Question Number 126683 by bramlexs22 last updated on 23/Dec/20 $$\:{solve}\:\left({D}^{\mathrm{2}} +{a}\right){y}\:=\:\mathrm{tan}\:{ax}\: \\ $$ Answered by liberty last updated on 23/Dec/20 $$\left(\mathrm{1}\right)\:{Homogenous}\:{solution} \\ $$$$\:{y}_{{h}} =\:{A}\:\mathrm{cos}\:{ax}\:+\:{B}\:\mathrm{sin}\:{ax}\: \\…
Question Number 61107 by arcana last updated on 29/May/19 $${solve}\:{Cauchy}'{s}\:{problem} \\ $$$${x}'=\:{t}\:+\:\frac{\left(\mu{x}\right)^{\mathrm{2}} }{\mathrm{1}+\left(\mu{x}\right)^{\mathrm{2}} },\:\mu\in\mathbb{R} \\ $$$${x}\left(\mathrm{0}\right)=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 126620 by fajri last updated on 22/Dec/20 $$\frac{{d}^{\mathrm{2}} {y}}{{dx}}\:−\:\mathrm{3}\:\frac{{dy}}{{dx}}\:+\:\mathrm{2}{y}\:=\:{e}^{\mathrm{4}{t}} \:,\:{y}\left(\mathrm{0}\right)\:=\:\mathrm{1},\:{y}'\left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$$${solve}\:{with}\:{Laplace}\:{Transform}! \\ $$ Answered by Olaf last updated on 22/Dec/20 $$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}}…
Question Number 126543 by pticantor last updated on 21/Dec/20 $$\:\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\sum}}\underset{{i}=\mathrm{1}} {\overset{{k}} {\prod}}{i}^{{k}} =??? \\ $$ Answered by MJS_new last updated on 21/Dec/20 $$=\underset{{k}=\mathrm{1}}…
Question Number 126503 by benjo_mathlover last updated on 21/Dec/20 $$\:{solve}\:{f}\:'\left({x}\right)={f}\left({x}+\frac{\pi}{\mathrm{2}}\right)\: \\ $$ Commented by Dwaipayan Shikari last updated on 21/Dec/20 $${f}\left({x}\right)={cosx} \\ $$ Terms of…
Question Number 126470 by BHOOPENDRA last updated on 20/Dec/20 $${yy}''−\left({y}'\right)^{\mathrm{2}} ={e}^{{ax}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 126460 by BHOOPENDRA last updated on 20/Dec/20 $${yy}''−\left({y}'\right)^{\mathrm{2}} =\mathrm{0} \\ $$ Commented by BHOOPENDRA last updated on 20/Dec/20 $${thanks}\:{alot}\:{sir} \\ $$ Answered by…
Question Number 60921 by necx1 last updated on 27/May/19 $${x}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:{x}\frac{{dy}}{{dx}}\:+\:{y}=\mathrm{0} \\ $$$${please}\:{solve}\:{this}\:{Euler}\:{equation} \\ $$ Answered by tanmay last updated on 27/May/19 $${x}={e}^{{t}}…
Question Number 191960 by Rupesh123 last updated on 04/May/23 Answered by a.lgnaoui last updated on 05/May/23 $$\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)\right)=\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)+\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}^{\mathrm{2}} \right)\right. \\ $$$$\:\:\Rightarrow\boldsymbol{\mathrm{f}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)\right)=\boldsymbol{\mathrm{f}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\right)−\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{f}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{y}}\right)+\boldsymbol{\mathrm{f}}^{−\mathrm{1}}…
Question Number 126408 by morarupaula last updated on 20/Dec/20 Answered by Dwaipayan Shikari last updated on 20/Dec/20 $$\overset{\bullet\bullet\bullet} {{x}}+\overset{\bullet} {{x}}=\mathrm{0}\:\:\:\:\:\:\:{x}={e}^{\lambda{t}} \\ $$$$\lambda^{\mathrm{3}} +\lambda=\mathrm{0}\Rightarrow\lambda=\mathrm{0}\:,\:\lambda=\pm{i} \\ $$$${x}=\Lambda+\Gamma{e}^{\lambda{ti}}…