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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} \\ $$ Commented by Joel577 last updated on…
Question Number 88492 by jagoll last updated on 11/Apr/20 $$\mathrm{y}''\:−\mathrm{4y}'+\mathrm{5y}\:=\:\mathrm{1}+\mathrm{8cos}\:\mathrm{x}+\mathrm{e}^{\mathrm{2x}} \\ $$ Commented by niroj last updated on 11/Apr/20 $$\:\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }−\mathrm{4}\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}+\mathrm{5}\boldsymbol{\mathrm{y}}=\:\mathrm{1}+\mathrm{8}\boldsymbol{\mathrm{cos}}\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{e}}^{\mathrm{2}\boldsymbol{\mathrm{x}}} \\ $$$$\:\:\left(\mathrm{D}^{\mathrm{2}} −\mathrm{4D}+\mathrm{5}\right)\mathrm{y}=\:\mathrm{1}+\mathrm{8cos}\:\mathrm{x}\:+\mathrm{e}^{\mathrm{2x}}…
Question Number 88238 by M±th+et£s last updated on 09/Apr/20 $${solve}\: \\ $$$$\left(\mathrm{3}{x}^{\mathrm{5}} {y}^{\mathrm{4}} +\mathrm{4}{y}\right){dx}+\left(\mathrm{2}{x}^{\mathrm{6}} {y}^{\mathrm{3}} +\mathrm{3}{x}\right){dy}=\mathrm{0} \\ $$ Answered by ajfour last updated on 09/Apr/20…
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Question Number 22535 by Arjun Daniel last updated on 19/Oct/17 $$\mathrm{H}{ow}\:{to}\:{solve}\:{this}\:{homogeneous}\:{equation}.\:{Can}\:{u}\:{help}\:{me}\:{plz} \\ $$$${Q}.\:{solve}\:\left({x}\:{sin}\:\frac{{y}}{{x}}\right){dy}\:−\left({y}\:{sin}^{−\mathrm{1}} \:\frac{{y}}{{x}}\right){dx}=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com