Question Number 89600 by jagoll last updated on 18/Apr/20 $$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{y}+\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}{\mathrm{x}}\: \\ $$ Commented by mr W last updated on 18/Apr/20 $$\frac{{dy}}{{dx}}=\frac{{y}}{{x}}+\sqrt{\mathrm{1}−\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} } \\…
Question Number 89243 by M±th+et£s last updated on 16/Apr/20 $${solve}\:{the}\:{following}\:{diffirntial}\:{equation} \\ $$$$\left.\mathrm{1}\right)\left(\mathrm{2}{x}+{y}\right){dx}+\left({x}+{y}\right){dy}=\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\left(\mathrm{3}{x}−{y}\right){dx}−\left({x}−{y}\right){dy}=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:\left({cos}\left({x}\right)+{y}\right){dx}\:+\:\left(\mathrm{2}{y}+{x}\right){dy}=\mathrm{0} \\ $$ Answered by TANMAY PANACEA. last updated on…
Question Number 89205 by jagoll last updated on 16/Apr/20 $$\mathrm{2}\:\frac{{dy}}{{dx}}\:=\:\frac{{y}}{{x}}\:+\:\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} \\ $$ Commented by john santu last updated on 16/Apr/20 $${y}\:=\:{vx}\:\Rightarrow\frac{{dy}}{{dx}}\:=\:{v}\:+\:{x}\:\frac{{dv}}{{dx}} \\ $$$$\mathrm{2}{v}\:+\mathrm{2}{x}\:\frac{{dv}}{{dx}}\:=\:{v}+{v}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}\:\frac{{dv}}{{dx}}\:=\:{v}^{\mathrm{2}}…
Question Number 89178 by necxxx last updated on 15/Apr/20 $${If}\:{z}\left({z}^{\mathrm{2}} +\mathrm{3}{x}\right)+\mathrm{3}{y}=\mathrm{0}\:{prove}\:{that}\: \\ $$$$\frac{\partial^{\mathrm{2}} {z}}{\partial{x}^{\mathrm{2}} }\:+\:\frac{\partial^{\mathrm{2}} {z}}{\partial{y}^{\mathrm{2}} }=\:\frac{\mathrm{2}{z}\left({x}−\mathrm{1}\right)}{\left({z}^{\mathrm{2}} +{x}\right)^{\mathrm{3}} } \\ $$$$ \\ $$$$ \\ $$$${please}\:{help}.…
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Question Number 89029 by Jidda28 last updated on 14/Apr/20 Commented by MJS last updated on 14/Apr/20 $${t}=\sqrt[{\mathrm{4}}]{\mathrm{tan}\:{x}}\:\mathrm{or}\:{t}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{tan}\:{x}}}\:\mathrm{and}\:\mathrm{then}\:\mathrm{decompose} \\ $$$$\frac{{f}\left({t}\right)}{{t}^{\mathrm{8}} +\mathrm{1}}\:\mathrm{which}\:\mathrm{is}\:\mathrm{possible}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{no}\:\mathrm{fun}\:\mathrm{dealing} \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{factors} \\ $$ Terms…
Question Number 23445 by tawa tawa last updated on 30/Oct/17 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}:\:\:\:\frac{\partial^{\mathrm{2}} \mathrm{u}}{\partial\mathrm{x}\partial\mathrm{y}}\:=\:\mathrm{sin}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{y}\right),\:\:\:\mathrm{subjected}\:\mathrm{to}\:\mathrm{the}\:\mathrm{boundary} \\ $$$$\mathrm{conditions}\:\mathrm{at}\:\:\:\mathrm{y}\:=\:\frac{\pi}{\mathrm{2}},\:\:\:\:\frac{\partial\mathrm{u}}{\partial\mathrm{x}}\:=\:\mathrm{2x}\:\:\:\:\mathrm{and}\:\:\:\:\:\mathrm{x}\:=\:\pi,\:\:\:\:\mathrm{u}\:=\:\mathrm{2sin}\left(\mathrm{y}\right) \\ $$ Answered by mrW1 last updated on 31/Oct/17 $$\:\frac{\partial^{\mathrm{2}} \mathrm{u}}{\partial\mathrm{x}\partial\mathrm{y}}\:=\:\mathrm{sin}\left(\mathrm{x}\right)\mathrm{cos}\left(\mathrm{y}\right)…
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Question Number 154254 by Engr_Jidda last updated on 16/Sep/21 $${form}\:{a}\:{PDE}\:{from}\:{Z}\left({x},{y}\right)={xf}_{\mathrm{1}} \left({x}−{y}\right)+\mathrm{2}{xf}_{\mathrm{2}} \left(\mathrm{2}{x}+{y}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 22975 by selestian last updated on 24/Oct/17 Commented by selestian last updated on 24/Oct/17 $${solve}\:{the}\:{clock} \\ $$ Commented by Joel577 last updated on…