Question Number 91211 by john santu last updated on 28/Apr/20 $${x}\:{dy}\:+\mathrm{5}{y}\:{dx}\:=\:\mathrm{2}{y}^{\mathrm{4}} {x}\:{dx} \\ $$ Commented by john santu last updated on 29/Apr/20 Answered by MWSuSon…
Question Number 91147 by john santu last updated on 28/Apr/20 $$\left({D}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} {y}\:=\:{x}^{\mathrm{2}} \mathrm{cos}\:{x}\: \\ $$ Commented by MWSuSon last updated on 28/Apr/20 I've corrected the mistake. Answered…
Question Number 91139 by john santu last updated on 28/Apr/20 $$\frac{{dy}}{{dx}}\:=\:\frac{{y}−{x}+\mathrm{1}}{{y}−{x}+\mathrm{5}} \\ $$ Commented by john santu last updated on 28/Apr/20 $$\frac{{dy}}{{dx}}\:=\:\frac{{y}−{x}+\mathrm{5}−\mathrm{4}}{{y}−{x}+\mathrm{5}}\:=\:\mathrm{1}−\frac{\mathrm{4}}{{y}−{x}+\mathrm{5}} \\ $$$$\left[\:{let}\:{Q}={y}−{x}\:\right]\: \\…
Question Number 91024 by john santu last updated on 27/Apr/20 $${x}^{\mathrm{2}} {y}''+\mathrm{3}{xy}'\:+\mathrm{2}{y}\:=\:\mathrm{4}{x}^{\mathrm{2}} \\ $$ Commented by jagoll last updated on 27/Apr/20 $${i}\:{can}\:{try} \\ $$$${let}\:{x}\:=\:{e}^{{z}} \\…
Question Number 91012 by jagoll last updated on 27/Apr/20 $$\frac{{dy}}{{dx}}\:+\:\mathrm{2}{xy}\:=\:{xe}^{−{x}^{\mathrm{2}} } {y}^{\mathrm{3}} \\ $$ Answered by john santu last updated on 27/Apr/20 Terms of Service…
Question Number 91010 by jagoll last updated on 27/Apr/20 $${solve}\:{the}\:{diff}\:{eq}\: \\ $$$${y}'''−{y}''+\mathrm{4}{y}'−\mathrm{4}{y}=\:{e}^{{x}} \\ $$ Commented by john santu last updated on 27/Apr/20 $${the}\:{characteristic}\:{equation}\: \\ $$$${w}^{\mathrm{3}}…
Question Number 90983 by jagoll last updated on 27/Apr/20 $${y}''−\mathrm{5}{y}'+\mathrm{6}{y}={x}^{\mathrm{2}} \\ $$ Commented by john santu last updated on 27/Apr/20 $$\mathrm{complementari}\:\mathrm{solution} \\ $$$$\mathrm{y}\:=\:\mathrm{Ae}^{\mathrm{q}_{\mathrm{1}} \mathrm{x}} \:+\:\mathrm{Be}^{\mathrm{q}_{\mathrm{2}}…
Question Number 90952 by jagoll last updated on 27/Apr/20 $$ \\ $$$${solve}\:{diff}\:{equation}\: \\ $$$${x}^{\mathrm{2}} {y}''\:+\:{xy}'\:+{y}\:=\:\mathrm{5}{x}^{\mathrm{2}} \:\:{x}\:>\mathrm{0} \\ $$ Answered by MWSuSon last updated on 27/Apr/20…
Question Number 90955 by jagoll last updated on 27/Apr/20 $${y}''−\mathrm{5}{y}'−\mathrm{24}{y}={e}^{\mathrm{3}{x}} \\ $$$$ \\ $$ Commented by jagoll last updated on 27/Apr/20 $${i}\:{got}\: \\ $$$${y}_{{h}} \:=\:{Ae}^{\mathrm{8}{x}}…
Question Number 90886 by jagoll last updated on 26/Apr/20 $$\frac{{dy}}{{dx}}\:−\frac{\mathrm{4}{y}}{{x}}\:=\:\mathrm{1}+\frac{\mathrm{2}}{{x}} \\ $$ Answered by john santu last updated on 26/Apr/20 Commented by john santu last…