Question Number 115968 by Engr_Jidda last updated on 29/Sep/20 Answered by 1549442205PVT last updated on 30/Sep/20 $$\mathrm{Put}\:\mathrm{u}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{1};\mathrm{v}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{y}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{The}\:\mathrm{condition}\:\mathrm{Cauchy}−\mathrm{Rieman}\:\mathrm{for} \\ $$$$\mathrm{the}\:\mathrm{analytical}\:\mathrm{is}\:\frac{\partial\mathrm{u}}{\partial\mathrm{x}}=\frac{\partial\mathrm{v}}{\partial\mathrm{y}}\left(\mathrm{1}\right),\frac{\partial\mathrm{u}}{\partial\mathrm{y}}=−\frac{\partial\mathrm{v}}{\partial\mathrm{x}}\left(\mathrm{2}\right) \\ $$$$\mathrm{We}\:\mathrm{have}\:\frac{\partial\mathrm{u}}{\partial\mathrm{x}}=\mathrm{2x},\frac{\partial\mathrm{u}}{\partial\mathrm{y}}=\mathrm{0},\frac{\partial\mathrm{v}}{\partial\mathrm{x}}=\mathrm{0},\frac{\partial\mathrm{v}}{\partial\mathrm{y}}=\mathrm{2y}…
Question Number 115869 by bobhans last updated on 29/Sep/20 $$\:\frac{{dy}}{{dx}}\:+\mathrm{2}{y}\:=\:{x}^{\mathrm{3}} .{e}^{−\mathrm{2}{y}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 115771 by bemath last updated on 28/Sep/20 $${y}''−{y}'−\mathrm{2}{y}={e}^{\mathrm{2}{x}} .\mathrm{cos}\:^{\mathrm{2}} {x} \\ $$ Answered by TANMAY PANACEA last updated on 28/Sep/20 Answered by mathmax…
Question Number 115721 by bemath last updated on 28/Sep/20 $$\left({D}^{\mathrm{2}} −\mathrm{6}{D}+\mathrm{9}\right){y}\:=\:\frac{{e}^{\mathrm{3}{x}} }{{x}^{\mathrm{2}} } \\ $$ Commented by mohammad17 last updated on 28/Sep/20 $${m}^{\mathrm{2}} −\mathrm{6}{m}+\mathrm{9}=\mathrm{0}\Rightarrow\left({m}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{m}_{\mathrm{1}}…
Question Number 115706 by Rio Michael last updated on 27/Sep/20 $$\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square},\:{A}\left({t}\right)\:\mathrm{is}\:\mathrm{increased} \\ $$$$\mathrm{at}\:\mathrm{a}\:\mathrm{rate}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{its}\:\mathrm{perimeter} \\ $$$$\mathrm{write}\:\mathrm{a}\:\mathrm{differential}\:\mathrm{equation}\:\mathrm{that}\:{A}\left({t}\right) \\ $$$$\mathrm{satisfy}\:,\:\mathrm{starting}\:\mathrm{from}\:\frac{{dA}}{{dt}}\:= \\ $$ Commented by Dwaipayan Shikari last updated…
Question Number 115539 by bemath last updated on 26/Sep/20 $$\frac{{dy}}{{dx}}\:=\:\frac{{e}^{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)} \:−{y}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$ Answered by bobhans last updated on 26/Sep/20 $$\frac{{dy}}{{dx}}\:+\:\frac{{y}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\:\frac{{e}^{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}…
Question Number 49678 by Rio Michael last updated on 09/Dec/18 Commented by Rio Michael last updated on 09/Dec/18 $${A}\:{closed}\:{Cylinder}\:{can}\:{is}\:{made}\:{from}\:{a}\:{fix}\:{amount}\:{A}\:{of}\:{thin} \\ $$$${metal}\:{sheet}.{Find}\:{the}\:{relation}\:{between}\:{its}\:{radius}\:{and}\:{height}, \\ $$$${if}\:{it}\:{is}\:{to}\:{contain}\:{the}\:{maximum}\:{possible}\:{volume} \\ $$$$\left({Note}:\:{Wastage}\:{of}\:{material}\:{ignored}\right).…
Question Number 114980 by srijachakraborty last updated on 22/Sep/20 Answered by Dwaipayan Shikari last updated on 22/Sep/20 $$\left({y}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} {y}\right)+\left({x}^{\mathrm{4}} −\mathrm{2}{xy}^{\mathrm{3}} \right)\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{2}{x}^{\mathrm{3}} {vx}−{v}^{\mathrm{4}}…
Question Number 114978 by srijachakraborty last updated on 22/Sep/20 $$ \\ $$$$\:\:{Solve}\: \\ $$$$\:{x}^{\mathrm{2}} {dy}\:+\:{y}\left({x}+{y}\right){dx}=\mathrm{0} \\ $$ Answered by soumyasaha last updated on 22/Sep/20 Answered…
Question Number 114722 by Dwaipayan Shikari last updated on 20/Sep/20 $${a}_{{n}} \frac{{d}^{{n}} \Psi}{{dt}^{{n}} }+{a}_{{n}−\mathrm{1}} \frac{{d}^{{n}−\mathrm{1}} \Psi}{{dt}^{{n}−\mathrm{1}} }+…..+{a}_{\mathrm{1}} \frac{{d}\Psi}{{dt}}+{a}_{\mathrm{0}} \Psi=\mathrm{0} \\ $$$${Is}\:{it}\:{solvable}??? \\ $$ Answered by…