Question Number 102515 by bemath last updated on 09/Jul/20 $$\left(\frac{{x}}{{y}}\right){y}'=\:\frac{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}}{{x}+\mathrm{1}} \\ $$ Answered by bobhans last updated on 09/Jul/20 $$\frac{{dy}}{{y}\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}\right)}\:=\:\frac{{dx}}{{x}\left({x}+\mathrm{1}\right)} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{1}}{{y}\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}\right)}\:=\:\frac{{A}}{{y}}\:+\:\frac{{By}+{C}}{\mathrm{2}{y}^{\mathrm{2}}…
Question Number 102517 by bemath last updated on 09/Jul/20 $$\left(\mathrm{1}+{x}^{\mathrm{2}} \right){dy}\:+\:\left(\mathrm{1}+{y}^{\mathrm{2}} \right){dx}\:=\:\mathrm{0} \\ $$ Commented by bemath last updated on 09/Jul/20 Terms of Service Privacy…
Question Number 168011 by Mastermind last updated on 31/Mar/22 $${Solve}\: \\ $$$$\left(\mathrm{2}{x}+\mathrm{5}{y}+\mathrm{1}\right){dx}\:−\:\left(\mathrm{5}{x}+\mathrm{2}{y}−\mathrm{1}\right){dy}=\mathrm{0} \\ $$$$ \\ $$$${Mastermind} \\ $$ Answered by mr W last updated on…
Question Number 102382 by Ar Brandon last updated on 08/Jul/20 $$\mathrm{x}^{\mathrm{2}} \centerdot\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}^{\mathrm{2}} +\mathrm{xy}+\mathrm{y}^{\mathrm{2}} \\ $$ Answered by PRITHWISH SEN 2 last updated on 08/Jul/20 $$\mathrm{put}\:\mathrm{y}=\mathrm{vx}…
Question Number 167917 by guddupari last updated on 29/Mar/22 $$\:_{} \: \\ $$$$\mathrm{Two}\:\mathrm{small}\:\mathrm{charged}\:\mathrm{spheresn} \\ $$$$\mathrm{Two}\:\mathrm{small}\:\mathrm{charged}\:\mathrm{spheresn} \\ $$$$\mathrm{cotain}\:\mathrm{charges}\:+\:\mathrm{q1}\:\mathrm{and}\:+\:\mathrm{q2r} \\ $$$$\mathrm{espectively}.\:\mathrm{A}\:\mathrm{charge}\:\mathrm{da}\:\mathrm{ism} \\ $$$$\mathrm{reoved}\:\mathrm{from}\:\mathrm{sphere}\:\mathrm{carryingh} \\ $$$$\mathrm{carge}\:\mathrm{q1}\:\mathrm{and}\:\mathrm{is}\:\mathrm{transferredtoh} \\ $$$$\mathrm{te}\:\mathrm{other}.\:\mathrm{Find}\:\mathrm{charge}\:\mathrm{on}\:\mathrm{eachs}…
Question Number 102298 by Ar Brandon last updated on 08/Jul/20 $$\mathrm{sinx}\centerdot\frac{\mathrm{dy}}{\mathrm{dx}}−\mathrm{ycosx}=\mathrm{y}^{\mathrm{3}} \mathrm{sin}^{\mathrm{2}} \mathrm{xcosx} \\ $$ Answered by john santu last updated on 08/Jul/20 $$\frac{{dy}}{{dx}}−{y}\mathrm{cot}\:{x}=\mathrm{sin}\:{x}\mathrm{cos}\:{x}\:.{y}^{\mathrm{3}} \\…
Question Number 102295 by 1549442205 last updated on 08/Jul/20 $$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{equation}}\:: \\ $$$$\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{lny}}−\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{y}}'=\boldsymbol{\mathrm{y}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 102283 by Ar Brandon last updated on 08/Jul/20 $$\mathrm{sinx}\frac{\mathrm{dy}}{\mathrm{dx}}−\mathrm{2ycosx}=\mathrm{3sinx} \\ $$ Answered by john santu last updated on 08/Jul/20 $$\frac{{dy}}{{dx}}\:−\mathrm{2cot}\:{x}.{y}\:=\:\mathrm{3} \\ $$$${IF}\:\Rightarrow{u}\left({x}\right)={e}^{\int\:−\mathrm{2cot}\:{x}\:{dx}\:} =\:{e}^{−\mathrm{2ln}\left(\mathrm{sin}\:{x}\right)}…
Question Number 101848 by bramlex last updated on 05/Jul/20 $$\left(\mathrm{cos}\:\mathrm{x}\right)\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\mathrm{y}\:\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{2x}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:, \\ $$$$\mathrm{y}\left(\frac{\pi}{\mathrm{4}}\right)\:=\:\frac{−\mathrm{15}\pi^{\mathrm{2}} \sqrt{\mathrm{2}}}{\mathrm{32}} \\ $$ Answered by john santu last updated on 05/Jul/20 $$\frac{{dy}}{{dx}}\:+\:{y}.\mathrm{tan}\:\left({x}\right)=\:\mathrm{2}{x}\:\mathrm{cos}\:{x}…
Question Number 101841 by bemath last updated on 05/Jul/20 $${xy}'\:+\:{y}\:=\:{y}^{\mathrm{2}} \\ $$ Answered by bobhans last updated on 05/Jul/20 $${xy}'\:=\:{y}^{\mathrm{2}} −{y}\:\Rightarrow\:\frac{{dy}}{{y}\left({y}−\mathrm{1}\right)}\:=\:\frac{{dx}}{{x}} \\ $$$$\int\:\frac{{dy}}{{y}−\mathrm{1}}\:−\:\int\:\frac{{dy}}{{y}}\:=\:\frac{{dx}}{{x}} \\ $$$$\mathrm{ln}\left(\frac{{y}−\mathrm{1}}{{y}}\right)\:=\:\mathrm{ln}\mid{Cx}\mid\:\Rightarrow\:\mathrm{1}−\frac{\mathrm{1}}{{y}}\:=\:\mid{Cx}\mid\:…