frac{{y}}{{x}}+\sqrt{\mathrm{1}−\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} } \\…
Question Number 89243 by M±th+et£s last updated on 16/Apr/20 $${solve}\:{the}\:{following}\:{diffirntial}\:{equation} \\ $$$$\left.\mathrm{1}\right)\left(\mathrm{2}{x}+{y}\right){dx}+\left({x}+{y}\right){dy}=\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\left(\mathrm{3}{x}−{y}\right){dx}−\left({x}−{y}\right){dy}=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:\left({cos}\left({x}\right)+{y}\right){dx}\:+\:\left(\mathrm{2}{y}+{x}\right){dy}=\mathrm{0} \\ $$ Answered by TANMAY PANACEA. last updated on…
Question Number 89205 by jagoll last updated on 16/Apr/20 $$\mathrm{2}\:\frac{{dy}}{{dx}}\:=\:\frac{{y}}{{x}}\:+\:\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} \\ $$ Commented by john santu last updated on 16/Apr/20 $${y}\:=\:{vx}\:\Rightarrow\frac{{dy}}{{dx}}\:=\:{v}\:+\:{x}\:\frac{{dv}}{{dx}} \\ $$$$\mathrm{2}{v}\:+\mathrm{2}{x}\:\frac{{dv}}{{dx}}\:=\:{v}+{v}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}\:\frac{{dv}}{{dx}}\:=\:{v}^{\mathrm{2}}…
Question Number 89178 by necxxx last updated on 15/Apr/20 $${If}\:{z}\left({z}^{\mathrm{2}} +\mathrm{3}{x}\right)+\mathrm{3}{y}=\mathrm{0}\:{prove}\:{that}\: \\ $$$$\frac{\partial^{\mathrm{2}} {z}}{\partial{x}^{\mathrm{2}} }\:+\:\frac{\partial^{\mathrm{2}} {z}}{\partial{y}^{\mathrm{2}} }=\:\frac{\mathrm{2}{z}\left({x}−\mathrm{1}\right)}{\left({z}^{\mathrm{2}} +{x}\right)^{\mathrm{3}} } \\ $$$$ \\ $$$$ \\ $$$${please}\:{help}.…