Question Number 98255 by john santu last updated on 12/Jun/20 $$\mathrm{solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{initial}\:\mathrm{values} \\ $$$$\mathrm{DEs}\:\mathrm{20y}''\:+\:\mathrm{4y}'\:+\mathrm{y}\:=\:\mathrm{0} \\ $$$$;\:\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{3}.\mathrm{2}\:\mathrm{and}\:\mathrm{y}'\left(\mathrm{0}\right)\:=\:\mathrm{0}\: \\ $$ Commented by bemath last updated on 13/Jun/20 $$\mathrm{auxilary}\:\mathrm{equation}\:\mathrm{euler}−\mathrm{cauchy}…
Question Number 98106 by bobhans last updated on 11/Jun/20 $$\mathrm{y}''+\mathrm{y}\:=\:\mathrm{cos}\:\mathrm{3x}−\mathrm{2sin}\:\mathrm{3x} \\ $$ Answered by bemath last updated on 11/Jun/20 $$\mathrm{homogenous}\:\mathrm{solution}\: \\ $$$$\lambda^{\mathrm{2}} +\mathrm{1}=\mathrm{0}\:\Rightarrow\lambda\:=\:\pm\:{i} \\ $$$$\mathrm{y}_{\mathrm{h}}…
Question Number 163591 by Ar Brandon last updated on 08/Jan/22 $$\mathrm{R}\acute {\mathrm{e}soudre}\:\:\:\:\:\:\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {u}}{\partial{y}^{\mathrm{2}} }=\mathrm{10}{e}^{\mathrm{2}{x}+{y}} \\ $$ Commented by mkam last updated on 08/Jan/22…
Question Number 97922 by bemath last updated on 10/Jun/20 $$\mathrm{y}''\:+\:\mathrm{y}\:=\:\mathrm{cot}\:{x}\: \\ $$ Answered by bobhans last updated on 10/Jun/20 Commented by bemath last updated on…
Question Number 97868 by bemath last updated on 10/Jun/20 $$\mathrm{3y}'\:=\:\mathrm{2x}+\mathrm{y}−\mathrm{1}\: \\ $$ Answered by bobhans last updated on 10/Jun/20 Commented by bemath last updated on…
Question Number 97866 by me2love2math last updated on 10/Jun/20 Commented by me2love2math last updated on 10/Jun/20 $$\mathrm{1}\:\mathrm{2}\:\mathrm{5}\:{and}\:\mathrm{6} \\ $$ Commented by bobhans last updated on…
Question Number 97847 by bemath last updated on 10/Jun/20 $$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}\right)\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:+\:\left(\mathrm{1}−\mathrm{2x}^{\mathrm{2}} \right)\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{1}\right)\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{1} \\ $$ Answered by niroj last updated on 10/Jun/20…
Question Number 97751 by bobhans last updated on 09/Jun/20 $$\left(\mathrm{y}^{\mathrm{2}} −\mathrm{xy}\right)\mathrm{dx}\:+\:\mathrm{2xy}\:\mathrm{dy}\:=\:\mathrm{0}\: \\ $$ Commented by bemath last updated on 09/Jun/20 $$\Leftrightarrow\:\mathrm{2xy}\:\mathrm{dy}\:=\:\left(\mathrm{xy}−\mathrm{y}^{\mathrm{2}} \right)\:\mathrm{dx}\: \\ $$$$\mathrm{set}\:\mathrm{y}\:=\:\mathrm{zx}\:\Leftrightarrow\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{z}\:+\:\mathrm{x}\:\frac{\mathrm{dz}}{\mathrm{dx}} \\…
Question Number 163113 by Lordose last updated on 03/Jan/22 $$\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:+\:\mathrm{ytan}\left(\mathrm{x}\right)\:=\:\mathrm{e}^{\mathrm{x}} \:\:\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{1},\:\mathrm{y}'\left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 97539 by bobhans last updated on 08/Jun/20 $${x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} +\mathrm{2}{xy}\:\frac{{dy}}{{dx}}\:=\:\mathrm{0} \\ $$ Answered by bemath last updated on 08/Jun/20 Commented by bobhans last…