Question Number 96610 by bemath last updated on 03/Jun/20 $$\mathrm{solve}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{x}+\mathrm{y}+\mathrm{4}}{\mathrm{x}−\mathrm{y}−\mathrm{6}} \\ $$ Commented by bobhans last updated on 03/Jun/20 $$\mathrm{set}\:\mathrm{x}+\mathrm{y}+\mathrm{4}\:=\:\mathrm{0}\:\&\:\mathrm{x}−\mathrm{y}−\mathrm{6}=\mathrm{0}\:,\:\mathrm{intercept} \\ $$$$\mathrm{at}\:\mathrm{point}\left(\mathrm{1},−\mathrm{5}\right).\:\mathrm{Let}\:\mathrm{x}=\mathrm{X}+\mathrm{1}\:,\:\mathrm{y}=\mathrm{Y}−\mathrm{5} \\ $$$$\mathrm{dX}=\mathrm{dx}\:,\:\mathrm{dY}=\mathrm{dy}.\:\mathrm{Let}\:\mathrm{Y}=\mathrm{XZ} \\…
Question Number 96596 by bemath last updated on 03/Jun/20 $${x}\left({x}+\mathrm{1}\right)\:\frac{{dy}}{{dx}}−\left(\mathrm{2}{x}+\mathrm{1}\right){y}\:=\:\mathrm{0} \\ $$ Answered by bobhans last updated on 03/Jun/20 $$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}}.{y}\:\Rightarrow\frac{{dy}}{{y}}\:=\:\frac{\left(\mathrm{2}{x}+\mathrm{1}\right){dx}}{{x}^{\mathrm{2}} +{x}} \\ $$$$\mathrm{ln}\mid\mathrm{y}\mid\:=\:\mathrm{ln}\mid\mathrm{C}\left({x}^{\mathrm{2}} +{x}\right)\mid\:\Rightarrow\:\mathrm{y}=\:\mathrm{C}\left({x}^{\mathrm{2}}…
Question Number 96392 by john santu last updated on 01/Jun/20 $$\mathrm{solve}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\left(\mathrm{4x}+\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} \:,\:\mathrm{when}\: \\ $$$$\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{1} \\ $$ Answered by bobhans last updated on 01/Jun/20 $$\mathrm{set}\:\mathrm{s}\:=\:\mathrm{4x}+\mathrm{y}+\mathrm{1}\:\Rightarrow\:\frac{\mathrm{ds}}{\mathrm{dx}}\:=\:\mathrm{4}\:+\frac{\mathrm{dy}}{\mathrm{dx}} \\…
Question Number 96217 by john santu last updated on 30/May/20 $$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\left(\frac{\mathrm{y}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} +\mathrm{y}}{\mathrm{x}}\right)\: \\ $$ Answered by bobhans last updated on 30/May/20 $$\mathrm{set}\:\mathrm{y}\:=\:{xz}\:\Rightarrow\:\frac{{dy}}{{dx}}\:=\:{z}\:+\:{x}\:\frac{{dz}}{{dx}} \\ $$$${z}\:+\:{x}\:\frac{{dz}}{{dx}}\:=\:\frac{{x}^{\mathrm{2}}…
Question Number 96182 by bemath last updated on 30/May/20 $${x}^{\mathrm{2}} {y}''−{xy}'+\mathrm{y}\:=\:\mathrm{0}\: \\ $$ Answered by john santu last updated on 30/May/20 $$\mathrm{let}\:{y}\:=\:{sx}\:\Rightarrow\mathrm{y}'=\:{s}+{xs}^{'} \\ $$$$\mathrm{y}''\:=\:{xs}''+\mathrm{2}{s}' \\…
Question Number 96125 by john santu last updated on 30/May/20 $$\mathrm{4x}^{\mathrm{2}} \mathrm{y}''\:+\mathrm{12xy}'\:+\:\mathrm{3y}\:=\:\mathrm{0} \\ $$ Answered by john santu last updated on 30/May/20 $$\mathrm{cauchy}\:−\:\mathrm{euler}\: \\ $$$$\mathrm{4r}^{\mathrm{2}}…
Question Number 96117 by bobhans last updated on 30/May/20 $$\left(\mathrm{1}−\mathrm{2}{xy}\right)\:{dx}\:+\:\left(\mathrm{4}{y}^{\mathrm{3}} −{x}^{\mathrm{2}} \right)\:{dy}\:=\:\mathrm{0}\: \\ $$ Answered by john santu last updated on 30/May/20 Commented by john…
Question Number 96065 by john santu last updated on 29/May/20 $$\mathrm{y}'\:+\:\mathrm{y}\:=\:\mathrm{x}\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{y}^{\mathrm{2}} } \\ $$ Answered by bobhans last updated on 29/May/20 $$\mathrm{Bernoulli}\:\mathrm{eq}\: \\ $$$$\mathrm{v}\:=\:\mathrm{y}^{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}} \:=\:\mathrm{y}^{\frac{\mathrm{1}}{\mathrm{3}}}…
Question Number 96051 by i jagooll last updated on 29/May/20 $$\left(\mathrm{x}−\mathrm{y}\right)\:\mathrm{dx}\:+\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\:\mathrm{dy}\:=\:\mathrm{0}\: \\ $$ Commented by bobhans last updated on 29/May/20 $$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{y}−\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:.\:\mathrm{set}\:\mathrm{y}\:=\:\mathrm{vx}\:…
Question Number 95933 by john santu last updated on 28/May/20 $$\mathrm{y}'''+\mathrm{2y}'−\mathrm{3y}=\:\mathrm{e}^{\mathrm{x}} \:\left(\mathrm{x}+\mathrm{3}\right)\: \\ $$ Commented by Sourav mridha last updated on 29/May/20 $$\mathrm{are}\:\mathrm{you}\:\mathrm{sure}\:\mathrm{it}\:\mathrm{is}\:\mathrm{3rd}\:\mathrm{order}\:\mathrm{ODE},\mathrm{i}\:\mathrm{think} \\ $$$$\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{2nd}\:\mathrm{order}.…