Question Number 103767 by bramlex last updated on 17/Jul/20 $${solve}\:{y}'−{y}\:=\:{y}^{\mathrm{4}} \:{at}\:{y}\left(\mathrm{0}\right)\:=\:\mathrm{1}\: \\ $$ Answered by bemath last updated on 17/Jul/20 $${substitute}\:{v}\:=\:{y}^{−\mathrm{3}} \\ $$$$\frac{{dv}}{{dx}}\:=\:−\mathrm{3}{y}^{−\mathrm{4}} \:\frac{{dy}}{{dx}}\:\Rightarrow\frac{{dy}}{{dx}}\:=\:−\frac{{y}^{\mathrm{4}} }{\mathrm{3}}\:\frac{{dv}}{{dx}}…
Question Number 103759 by bemath last updated on 17/Jul/20 $$\left({x}^{\mathrm{5}} +\mathrm{3}{y}\right)\:{dx}\:−{x}\:{dy}\:=\:\mathrm{0}\: \\ $$ Answered by MAB last updated on 17/Jul/20 $${x}^{\mathrm{5}} +\mathrm{3}{y}−{xy}'=\mathrm{0} \\ $$$${xy}'−\mathrm{3}{y}={x}^{\mathrm{5}} \\…
Question Number 169287 by cortano1 last updated on 28/Apr/22 $$\:\:{solve}\::\:{x}^{\mathrm{3}} \:\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }\:+\:\mathrm{2}{x}^{\mathrm{2}} \:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\mathrm{2}{y}\:=\:\mathrm{10}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$ Commented by infinityaction last updated on 01/May/22…
Question Number 103664 by byaw last updated on 16/Jul/20 Answered by bramlex last updated on 16/Jul/20 $$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\mathrm{dx}\:=\:\mathrm{2xy}\:\mathrm{dy}\: \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }{\mathrm{2xy}}\:;\:\mathrm{set}\:\mathrm{y}\:=\:\mathrm{zx}\: \\ $$$$\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{z}\:+\:\mathrm{x}\:\frac{\mathrm{dz}}{\mathrm{dx}}…
Question Number 103665 by byaw last updated on 16/Jul/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 169155 by Mastermind last updated on 27/Apr/22 $$\left(\mathrm{3}\right)\:\:{Consider}\:{the}\:{boundary}−{value}\: \\ $$$${problem}\:{y}''+\mathrm{2}{y}'+\mathrm{2}{y}=\mathrm{0},\:{y}\left({a}\right)={c}, \\ $$$${y}\left({b}\right)={d}.\:\left({i}\right)\:{if}\:{this}\:{problem}\:{has}\:{a}\: \\ $$$${unique}\:{solution},\:{how}\:{are}\:{a}\:{and}\:{b}\: \\ $$$${related}?\:\left({ii}\right)\:{if}\:{this}\:{problem}\:{has}\:{no} \\ $$$${solution},\:{how}\:{are}\:{a},{b},{c}\:{and}\:{d}\:{related}? \\ $$ Terms of Service…
Question Number 103597 by jimi last updated on 16/Jul/20 $$\boldsymbol{{pls}}\:\boldsymbol{{help}}\:\boldsymbol{{solve}}\:\boldsymbol{{this}}\:\boldsymbol{{differential}}\:\boldsymbol{{equation}} \\ $$$$\left(\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} \mathrm{sin}\:\left(\frac{\mathrm{1}}{\boldsymbol{{x}}}\right)\:+\:\boldsymbol{{y}}\right)\boldsymbol{{dx}}\:=\:\boldsymbol{{xcos}}\left(\frac{\mathrm{1}}{\boldsymbol{{x}}}\right)\:−\boldsymbol{{xdy}} \\ $$ Commented by OlafThorendsen last updated on 16/Jul/20 $$\mathrm{something}\:\mathrm{is}\:\mathrm{missing}\:\mathrm{after}\:{x}\mathrm{cos}\frac{\mathrm{1}}{{x}}\:? \\ $$…
Question Number 103553 by byaw last updated on 15/Jul/20 Answered by mathmax by abdo last updated on 16/Jul/20 $$\mathrm{2y}^{''} −\mathrm{4y}^{'} −\mathrm{6y}\:=\mathrm{0}\:\Rightarrow\mathrm{y}^{''} −\mathrm{2y}^{'} \:−\mathrm{3y}\:=\mathrm{0} \\ $$$$\rightarrow\mathrm{r}^{\mathrm{2}}…
Question Number 169053 by Mastermind last updated on 23/Apr/22 $${Solve}\:{the}\:{ODE}\: \\ $$$${y}'\:+\:\mathrm{2}{xy}\:=\:{xe}^{−{x}^{\mathrm{2}} } ,\:{with}\:{y}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$ \\ $$$${Mastermind} \\ $$ Answered by haladu last updated…
Question Number 169051 by Mastermind last updated on 23/Apr/22 $${Solve}\:{the}\:{ODE} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{2}\right){y}'\:+\:{xy}\:=\:\mathrm{0},\:{with}\:{y}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$ \\ $$$${Mastermind} \\ $$ Commented by haladu last updated on…