Question Number 105564 by Ar Brandon last updated on 30/Jul/20 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}; \\ $$$$\mathrm{y}''+\mathrm{4y}'+\mathrm{5y}=\mathrm{xe}^{−\mathrm{2x}} \mathrm{sinx} \\ $$ Answered by bobhans last updated on 30/Jul/20 $${Homogenous}\:{equation} \\…
Question Number 170998 by 2407 last updated on 06/Jun/22 Answered by ali009 last updated on 06/Jun/22 $${r}^{\mathrm{2}} +\mathrm{3}{r}+\mathrm{2}=\mathrm{0} \\ $$$${r}=−\mathrm{1}\:\:\:\:{r}=−\mathrm{2} \\ $$$${y}_{{h}} ={C}_{\mathrm{1}} {e}^{−{x}} ×{C}_{\mathrm{2}}…
Question Number 105361 by bemath last updated on 28/Jul/20 $${solve}\:{by}\:{Frobenius}\:{method} \\ $$$${x}^{\mathrm{2}} {y}''−{x}\left(\mathrm{2}{x}−\mathrm{1}\right){y}'+\left({x}+\mathrm{1}\right){y}=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 170847 by cortano1 last updated on 01/Jun/22 Commented by ali009 last updated on 01/Jun/22 $${i}\:{think}\:{it}'{s}\:\frac{{dx}}{{dy}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 105283 by bemath last updated on 27/Jul/20 $$\left(\mathrm{1}\right)\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2}{xy}}{\mathrm{4}{x}^{\mathrm{2}} −{y}^{\mathrm{3}} } \\ $$$$\left(\mathrm{2}\right)\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{{y}\left(\mathrm{2ln}\:{y}\:+\:\mathrm{1}\right)} \\ $$ Answered by bobhans last updated on 27/Jul/20 $$\left(\mathrm{2}\right)\:{y}\left(\mathrm{2ln}\:{y}+\mathrm{1}\right)\:{dy}\:=\:\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)\:{dx} \\…
Question Number 170795 by ali009 last updated on 31/May/22 $${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}} \\…