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Question-194735

Question Number 194735 by sonukgindia last updated on 14/Jul/23 Answered by TheHoneyCat last updated on 14/Jul/23 $$\mathrm{If}\:{x}\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant},\:\left({i}.{e}.\:{x}\left(\mathrm{2}\right)={x}×\mathrm{2}\:\right) \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{is}: \\ $$$${x}\left(\mathrm{2}\right)=\frac{\mathrm{4}\sqrt{\mathrm{7}}−\mathrm{2}\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{7}}+\sqrt{\mathrm{5}}} \\ $$$$ \\ $$$$\mathrm{But}\:\mathrm{I}'\mathrm{m}\:\mathrm{guessing}\:{x}\:\mathrm{is}\:\mathrm{here}\:\mathrm{a}\:\mathrm{function}……

Question-194654

Question Number 194654 by sonukgindia last updated on 12/Jul/23 Answered by MM42 last updated on 12/Jul/23 $$\left(\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{{tanx}} \right)^{{tanx}+\mathrm{2}} ={u} \\ $$$$\Rightarrow{u}+\frac{\mathrm{1}}{{u}}=\mathrm{6}\Rightarrow{u}^{\mathrm{2}} −\mathrm{6}{u}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{u}=\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}} \\…

Equation-J-1-z-Y-z-J-z-Y-1-z-2-piz-plz-Solve-this-Equation-J-z-is-First-Kind-Bessel-Function-Y-z-is-Second-Kind-Bessel-Function-aka-Neuman-Func

Question Number 194568 by MathedUp last updated on 10/Jul/23 $$\mathrm{Equation}.. \\ $$$${J}_{\boldsymbol{\mu}} ^{\left(\mathrm{1}\right)} \left({z}\right){Y}_{\boldsymbol{\mu}} \left({z}\right)−{J}_{\boldsymbol{\mu}} \left({z}\right){Y}_{\boldsymbol{\mu}} ^{\left(\mathrm{1}\right)} \left({z}\right)=−\frac{\mathrm{2}}{\pi{z}} \\ $$$$\mathrm{plz}……\mathrm{Solve}\:\mathrm{this}\:\mathrm{Equation}……. \\ $$$${J}_{\boldsymbol{\mu}} \left({z}\right)\:\mathrm{is}\:\mathrm{First}\:\mathrm{Kind}\:\mathrm{Bessel}\:\mathrm{Function} \\ $$$${Y}_{\boldsymbol{\mu}}…