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Author: Tinku Tara

Question-212790

Question Number 212790 by RojaTaniya last updated on 24/Oct/24 Answered by A5T last updated on 24/Oct/24 $${x}−{y}=\left({y}−\mathrm{1}+{x}−\mathrm{1}\right)\left({y}−\mathrm{1}−{x}+\mathrm{1}\right) \\ $$$$\Rightarrow{x}={y}\:{or}\:−\mathrm{1}={x}+{y}−\mathrm{2}\Rightarrow{x}+{y}=\mathrm{1} \\ $$$$\Rightarrow{x}+\mathrm{1}={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\:{or}\:{x}+\mathrm{1}={x}^{\mathrm{2}} \Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={x}+{y}+\mathrm{2}=\mathrm{3}…

Question-212816

Question Number 212816 by Akayx last updated on 24/Oct/24 Answered by mahdipoor last updated on 24/Oct/24 $${f}\left({t}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}\left({t}−\mathrm{4}{k}\right)\left({u}_{\mathrm{4}{k}} −{u}_{\mathrm{4}{k}+\mathrm{2}} \right)= \\ $$$$\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\left({t}−\mathrm{4}{k}\right){u}_{\mathrm{4}{k}}…

i-generalized-Bessel-function-s-Laplace-Transform-L-TJ-z-s-s-2-1-s-2-1-s-0-R-L-T-Y-z-cot-pi-s-s-2-1-s-2-1-csc-pi-s-s-2-1

Question Number 212765 by issac last updated on 23/Oct/24 $$\mathrm{i}\:\:\mathrm{generalized}\:\boldsymbol{\mathrm{Bessel}}\:\boldsymbol{\mathrm{function}}'\mathrm{s} \\ $$$$\mathrm{Laplace}\:\mathrm{Transform} \\ $$$$\mathrm{L}.\mathrm{T}{J}_{\nu} \left({z}\right)=\frac{\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}\:,\:{s}\in\left[\mathrm{0},\infty\right)\:,\:\nu\in\mathbb{R} \\ $$$$\mathrm{L}.\mathrm{T}\:{Y}_{\nu} \left({z}\right)=\frac{\mathrm{cot}\left(\pi\nu\right)\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}−\frac{\mathrm{csc}\left(\pi\nu\right)\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{\nu}…