Question Number 92114 by mhmd last updated on 04/May/20 Answered by mr W last updated on 05/May/20 $${u}={y}' \\ $$$${u}\frac{{du}}{{dy}}+\mathrm{3}{y}^{\mathrm{2}} {u}=\mathrm{0} \\ $$$$\Rightarrow{u}=\mathrm{0}\:\Rightarrow\frac{{dy}}{{dx}}=\mathrm{0}\:\Rightarrow{y}={C} \\ $$$$\Rightarrow\frac{{du}}{{dy}}+\mathrm{3}{y}^{\mathrm{2}}…
Question Number 92115 by mhmd last updated on 04/May/20 Answered by mr W last updated on 05/May/20 $${u}=\frac{{dy}}{{dx}} \\ $$$$\frac{{du}}{{dx}}+\mathrm{2}{x}\left(\mathrm{1}+{u}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\frac{{du}}{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }=−\mathrm{2}{xdx} \\…
Question Number 92041 by mhmd last updated on 04/May/20 $${solve}\::\left[\:\:\mathrm{2}{yy}''=\mathrm{1}+\left({y}'\right)^{\mathrm{2}} \:\right]\:{if}\:{y}\left(\mathrm{0}\right)=\mathrm{2}\:,\:{y}'\left(\mathrm{0}\right)=−\mathrm{1} \\ $$$$ \\ $$ Answered by mr W last updated on 04/May/20 $$\left({see}\:{also}\:{Q}\mathrm{92040}\right) \\…
Question Number 92040 by mhmd last updated on 04/May/20 $${y}^{''} +\left({y}'\right)^{\mathrm{2}} +{y}=\mathrm{0}\:\:\:{y}\left(\mathrm{0}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\:,\:{y}'\left(\mathrm{0}\right)=−\mathrm{1} \\ $$$${help}\:{me}\:{sir}\: \\ $$ Answered by mr W last updated on 04/May/20 $${let}\:{u}={y}'=\frac{{dy}}{{dx}}…
Question Number 26447 by yesaditya22@gmail.com last updated on 25/Dec/17 Commented by kaivan.ahmadi last updated on 25/Dec/17 $$\mathrm{y}=\mathrm{x}^{\mathrm{x}^{\mathrm{x}} } \Rightarrow\mathrm{lny}=\mathrm{x}^{\mathrm{x}} \mathrm{lnx}\Rightarrow\frac{\mathrm{y}^{'} }{\mathrm{y}}=\left(\mathrm{x}^{\mathrm{x}} \right)^{'} \mathrm{lnx}+\frac{\mathrm{x}^{\mathrm{x}} }{\mathrm{x}}\Rightarrow \\…
Question Number 157382 by cortano last updated on 22/Oct/21 Answered by mr W last updated on 23/Oct/21 $${g}\left({x}\right)={x}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$${g}'\left({x}\right)=\mathrm{1}−\frac{\mathrm{2}}{{x}^{\mathrm{3}} }=\mathrm{0}\:\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\in\left[\mathrm{1},\mathrm{2}\right] \\ $$$${g}\left({x}\right)_{{min}} =\frac{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{2}}}{\:\mathrm{2}}={k}…
Question Number 26288 by sakshiraj@gmail.com last updated on 23/Dec/17 $$\frac{\boldsymbol{{d}}}{\boldsymbol{{dx}}}\:\frac{\boldsymbol{{x}}−\mathrm{4}}{\mathrm{2}\sqrt{\boldsymbol{{x}}}} \\ $$ Answered by Joel578 last updated on 24/Dec/17 $$\mathrm{Use}\:\mathrm{quotient}\:\mathrm{rule} \\ $$$${u}\:=\:{x}\:−\:\mathrm{4}\:\:\rightarrow\:\:{u}'\:=\:\mathrm{1} \\ $$$${v}\:=\:\mathrm{2}\sqrt{{x}}\:\:\:\:\:\:\rightarrow\:\:{v}'\:=\:\frac{\mathrm{1}}{\:\sqrt{{x}}} \\…
Question Number 91812 by niroj last updated on 03/May/20 $$\:\mathrm{Solve}\:\mathrm{clairaut}'\mathrm{s}\:\mathrm{equation}\:\mathrm{and}\:\mathrm{find} \\ $$$$\:\mathrm{general}\:\mathrm{and}\:\mathrm{singular}\:\mathrm{solution}: \\ $$$$\left(\mathrm{i}\right)\:\mathrm{y}=\mathrm{px}+\mathrm{p}^{\mathrm{n}} \\ $$$$\:\left(\mathrm{ii}\right)\:\left(\mathrm{y}+\mathrm{1}\right)\mathrm{p}−\mathrm{xp}^{\mathrm{2}} +\mathrm{2}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$ Commented by…
Question Number 157278 by mnjuly1970 last updated on 21/Oct/21 Answered by mindispower last updated on 21/Oct/21 $${tanh}^{−} \left({x}\right)=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{{x}^{\mathrm{2}{m}+\mathrm{1}} }{\mathrm{2}{m}+\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−\sqrt{{x}}\right){tanh}^{−} \left(\sqrt{{x}}\right)}{{x}}{dx}…
Question Number 157247 by john_santu last updated on 21/Oct/21 $${F}\left({x},{y}\right)={x}^{\mathrm{2}} −\mathrm{2}{xy}+\mathrm{6}{y}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{2}{y}+\mathrm{45} \\ $$$${find}\:{x}\:\&{y}\:{such}\:{that}\:{F}\left({x},{y}\right)\:{minimum} \\ $$ Answered by FongXD last updated on 21/Oct/21 $$\mathrm{Given}:\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{2xy}+\mathrm{6y}^{\mathrm{2}}…