Question Number 119235 by bemath last updated on 23/Oct/20 $$\:\left({D}^{\mathrm{3}} +{D}^{\mathrm{2}} −\mathrm{4}{D}−\mathrm{4}\right){y}\:=\:{e}^{\mathrm{4}{x}} \\ $$ Answered by benjo_mathlover last updated on 23/Oct/20 $$\left({D}+{I}\right)\left({D}−\mathrm{2}{I}\right)\left({D}+\mathrm{2}{I}\right)\left({y}\right)\:=\:{e}^{\mathrm{4}{x}} \\ $$$${let}\:\left({D}−\mathrm{2}{I}\right)\left({D}+\mathrm{2}{I}\right)\left({y}\right)\:=\:{z} \\…
Question Number 119233 by bemath last updated on 23/Oct/20 $$\:\left({D}^{\mathrm{2}} −\mathrm{5}{D}+\mathrm{6}\right){y}\:=\:{e}^{\mathrm{3}{x}} \\ $$ Answered by benjo_mathlover last updated on 23/Oct/20 $$\:\left({D}−\mathrm{2}{I}\right)\left({D}−\mathrm{3}{I}\right)\left({y}\right)\:=\:{e}^{\mathrm{3}{x}} \:;\:{I}\left({y}\right)={y} \\ $$$${let}\:\left({D}−\mathrm{3}{I}\right)\left({y}\right)\:=\:{z} \\…
Question Number 119065 by benjo_mathlover last updated on 21/Oct/20 $${solve}\:\left({D}^{\mathrm{2}} −\mathrm{2}{D}+\mathrm{1}\right){y}\:=\:{e}^{{x}} \:\mathrm{ln}\:{x}\: \\ $$$${by}\:{using}\:{the}\:{method}\:{of}\:{variation} \\ $$$${of}\:{parameters}. \\ $$ Answered by Olaf last updated on 22/Oct/20…
Question Number 119002 by benjo_mathlover last updated on 21/Oct/20 $$\frac{{dy}}{{dx}}\:=\:\frac{{xy}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{and}\:{y}\left(\sqrt{\mathrm{15}}\right)\:=\:\mathrm{2} \\ $$ Answered by bramlexs22 last updated on 21/Oct/20 $$\:\frac{{dy}}{{y}}\:=\:\frac{{x}\:{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow\:\int\:\frac{{dy}}{{y}}\:=\:\int\:\frac{{x}\:{dx}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{ln}\:\left({y}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}}…
Question Number 118975 by benjo_mathlover last updated on 21/Oct/20 $$\:\left(\mathrm{3}{x}−\mathrm{5}{y}\right)\:{dx}\:+\:\left({x}+{y}\right)\:{dy}\:=\:\mathrm{0} \\ $$ Answered by 1549442205PVT last updated on 21/Oct/20 $$\:\left(\mathrm{3}{x}−\mathrm{5}{y}\right)\:{dx}\:+\:\left({x}+{y}\right)\:{dy}\:=\:\mathrm{0}\left(\mathrm{1}\right) \\ $$$$\mathrm{Put}\:\mathrm{y}=\mathrm{xt}\Rightarrow\mathrm{dy}=\mathrm{xdt}+\mathrm{tdx} \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\left(\mathrm{3x}−\mathrm{5xt}\right)\mathrm{dx}+\left(\mathrm{x}+\mathrm{xt}\right)\left(\mathrm{xdt}+\mathrm{tdx}\right) \\…
Question Number 118962 by benjo_mathlover last updated on 21/Oct/20 $$\:{x}\:{dy}\:=\:\left({y}+\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:\right)\:{dx}\:;\:{y}\left(\sqrt{\mathrm{3}}\right)\:=\:\mathrm{1} \\ $$ Answered by TANMAY PANACEA last updated on 21/Oct/20 $$\frac{{xdy}−{ydx}}{{x}^{\mathrm{2}} }=\frac{\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}}…
Question Number 118954 by bramlexs22 last updated on 21/Oct/20 $${solve}\:\left(\mathrm{6}{x}^{\mathrm{2}} \:+\:\mathrm{3}{y}^{\mathrm{2}} \right)\:{dx}\:=\:\mathrm{2}{xy}\:{dy} \\ $$ Answered by john santu last updated on 21/Oct/20 $$\:{set}\:{y}\:=\:{gx}\:\Rightarrow\frac{{dy}}{{dx}}\:=\:{g}\:+\:{x}\:\frac{{dg}}{{dx}} \\ $$$${the}\:{differential}\:{equation}\:{can}\:{be}\:{we}…
Question Number 118917 by htet last updated on 20/Oct/20 $${x}\left({x}+{y}\right){dy}−{x}^{\mathrm{2}} {dx} \\ $$ Commented by Dwaipayan Shikari last updated on 20/Oct/20 $${Q}\:\mathrm{118588} \\ $$ Terms…
Question Number 118813 by john santu last updated on 19/Oct/20 $$\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:−\mathrm{4}{x}\:\frac{{dy}}{{dx}}\:+\:{y}\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}\right)=\:\mathrm{0}\: \\ $$ Answered by bemath last updated on 20/Oct/20 Terms of…
Question Number 118588 by bramlexs22 last updated on 18/Oct/20 $${x}\left({x}+{y}\right)\:{dy}\:−{x}^{\mathrm{2}} \:{dx}\:=\:\mathrm{0}\: \\ $$ Commented by Dwaipayan Shikari last updated on 18/Oct/20 $${x}\left({x}+{y}\right)\frac{{dy}}{{dx}}−{x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}\left({x}+{y}\right)={x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…