Question Number 104240 by bemath last updated on 20/Jul/20 $$\frac{{dy}}{{dx}}\:=\:\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left({xy}\right)^{\mathrm{2}} \\ $$ Commented by bemath last updated on 20/Jul/20 $${thank}\:{you}\:{all}.\:{correct} \\ $$ Answered…
Question Number 104051 by bemath last updated on 19/Jul/20 $$\left({x}^{\mathrm{2}} −\mathrm{1}\right)\:\frac{{dy}}{{dx}}\:+\:\mathrm{2}{y}\:=\:\left({x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$ Answered by bramlex last updated on 19/Jul/20 Commented by bramlex last updated…
Question Number 169558 by greougoury555 last updated on 03/May/22 $$\:\:\:\:\:\:\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2}{x}−{y}+\mathrm{1}}{{x}−\mathrm{4}{y}+\mathrm{3}}\:\:\: \\ $$ Answered by ali009 last updated on 03/May/22 $${a}\mathrm{1}=\mathrm{2}\:\:{a}\mathrm{2}=\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}\mathrm{1}/{a}\mathrm{2}=\mathrm{2} \\ $$$${b}\mathrm{1}=−\mathrm{1}\:\:{b}\mathrm{2}=−\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:{b}\mathrm{1}/{b}\mathrm{2}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${so} \\…
Question Number 103958 by bemath last updated on 18/Jul/20 $${what}\:{is}\:{integrating}\:{factor} \\ $$$${of}\:\left({xy}^{\mathrm{2}} −{y}\right)\:{dx}\:−\:{x}\:{dy}\:=\:\mathrm{0} \\ $$ Answered by bramlex last updated on 18/Jul/20 $$\left({xy}^{\mathrm{2}} −{y}\right)\:{dx}\:=\:{x}\:{dy}\: \\…
Question Number 103931 by bemath last updated on 18/Jul/20 $$\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}^{'}…