Question Number 117728 by syamil last updated on 13/Oct/20 $${Solution}\:\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{3}\frac{{dy}}{{dx}}\:−\:\mathrm{4}{y}\:=\:{x}^{\mathrm{2}} \\ $$ Answered by TANMAY PANACEA last updated on 13/Oct/20 $${y}={e}^{{mx}} \\ $$$$\left({m}^{\mathrm{2}}…
Question Number 117632 by syamil last updated on 12/Oct/20 $${Solution}\:{from}\:\:\:\mathrm{2}{xy}\:{dy}\:=\:\left({x}^{\mathrm{2}\:} \:−\:{y}^{\mathrm{2}} \right){dx} \\ $$ Answered by 1549442205PVT last updated on 13/Oct/20 $$\:\mathrm{2}{xy}\:{dy}\:=\:\left({x}^{\mathrm{2}\:} \:−\:{y}^{\mathrm{2}} \right){dx} \\…
Question Number 117608 by TANMAY PANACEA last updated on 12/Oct/20 $${solve} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }+{w}^{\mathrm{2}} {x}=\mathrm{0} \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 117602 by TANMAY PANACEA last updated on 12/Oct/20 $$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{a}^{\mathrm{2}} {y}={cosax} \\ $$ Answered by TANMAY PANACEA last updated on 12/Oct/20 $${y}={e}^{{mx}}…
Question Number 117585 by syamil last updated on 12/Oct/20 $${solution}\:\:\:\:\frac{{dy}}{{dx}}\:=\:{sin}\:{x}\:+\:{e}^{\mathrm{2}{x}} \:+\:{x}^{\mathrm{2}} \\ $$ Commented by TANMAY PANACEA last updated on 12/Oct/20 $${i}\:{can}\:{not}\:{post}\:{question} \\ $$$${so}\:{here}\:{i}\:{am}\:{posting}?{question} \\…
Question Number 182936 by CHAGGY last updated on 17/Dec/22 $$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 51840 by 33 last updated on 31/Dec/18 $${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}}…