Question Number 103931 by bemath last updated on 18/Jul/20 $$\left({y}^{\mathrm{2}} +\mathrm{2}\right)\:{dx}\:=\:\left({xy}+\mathrm{2}{y}+{y}^{\mathrm{3}} \right)\:{dy} \\ $$ Answered by bobhans last updated on 18/Jul/20 $$\frac{{dy}}{{dx}}\:=\:\frac{{y}^{\mathrm{2}} +\mathrm{2}}{{y}\left({x}+\mathrm{2}+{y}^{\mathrm{2}} \right)} \\…
Question Number 103790 by ugwu Kingsley last updated on 17/Jul/20 Commented by ugwu Kingsley last updated on 17/Jul/20 $${i}\:{need}\:{help}\:{with}\:{this}\:{asap} \\ $$ Answered by bemath last…
Question Number 103788 by bemath last updated on 17/Jul/20 $${y}''−{y}\:=\:\mathrm{cot}\:{x}\: \\ $$ Answered by mathmax by abdo last updated on 18/Jul/20 $$\mathrm{y}^{''} −\mathrm{y}\:=\frac{\mathrm{cosx}}{\mathrm{sinx}} \\ $$$$\mathrm{h}\rightarrow\mathrm{r}^{\mathrm{2}}…
Question Number 103776 by bemath last updated on 17/Jul/20 $${y}'\:−\:\frac{{y}}{{x}^{\mathrm{2}} −\mathrm{1}}\:=\:{y}^{\mathrm{2}} \\ $$ Answered by bemath last updated on 17/Jul/20 $${set}\:{v}\:=\:{y}^{−\mathrm{1}} \:\Rightarrow\frac{{dv}}{{dx}}\:=\:−{y}^{−\mathrm{2}} \:\frac{{dy}}{{dx}} \\ $$$$\frac{{dy}}{{dx}}\:=\:−{y}^{\mathrm{2}}…
Question Number 103771 by bobhans last updated on 17/Jul/20 $${y}''−{y}'+{y}\:=\:\mathrm{cos}\:\mathrm{3}{x} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 103769 by bemath last updated on 17/Jul/20 $$\mathrm{2}{y}''−{y}'+{y}\:=\:\mathrm{cos}\:\mathrm{3}{x} \\ $$ Answered by mathmax by abdo last updated on 17/Jul/20 $$\mathrm{let}\:\mathrm{solve}\:\mathrm{by}\:\mathrm{laplace}\:\mathrm{transform} \\ $$$$\mathrm{e}\:\Rightarrow\mathrm{2L}\left(\mathrm{y}^{''} \right)−\mathrm{L}\left(\mathrm{y}^{'}…
Question Number 103767 by bramlex last updated on 17/Jul/20 $${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}}…