Question Number 54663 by Mikael_Marshall last updated on 08/Feb/19 $${y}\:=\:{cos}\mathrm{2}{t}×{sen}\mathrm{2}{t} \\ $$$$ \\ $$$${y}'\:=\:? \\ $$ Commented by maxmathsup by imad last updated on 08/Feb/19…
Question Number 120058 by bramlexs22 last updated on 29/Oct/20 $$\:\frac{{d}^{\mathrm{2}} {y}}{{dx}}\:+{x}\:\frac{{dy}}{{dx}}\:−{y}=\mathrm{0} \\ $$ Answered by Olaf last updated on 29/Oct/20 $$ \\ $$$${y}''+{xy}'−{y}\:=\:\mathrm{0}\:\left(\mathrm{1}\right) \\ $$$${y}\:=\:{xu}…
Question Number 119933 by bramlexs22 last updated on 28/Oct/20 $$\:\frac{{dx}}{{dy}}\:−{y}\:=\:{xy}^{\mathrm{2}} \\ $$ Answered by bobhans last updated on 28/Oct/20 $$\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{1}}{{y}+{xy}^{\mathrm{2}} }\:=\:\frac{\frac{\mathrm{1}}{{y}^{\mathrm{2}} }}{\frac{\mathrm{1}}{{y}}+{x}} \\ $$$${let}\:\frac{\mathrm{1}}{{y}}=\:{u}\:\Rightarrow\frac{{du}}{{dx}}\:=\:−\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\:\frac{{dy}}{{dx}}…
Question Number 119567 by abdul88 last updated on 25/Oct/20 $$ \\ $$$$\left(\mathrm{1}\:−{x}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}\:} }\:+\:{x}\frac{{dy}}{{dx}}\:−\:{xy}\:=\:\frac{\mathrm{1}}{\mathrm{1}\:−\:{x}}\:,\:{x}\neq\mathrm{1} \\ $$$${has}\:{power}\:{series}\:{solution}\:{for}\:\mid{x}\mid<\mathrm{1}\: \\ $$ Answered by mathmax by abdo last updated…
Question Number 119510 by syamil last updated on 25/Oct/20 $${Differential}\:{Equation}\: \\ $$$$\left(\mathrm{1}\:−\:{x}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:{x}\frac{{dy}}{{dx}}\:−\:{xy}\:=\:\frac{\mathrm{1}}{\mathrm{1}\:−\:{x}}\:,\:{x}\:\neq\:\mathrm{1} \\ $$$${has}\:{the}\:{power}\:{series}\:{solution}\:{for}\:\mid{x}\mid<\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 119472 by Engr_Jidda last updated on 24/Oct/20 Answered by Sach last updated on 24/Oct/20 $$\begin{cases}{{y}''+{xy}'−{y}={x}^{\mathrm{2}} +\mathrm{1}}\\{{y}\left(\mathrm{0}\right)=\mathrm{1}}\\{{y}'\left(\mathrm{0}\right)=\mathrm{2}}\end{cases} \\ $$$$\mathrm{1}^{\mathrm{rst}} \mathrm{step}:\:{let}'{s}\:{supose}\:{there}\:{is}\:{such}\:{a}\:{solution}\:{and}\:{that}\:{it}\:{is}\:{a}\:{serie} \\ $$$$\left({sorry}\:{for}\:{the}\:{spelling}\:{I}'{m}\:{french}\:{so}\:{I}\:{hope}\:{you}\:{understand}\:{what}\:{I}\:{wright}\:{and}\:{that}\:{I}\:{did}\:{not}\:{misunderstand}\:{your}\:{question}\right. \\ $$$${let}'{s}\:{call}\:\left({a}_{{k}}…
Question Number 119464 by Engr_Jidda last updated on 24/Oct/20 $$\mathrm{Expand}\:\mathrm{as}\:\mathrm{far}\:\mathrm{as}\:\mathrm{the}\:\mathrm{term}\:\mathrm{in}\:\mathrm{x}^{\mathrm{3}} \\ $$$$\left(\mathrm{1}\right)\:\left(\mathrm{3}−\mathrm{2x}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{9}} \\ $$$$\left(\mathrm{2}\right)\:\left(\mathrm{1}−\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{8}} \\ $$ Answered by TANMAY PANACEA last updated on…
Question Number 184937 by MikeH last updated on 14/Jan/23 $$\mathrm{find}\:\mathrm{the}\:\mathrm{laplace}\:\mathrm{transform}\:\mathrm{of}\:\mathrm{the}\:\mathrm{differential} \\ $$$$\mathrm{equation}\:\mathrm{below} \\ $$$$\frac{{dy}}{{dt}}\:+\:\mathrm{5}{y}\left({t}\right)\:+\:\mathrm{6}\int_{\mathrm{0}} ^{{t}} {y}\left(\tau\right){d}\tau\:=\:{u}\left({t}\right)\:\mathrm{where}\:{y}\left(\mathrm{0}\right)\:=\:\mathrm{2} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 184936 by MikeH last updated on 14/Jan/23 $$\mathrm{Use}\:\mathrm{Laplace}\:\mathrm{transform}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{differential} \\ $$$$\mathrm{equation} \\ $$$$\:\frac{{d}^{\mathrm{2}} {v}\left({t}\right)}{{dt}^{\mathrm{2}} }\:+\mathrm{6}\frac{{dv}\left({t}\right)}{{dt}}\:+\:\mathrm{8}{v}\left({t}\right)\:=\:\mathrm{2}{u}\left({t}\right)\:\: \\ $$$$\mathrm{when}\:{v}\left(\mathrm{0}\right)\:=\:\mathrm{1}\:\mathrm{and}\:\overset{\bullet} {{v}}\left(\mathrm{0}\right)\:=\:−\mathrm{2} \\ $$ Answered by hmr last…
Question Number 119290 by bobhans last updated on 23/Oct/20 $$\:\:\left(\mathrm{3}{x}−{y}+\mathrm{1}\right)\:{dx}\:+\left(\mathrm{6}{x}+\mathrm{2}{y}−\mathrm{3}\right)\:{dy}\:=\:\mathrm{0}\: \\ $$ Answered by TANMAY PANACEA last updated on 23/Oct/20 $${dy}\left(\mathrm{6}{x}+\mathrm{2}{y}−\mathrm{3}\right)=\left(−\mathrm{3}{x}+{y}−\mathrm{1}\right){dx} \\ $$$$\frac{{dy}}{{dx}}=\frac{−\mathrm{3}{x}+{y}−\mathrm{1}}{\mathrm{6}{x}+\mathrm{2}{y}−\mathrm{3}} \\ $$$${x}={X}+{h}…