Question Number 120811 by fajri last updated on 03/Nov/20 $${Find}\:{the}\:{singular}\:{point}\:{in}\:{the} \\ $$$${differential}\:{equation}\:: \\ $$$$ \\ $$$$\left({x}^{\mathrm{3}} \:−\:{x}^{\mathrm{2}} \:−\:\mathrm{9}{x}\:+\:\mathrm{9}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\mathrm{2}{x}\frac{{dy}}{{dx}}\:+\:\left({x}\:−\:\mathrm{3}\right){y}\:=\:\mathrm{0} \\ $$ Answered by 675480065…
Question Number 120810 by fajri last updated on 03/Nov/20 $${Determine}\:{the}\:{convergence}\:{intervval}\:{of}\:: \\ $$$$\underset{{n}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \:\left({x}\:−\:\mathrm{1}\right)^{{n}} \\ $$ Answered by 675480065 last updated on 03/Nov/20 $$\underset{\mathrm{n}\rightarrow\infty}…
Question Number 120595 by fajri last updated on 01/Nov/20 $$ \\ $$$${Solution}\:{from}\:{Differential}\:{Equation}\:{System}\:: \\ $$$$ \\ $$$$\left\{_{{x}_{\mathrm{2}} ^{'} \:=\:\:\mathrm{2}{x}_{\mathrm{1}} \:−\:\mathrm{3}{x}_{\mathrm{2}} \:\:,\:{x}_{\mathrm{2}} \:\left(\mathrm{0}\right)\:=\:−\mathrm{1}} ^{{x}_{\mathrm{1}} ^{'} \:=\:\mathrm{2}{x}_{\mathrm{1}} −\:\mathrm{2}{x}_{\mathrm{2}}…
Question Number 185833 by normans last updated on 28/Jan/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 54663 by Mikael_Marshall last updated on 08/Feb/19 $${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}}…