Question Number 170795 by ali009 last updated on 31/May/22 $${solve} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}\right){dx}+{x}\left({x}−\mathrm{2}{y}\right){dy}=\mathrm{0} \\ $$ Answered by LEKOUMA last updated on 31/May/22 $${x}^{\mathrm{2}} +{y}^{\mathrm{2}}…
Question Number 105257 by bemath last updated on 27/Jul/20 $${y}''−\mathrm{2}{y}'+{y}\:=\:{xe}^{{x}} \mathrm{sin}\:{x}\: \\ $$ Answered by john santu last updated on 27/Jul/20 $${homogenous}\:{solution} \\ $$$$\lambda^{\mathrm{2}} −\mathrm{2}\lambda+\mathrm{1}\:=\:\mathrm{0}\:\Rightarrow\lambda\:=\:\mathrm{1};\mathrm{1}…
Question Number 105238 by bemath last updated on 27/Jul/20 $$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{3}{e}^{{y}} −\mathrm{2}{x}}\:? \\ $$ Answered by john santu last updated on 27/Jul/20 $$\frac{{dx}}{{dy}}\:=\:\mathrm{3}{e}^{{y}} −\mathrm{2}{x}\: \\ $$$$\frac{{dx}}{{dy}}\:+\:\mathrm{2}{x}\:=\:\mathrm{3}{e}^{{y}}…
Question Number 105163 by bemath last updated on 26/Jul/20 $${x}^{\mathrm{2}} {y}''+{xy}'−{y}=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$ Answered by bramlex last updated on 27/Jul/20 $${we}\:{solve}\:{homogenous}\:{equation} \\ $$$${This}\:{is}\:{an}\:{Euler}−{Cauchy}\: \\…
Question Number 105118 by john santu last updated on 26/Jul/20 $${x}^{\mathrm{2}} {y}''+{xy}'−{y}\:=\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\: \\ $$ Answered by john santu last updated on 26/Jul/20 $$\Rightarrow\:{x}^{\mathrm{2}} {y}''+\mathrm{2}{xy}'−{xy}'\:−\:{y}\:=\:\frac{\mathrm{1}}{{x}+\mathrm{1}} \\…
Question Number 104893 by bramlex last updated on 24/Jul/20 $$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{tan}\:{x}\:\frac{{dy}}{{dx}}\:=\:\mathrm{sec}\:{x}\:+\:\mathrm{cot}\:{x} \\ $$ Answered by mathmax by abdo last updated on 24/Jul/20 $$\mathrm{y}^{''} \:+\mathrm{tanx}\:\mathrm{y}^{'}…
Question Number 170410 by Mastermind last updated on 23/May/22 $${Given}\:{y}={xe}^{−{x}} ,\:{show}\:{that} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{2}\frac{{dy}}{{dx}}+{y}=\mathrm{0} \\ $$$$ \\ $$$${Mastermind} \\ $$ Answered by floor(10²Eta[1]) last…
Question Number 104845 by john santu last updated on 24/Jul/20 $${solve}\:{y}'\:=\:{y}−{x}−\mathrm{1}+\left({x}−{y}+\mathrm{2}\right)^{−\mathrm{1}} \\ $$ Answered by john santu last updated on 24/Jul/20 $$\frac{{dy}}{{dx}}\:=\:−\left({x}−{y}+\mathrm{2}\right)+\mathrm{1}+\left({x}−{y}+\mathrm{2}\right)^{−\mathrm{1}} \\ $$$${set}\:{x}−{y}+\mathrm{2}\:=\:{m}\: \\…
Question Number 170329 by ali009 last updated on 21/May/22 $${find}\:{the}\:{general}\:{solution}\:{of}\: \\ $$$$\left.\mathrm{1}\right){y}'''−\mathrm{7}{y}'+\mathrm{6}{y}={x} \\ $$$$\left.\mathrm{2}\right){y}'''−\mathrm{3}{y}''+\mathrm{2}{y}'=\frac{{e}^{{x}} }{\mathrm{1}+{e}^{−{x}} } \\ $$ Answered by LEKOUMA last updated on 22/May/22…
Question Number 104669 by bramlex last updated on 23/Jul/20 $$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\:{dy}\:=\:{xy}\:{dx}\: \\ $$ Answered by bemath last updated on 23/Jul/20 $${set}\:{y}\:=\:{zx}\: \\ $$$$\frac{{dy}}{{dx}}\:=\:{z}\:+\:{x}\:\frac{{dz}}{{dx}} \\…