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}} \\ $$$$\lambda^{\mathrm{3}} +\lambda=\mathrm{0}\Rightarrow\lambda=\mathrm{0}\:,\:\lambda=\pm{i} \\ $$$${x}=\Lambda+\Gamma{e}^{\lambda{ti}}…
Question Number 126285 by bramlexs22 last updated on 19/Dec/20 $$\:\:\Rightarrow{solve}\:{x}^{\mathrm{2}} {y}\:=\:\int_{\mathrm{1}} ^{\:{x}^{\mathrm{2}} } {f}\left(\sqrt{{t}}\right){dx}+{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}}+\mathrm{1} \\ $$$${y}={f}\left({x}\right)=? \\ $$ Answered by liberty last updated on…
Question Number 60694 by maxmathsup by imad last updated on 24/May/19 $${solve}\:\:\left({x}^{\mathrm{3}} −{x}\right){y}^{''} \:\:\:−\mathrm{2}{x}\:{y}^{'} \:+\mathrm{3}{y}\:={xln}\left(\mathrm{1}+{x}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 126229 by MathSh last updated on 18/Dec/20 $$\left(\mathrm{1}+{y}\right){x}'={x} \\ $$ Answered by Dwaipayan Shikari last updated on 18/Dec/20 $$\left(\mathrm{1}+{y}\right)\frac{{dx}}{{dy}}={x}\:\Rightarrow\int\frac{{dy}}{\mathrm{1}+{y}}=\int\frac{{dx}}{{x}}\:\Rightarrow\mathrm{1}+{y}={Cx} \\ $$ Answered by…
Question Number 60693 by maxmathsup by imad last updated on 24/May/19 $${solve}\:\:\left(\mathrm{2}+{e}^{−{x}} \right){y}^{'} \:\:+\left(\mathrm{2}{x}+{e}^{{x}} \right){y}\:={e}^{{x}} {sinx}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 126223 by liberty last updated on 18/Dec/20 $$\:\:\frac{{dy}}{{dx}}\:+\:\frac{{x}^{\mathrm{2}} {y}}{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} }\:=\:\mathrm{0} \\ $$ Commented by bramlexs22 last updated on 18/Dec/20 $$\:{v}+{x}\frac{{dv}}{{dx}}\:=\:\frac{{x}^{\mathrm{3}} {v}}{{x}^{\mathrm{3}} \left(\mathrm{1}+{v}^{\mathrm{3}}…