Question Number 106065 by bemath last updated on 02/Aug/20 $$\mathrm{x}^{\mathrm{2}} \:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:−\mathrm{x}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\mathrm{y}\:=\:\mathrm{x}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 105879 by john santu last updated on 01/Aug/20 $${y}''+{y}'−\mathrm{6}{y}\:=\:{x}\: \\ $$ Answered by bemath last updated on 01/Aug/20 Answered by mathmax by abdo…
Question Number 105878 by bemath last updated on 01/Aug/20 $$\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }\:+\:\mathrm{3}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{2}\:\frac{{dy}}{{dx}}\:=\:{x}^{\mathrm{2}} \\ $$ Answered by bobhans last updated on 01/Aug/20 $$\mathrm{HE}\::\:\flat^{\mathrm{3}} +\mathrm{3}\flat^{\mathrm{2}}…
Question Number 105877 by john santu last updated on 01/Aug/20 Answered by bemath last updated on 01/Aug/20 $$\frac{{dy}}{{dx}}.\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:+\:\frac{\mathrm{1}+{y}^{\mathrm{2}} }{{y}^{\mathrm{4}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:=\:\mathrm{0} \\ $$$$\frac{{y}^{\mathrm{4}} \:{dy}}{\mathrm{1}+{y}^{\mathrm{2}}…
Question Number 105854 by bobhans last updated on 01/Aug/20 $$\mathrm{sin}\:\mathrm{2}{x}\:\frac{{dy}}{{dx}}\:−{y}\:=\:\mathrm{tan}\:{x} \\ $$ Answered by bemath last updated on 01/Aug/20 $$\frac{{dy}}{{dx}}−\mathrm{csc}\:\mathrm{2}{x}.{y}\:=\:\mathrm{csc}\:\mathrm{2}{x}.\mathrm{tan}\:{x} \\ $$$${integrating}\:{factor}\: \\ $$$${u}\left({x}\right)=\:{e}^{\int\:\mathrm{csc}\:\mathrm{2}{x}\:{dx}} \:=\:{e}^{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{tan}\:{x}\mid}…
Question Number 171336 by cortano1 last updated on 13/Jun/22 Answered by mr W last updated on 13/Jun/22 $${k}=\frac{{y}}{{x}} \\ $$$${x}^{\mathrm{2}} −\mathrm{10}{x}+{k}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{10}{kx}+\mathrm{41}=\mathrm{0} \\ $$$$\left(\mathrm{1}+{k}^{\mathrm{2}}…
Question Number 105738 by bobhans last updated on 31/Jul/20 $$\left({x}\:\mathrm{tan}\:\left(\frac{{y}}{{x}}\right)−{y}\:\mathrm{sec}\:^{\mathrm{2}} \left(\frac{{y}}{{x}}\right)\right)\:{dx}−{x}\:\mathrm{sec}\:^{\mathrm{2}} \left(\frac{{y}}{{x}}\right){dy}=\mathrm{0} \\ $$ Answered by john santu last updated on 01/Aug/20 $${set}\:\frac{{y}}{{x}}\:=\:\vartheta\:\Rightarrow{y}=\vartheta{x} \\ $$$$\frac{{dy}}{{dx}}\:=\:\vartheta\:+\:{x}\:\frac{{d}\vartheta}{{dx}}…
Question Number 105734 by bemath last updated on 31/Jul/20 $$\left(\mathrm{2}{x}+\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\mathrm{cot}\:{x}\right)\:{dx}\:+\mathrm{2}{y}\:{dy}\:=\:\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 105710 by john santu last updated on 31/Jul/20 $$\left(\mathrm{2}{xy}+\mathrm{cos}\:{y}\right)\:{dx}\:+\left({x}^{\mathrm{2}} −{x}\mathrm{sin}{y}−\mathrm{2}{y}\right){dy}=\mathrm{0} \\ $$ Answered by Smail last updated on 31/Jul/20 $${let}\:{M}=\mathrm{2}{xy}+{cosy}\:{and}\:\:{N}={x}^{\mathrm{2}} −{xsiny}−\mathrm{2}{y} \\ $$$$\frac{{dM}}{{dy}}=\mathrm{2}{x}−{siny}…
Question Number 105659 by bemath last updated on 30/Jul/20 $$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\frac{{dy}}{{dx}}−\mathrm{2}{y}=\mathrm{2}{x}\: \\ $$ Answered by john santu last updated on 30/Jul/20 Commented by bramlex…