Question Number 84830 by jagoll last updated on 16/Mar/20 $$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}−\mathrm{x}−\mathrm{y}}{\mathrm{x}+\mathrm{y}} \\ $$ Answered by john santu last updated on 16/Mar/20 Terms of Service Privacy Policy…
Question Number 19199 by priyankavarma094@gmail.com last updated on 06/Aug/17 $$\mathrm{y}=\mathrm{tan}\:\mathrm{x}^{\mathrm{tan}\:\mathrm{x}^{\mathrm{tan}\:\mathrm{x}} } \\ $$ Answered by NEC last updated on 06/Aug/17 $${let}\:{u}=\mathrm{tan}\:{x} \\ $$$$ \\ $$$${y}={u}^{{u}^{{u}}…
Question Number 84544 by M±th+et£s last updated on 14/Mar/20 $${solve}\: \\ $$$${xy}^{''} ={y}^{'} \left({e}^{{y}} −\mathrm{1}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 84477 by jagoll last updated on 13/Mar/20 $$\left(\mathrm{ycos}\:\mathrm{x}+\mathrm{2xe}^{\mathrm{y}} \right)\mathrm{dx}+\left(\mathrm{sin}\:\mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{e}^{\mathrm{y}} −\mathrm{1}\right)\mathrm{dy}=\mathrm{0} \\ $$ Answered by john santu last updated on 13/Mar/20 $$\frac{\partial\mathrm{M}}{\partial\mathrm{y}}\:=\:\mathrm{cos}\:\mathrm{x}+\mathrm{2xe}^{\mathrm{y}} \\…
Question Number 84407 by jagoll last updated on 12/Mar/20 $$\mathrm{dy}+\mathrm{2xy}\:\mathrm{dx}\:=\:\mathrm{xe}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{y}^{\mathrm{3}} \:\mathrm{dx} \\ $$$$ \\ $$ Answered by john santu last updated on 13/Mar/20…
Question Number 18800 by tawa tawa last updated on 29/Jul/17 $$\mathrm{Solve}:\:\:\mathrm{x}\:\mathrm{sin}\left(\mathrm{y}\right)\mathrm{dx}\:+\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}\:\mathrm{cos}\left(\mathrm{y}\right)\mathrm{dy}\:=\:\mathrm{0},\:\:\mathrm{subject}\:\mathrm{to}\:\mathrm{y}\left(\mathrm{1}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 84232 by niroj last updated on 10/Mar/20 $$\:\:\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{following}}\:\boldsymbol{\mathrm{differential}}\:\boldsymbol{\mathrm{equation}}. \\ $$$$\:\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }\:+\:\frac{\boldsymbol{\mathrm{x}}}{\mathrm{1}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}−\:\frac{\boldsymbol{\mathrm{y}}}{\mathrm{1}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }=\:\boldsymbol{\mathrm{x}}\sqrt{\mathrm{1}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \\ $$ Commented by mind is power last…
Question Number 149757 by iloveisrael last updated on 07/Aug/21 Commented by amin96 last updated on 07/Aug/21 $${x}\frac{{dy}}{{dx}}+\mathrm{2}{y}=\frac{{sinx}}{{x}}\:\:\:\:\:\:\frac{{dy}}{{dx}}=\frac{\mathrm{sin}\:{x}−\mathrm{2}{xy}}{{x}^{\mathrm{2}} } \\ $$$$\left(\mathrm{sin}\:{x}−\mathrm{2}{xy}\right){dx}+\left(−{x}^{\mathrm{2}} \right){dy}=\mathrm{0} \\ $$$$\frac{\partial{M}}{\partial{y}}=−\mathrm{2}{x}\:\:\:\:\:\:\:\frac{\partial{N}}{\partial{x}}=−\mathrm{2}{x}\:\:\:\:\:\:{M}_{{y}} ={N}_{{x}} \:\:\:…
Question Number 84212 by jagoll last updated on 10/Mar/20 $$\mathrm{y}'\:=\:\left(\mathrm{2x}+\mathrm{3y}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\: \\ $$ Commented by niroj last updated on 10/Mar/20 $$\:\:\frac{\mathrm{dy}}{\mathrm{dx}}=\:\left(\mathrm{2x}+\mathrm{3y}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\mathrm{put},\:\:\mathrm{2x}+\mathrm{3y}+\mathrm{1}=\mathrm{v} \\…
Question Number 84085 by M±th+et£s last updated on 09/Mar/20 $${solve} \\ $$$$\frac{{d}}{{dr}}\left({r}\frac{{d}\theta}{{dr}}\right)=\mathrm{0} \\ $$ Answered by TANMAY PANACEA last updated on 09/Mar/20 $$\frac{{d}}{{dr}}\left({r}\frac{{d}\theta}{{dr}}\right)=\mathrm{0}=\frac{{dA}}{{dr}}\:\:{A}={constant} \\ $$$${r}\frac{{d}\theta}{{dr}}={A}…