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}} } \\ $$ Answered by john santu last updated on 19/Apr/20 Answered by mr…
Question Number 89776 by john santu last updated on 19/Apr/20 $${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}^{'}…