Question Number 214678 by issac last updated on 16/Dec/24 $$\mathrm{solve} \\ $$$$\mathrm{partial}\:\mathrm{differantial}\:\mathrm{equation} \\ $$$${x}\frac{\partial{f}\left({x},{y}\right)}{\partial{x}}+{y}\frac{\partial{f}\left({x},{y}\right)}{\partial{y}}={f}\left({x},{y}\right)\mathrm{ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$$\frac{\partial^{\mathrm{2}} {f}\left({x},{y}\right)}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {f}\left({x},{y}\right)}{\partial{y}^{\mathrm{2}} }=\mathrm{0} \\ $$ Answered…
Question Number 214667 by issac last updated on 16/Dec/24 $$\mathrm{let}'\mathrm{s}\:\mathrm{define}\:\mathrm{linear}\:\mathrm{differantial}\:\mathrm{operator}\:\mathcal{D} \\ $$$$\mathrm{as}\:\mathcal{D}={z}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}\left({z}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}\right)+{z}\left(\mathrm{1}−\left(\frac{\alpha}{{z}}\right)^{\mathrm{2}} \right) \\ $$$$\mathrm{when} \\ $$$$\mathcal{D}{f}\left({z}\right)=\left\{{z}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}\left({z}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}\right)+{z}\left(\mathrm{1}−\left(\frac{\alpha}{{z}}\right)^{\mathrm{2}} \right)\right\}{f}\left({z}\right)=\mathrm{0} \\ $$$${f}\left({z}\right)=? \\ $$ Terms of Service…
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Question Number 214631 by issac last updated on 14/Dec/24 $$\mathrm{why}\:\int\int_{\partial\mathrm{D}} \overset{\rightarrow} {\boldsymbol{\mathrm{B}}}\centerdot\mathrm{d}\overset{\rightarrow} {\boldsymbol{\mathrm{a}}}=\mathrm{0} \\ $$$$\overset{\rightarrow} {\boldsymbol{\mathrm{B}}}\:\mathrm{ia}\:\mathrm{magnetic}\:\mathrm{Field} \\ $$ Answered by MrGaster last updated on 24/Dec/24…
Question Number 214587 by issac last updated on 13/Dec/24 $$\mathrm{Does}\:\mathrm{Magnetic}\:\mathrm{Monopole}\:\mathrm{really}\:\mathrm{not}\:\mathrm{exist}? \\ $$$$\mathrm{In}\:\mathrm{maxwell}\:\mathrm{Equation} \\ $$$$\int\int_{\:\partial{V}} \:\overset{\rightarrow} {\boldsymbol{\mathrm{B}}}\centerdot\:\mathrm{d}\boldsymbol{\mathrm{a}}=\mathrm{0}\:\:,\overset{\rightarrow} {\boldsymbol{\mathrm{B}}}\:\mathrm{is}\:\mathrm{Magnetic}\:\mathrm{field} \\ $$ Answered by wrya last updated on…
Question Number 214556 by MathematicalUser2357 last updated on 12/Dec/24 $$\mathrm{Hey}\:\mathrm{Tinku}\:\mathrm{Tara}, \\ $$$$\mathrm{I}\:\mathrm{got}\:\mathrm{some}\:\mathrm{plot}\:\mathrm{issues}. \\ $$$$\mathrm{When}\:\mathrm{I}\:\mathrm{click}\:\mathrm{the}\:\mathrm{plot}\:\mathrm{button},\:\mathrm{it}\:\mathrm{displays}\:\mathrm{the}\:\mathrm{error}\:\mathrm{message} \\ $$$$“\mathrm{Check}\:\mathrm{if}\:\mathrm{the}\:\mathrm{variable}\:\mathrm{name}\:\mathrm{is}\:{x}\:\mathrm{and}\:\mathrm{you}\:\mathrm{are}\:\mathrm{logged}\:\mathrm{in}.''. \\ $$ Commented by MathematicalUser2357 last updated on 12/Dec/24…
Question Number 214563 by MathematicalUser2357 last updated on 12/Dec/24 $$ \\ $$ Commented by mr W last updated on 12/Dec/24 $${you}\:{don}'{t}\:{need}\:{to}\:{make}\:{problems}, \\ $$$$\left.{but}\:{just}\:{to}\:{solve}\:{them}\::\right) \\ $$…
Question Number 214511 by issac last updated on 11/Dec/24 $$\mathrm{Let}'\mathrm{s}\:{R}\left({z}\right)\:\mathrm{define}\:\mathrm{as}\: \\ $$$${R}\left({z}\right)=\frac{\pi\:\int_{\mathrm{0}} ^{\:{z}} \:{f}^{\mathrm{2}} \left({t}\right)\mathrm{d}{t}}{\mathrm{2}\pi\:\int_{\mathrm{0}} ^{\:{z}} \:{f}\left({t}\right)\sqrt{\mathrm{1}+\left({f}^{\left(\mathrm{1}\right)} \left({t}\right)\right)^{\mathrm{2}} }\mathrm{d}{t}} \\ $$$$\mathrm{and}\:\mathrm{both}\:\mathrm{integral} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:{f}^{\:\mathrm{2}}…
Question Number 214516 by MathematicalUser2357 last updated on 11/Dec/24 $$\mathrm{Hey}\:\mathrm{Tinku}\:\mathrm{Tara}, \\ $$$$\mathrm{I}\:\mathrm{got}\:\mathrm{some}\:\mathrm{plot}\:\mathrm{issues}. \\ $$$$\mathrm{When}\:\mathrm{I}\:\mathrm{click}\:\mathrm{the}\:\mathrm{plot}\:\mathrm{button},\:\mathrm{it}\:\mathrm{displays}\:\mathrm{the}\:\mathrm{error}\:\mathrm{message} \\ $$$$“\mathrm{Check}\:\mathrm{if}\:\mathrm{the}\:\mathrm{variable}\:\mathrm{name}\:\mathrm{is}\:{x}\:\mathrm{and}\:\mathrm{you}\:\mathrm{are}\:\mathrm{logged}\:\mathrm{in}.''. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 214471 by issac last updated on 09/Dec/24 $${d}^{\mathrm{2}} −{d}+\mathrm{2}=\mathrm{0} \\ $$$$\underset{{k};\:{d}^{\mathrm{2}} −{d}+\mathrm{2}=\mathrm{0}} {\sum}\:\frac{\mathrm{1}}{{k}}=?? \\ $$ Answered by mr W last updated on 09/Dec/24…