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solve-the-EDP-1-x-f-x-y-f-y-0-2-x-f-x-f-y-z-2-x-3-x-2-2-f-x-2-y-2-2-f-y-2-0-

Question Number 198355 by lapache last updated on 18/Oct/23 $${solve}\:{the}\:{EDP} \\ $$$$\mathrm{1}−\:\:\:{x}\frac{\partial{f}}{\partial{x}}−{y}\frac{\partial{f}}{\partial{y}}=\mathrm{0} \\ $$$$\mathrm{2}−\:\:\:{x}\frac{\partial{f}}{\partial{x}}−\frac{\partial{f}}{\partial{y}}=\frac{{z}^{\mathrm{2}} }{{x}} \\ $$$$ \\ $$$$\mathrm{3}−\:\:{x}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }−{y}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {f}}{\partial{y}^{\mathrm{2}} }=\mathrm{0}…

Question-198377

Question Number 198377 by sonukgindia last updated on 18/Oct/23 Answered by witcher3 last updated on 19/Oct/23 $$=\int_{\mathrm{0}} ^{\infty} \mathrm{xe}^{−\mathrm{4x}^{\mathrm{2}} } .\mathrm{xe}^{\frac{−\mathrm{9}}{\mathrm{x}^{\mathrm{2}} }} \mathrm{dx}\:\:\:\mathrm{IBP} \\ $$$$=\left[−\frac{\mathrm{e}^{−\mathrm{4x}^{\mathrm{2}}…

Question-198302

Question Number 198302 by sonukgindia last updated on 17/Oct/23 Answered by mr W last updated on 17/Oct/23 $${z}_{\mathrm{1}} =\mathrm{1}+{i}\sqrt{\mathrm{3}}=\mathrm{2}{e}^{{i}\left(\frac{\pi}{\mathrm{3}}\right)} \\ $$$${z}_{\mathrm{2}} =\mathrm{2}{e}^{{i}\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)} =\mathrm{2}{e}^{{i}\pi} =−\mathrm{2} \\…

Question-198313

Question Number 198313 by sonukgindia last updated on 17/Oct/23 Commented by AST last updated on 17/Oct/23 $${n}^{\mathrm{2}} \equiv\mathrm{34}\left({mod}\:\mathrm{15}\right)?\Rightarrow{n}^{\mathrm{2}} \equiv\mathrm{4}\left({mod}\:\mathrm{15}\right) \\ $$$$\Rightarrow{n}=\mathrm{2},\mathrm{7},\mathrm{8},\mathrm{13}\left({mod}\:\mathrm{15}\right) \\ $$ Terms of…