Question Number 156395 by alcohol last updated on 10/Oct/21 $${A}\:{car}\:{is}\:{moving}\:{along}\:{a}\:{straight}\:{road}\:{level}\:{when}\: \\ $$$${the}\:{driver}\:{see}\:{a}\:{boy}\:{crossing}\:{the}\:{road}\:\mathrm{1}.\mathrm{5}{m}\:{when}\:{the}\:{driver} \\ $$$${immediately}\:{applies}\:{the}\:{brakes}\:{which}\:{produces} \\ $$$$\:{constant}\:{retardation}\:{in}\:{the}\:{first}\:{second}.\:{After}\:{applying}\:{the}\:{brakes} \\ $$$${the}\:{car}\:{travels}\:\mathrm{25}{m}\:{and}\:{in}\:{the}\:{next}\:{second}\:{it} \\ $$$${traveles}\:\mathrm{15}{m}.\: \\ $$$$\left.{i}\right)\:{find}\:{the}\:{retardation}\:{in}\:{m}/{s}^{\mathrm{2}} \\ $$$$\left.{ii}\right)\:{show}\:{that}\:{the}\:{car}\:{comes}\:{to}\:{rest}\:{at}\: \\…
Question Number 90863 by ajfour last updated on 26/Apr/20 $$\frac{{dy}}{{dx}}−{a}\left(\frac{{y}}{{x}}\right)=\mathrm{1}+\frac{\mathrm{1}}{{x}} \\ $$ Answered by mr W last updated on 26/Apr/20 $${p}\left({x}\right)=−\frac{{a}}{{x}} \\ $$$$\int{p}\left({x}\right){dx}=−\int\frac{{a}}{{x}}{dx}=−{a}\mathrm{ln}\:{x}=\mathrm{ln}\:{x}^{−{a}} \\ $$$${u}\left({x}\right)={e}^{\int{p}\left({x}\right){dx}}…
Question Number 90851 by john santu last updated on 26/Apr/20 $${x}\:\frac{{dy}}{{dx}}\:=\:{y}\:+\:\mathrm{3}{x}^{\mathrm{4}} \:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{{y}}{{x}}\right)\: \\ $$ Answered by jagoll last updated on 26/Apr/20 Commented by john…
Question Number 90810 by john santu last updated on 26/Apr/20 $$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{xy}}\: \\ $$ Commented by peter frank last updated on 26/Apr/20 $${help}\:{Qn}\:\:\mathrm{90812}…
Question Number 90770 by ajfour last updated on 26/Apr/20 $$\left({ax}+{by}+{c}\right){dx}+\left({px}+{qy}+{r}\right){dy}=\mathrm{0} \\ $$ Answered by mr W last updated on 26/Apr/20 $$\frac{{dy}}{{dx}}=−\frac{{ax}+{by}+{c}}{{px}+{qy}+{r}} \\ $$$${let}\:{x}={u}+{u}_{\mathrm{0}} ,\:{y}={v}+{v}_{\mathrm{0}} \\…
Question Number 25109 by ifcrna380w last updated on 04/Dec/17 Answered by mrW1 last updated on 04/Dec/17 $$\frac{{dy}}{{y}^{\mathrm{2}} }=\frac{{dx}}{{e}^{{x}} +{e}^{−{x}} }=\frac{\mathrm{1}}{\left({e}^{{x}} \right)^{\mathrm{2}} +\mathrm{1}}{d}\left({e}^{{x}} \right)=\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}} \\…
Question Number 90641 by Tony Lin last updated on 25/Apr/20 $${Solve}\:{x}^{\mathrm{2}} {y}''+{xy}'+{x}^{\mathrm{2}} {y}=\mathrm{0} \\ $$ Commented by jagoll last updated on 25/Apr/20 $${Euler}\:−\:{Cauchy}? \\ $$…
Question Number 25094 by ifcrna380w last updated on 03/Dec/17 Answered by mrW1 last updated on 04/Dec/17 $${e}^{−\mathrm{2}{y}} {dy}={e}^{\mathrm{3}{x}} {dx} \\ $$$$\int{e}^{−\mathrm{2}{y}} {dy}=\int{e}^{\mathrm{3}{x}} {dx} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\int{e}^{−\mathrm{2}{y}}…
Question Number 155965 by john_santu last updated on 06/Oct/21 $$\left({x}\:\mathrm{sin}\:\frac{{y}}{{x}}−{y}\:\mathrm{cos}\:\frac{{y}}{{x}}\right){dx}+{x}\:\mathrm{cos}\:\frac{{y}}{{x}}\:{dy}=\mathrm{0} \\ $$ Answered by mindispower last updated on 07/Oct/21 $$\left({tg}\left(\frac{{y}}{{x}}\right)−\frac{{y}}{{x}}\right){dx}+{dy}=\mathrm{0} \\ $$$$\frac{{y}}{{x}}={u} \\ $$$${dy}={xdu}+{udx} \\…
Question Number 90394 by jagoll last updated on 23/Apr/20 $$\left(\mathrm{3x}^{\mathrm{2}} +\mathrm{9xy}+\mathrm{5y}^{\mathrm{2}} \right)\mathrm{dx}\:=\:\left(\mathrm{6x}^{\mathrm{2}} +\mathrm{4xy}\right)\mathrm{dy} \\ $$ Commented by john santu last updated on 23/Apr/20 $$\left(\mathrm{3}+\mathrm{9}\left(\frac{{y}}{{x}}\right)+\mathrm{5}\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} \right)\:=\:\left(\mathrm{6}+\mathrm{4}\left(\frac{{y}}{{x}}\right)\right)\:{dy}…