Question Number 125693 by fajri last updated on 13/Dec/20 $${find}\:: \\ $$$$ \\ $$$$\frac{{d}^{\mathrm{4}} {y}}{{dx}^{\mathrm{4}} }\:+\:\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }\:−\:\mathrm{7}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:−\:\frac{{dy}}{{dx}\:\:}\:+\:\mathrm{6}{y}\:=\:\mathrm{0} \\ $$$$ \\ $$$${for}\:{y}\left(\mathrm{0}\right)\:=\:\mathrm{1},\:{y}'\left(\mathrm{0}\right)\:=\:\mathrm{0},\:{y}''\left(\mathrm{0}\right)\:=\:−\mathrm{2}\:,\:{y}'''\left(\mathrm{0}\right)\:=\:−\mathrm{1} \\…
Question Number 125692 by fajri last updated on 13/Dec/20 $${find}\:{soultion}\:: \\ $$$$ \\ $$$${x}'\:=\:\begin{pmatrix}{−\mathrm{2}\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:−\mathrm{2}}\end{pmatrix}\:{x}\:+\:\begin{pmatrix}{\mathrm{2}{e}^{−{n}} }\\{\mathrm{3}{n}}\end{pmatrix} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 125691 by fajri last updated on 13/Dec/20 $${find}\:{solution}\:: \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}\:} }\:+\:\mathrm{4}\frac{{dy}}{{dx}}\:=\:\mathrm{3}\:{cosec}\:\theta \\ $$ Answered by mathmax by abdo last updated on 14/Dec/20…
Question Number 125664 by fajri last updated on 12/Dec/20 $$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{4}\frac{{dy}}{{dx}}\:\:=\:\mathrm{3}\:{cosec}\:\theta \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 59907 by Sardor2211 last updated on 15/May/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 59872 by necx1 last updated on 15/May/19 $${solve}\:{the}\:{o}\:{d}\:{e} \\ $$$$\left(\mathrm{1}+{siny}\right){dx}=\left\{\mathrm{2}{y}\mathrm{cos}\:{y}−{x}\left({secy}+{tany}\right)\right\}{dy} \\ $$ Answered by ajfour last updated on 15/May/19 $$\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{x}\left(\mathrm{sec}\:\mathrm{y}+\mathrm{tan}\:\mathrm{y}\right)=\mathrm{2ycos}\:\mathrm{y} \\ $$$$\mathrm{e}^{\int\frac{\:\mathrm{1}+\mathrm{sin}\:\mathrm{y}}{\mathrm{cos}\:\mathrm{y}}\mathrm{dy}} =\mathrm{e}^{\int\mathrm{tan}\:\frac{\mathrm{y}}{\mathrm{2}}\mathrm{dy}}…
Question Number 190851 by Rupesh123 last updated on 13/Apr/23 Answered by mr W last updated on 13/Apr/23 $$\mathrm{cos}\:{y}'=\mathrm{sin}\:{y} \\ $$$$\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−{y}'\right)=\mathrm{sin}\:{y} \\ $$$$\frac{\pi}{\mathrm{2}}−{y}'={n}\pi+\left(−\mathrm{1}\right)^{{n}} {y} \\ $$$$−{y}'=\left(−\mathrm{1}\right)^{{n}}…
Question Number 59720 by Tawa1 last updated on 13/May/19 $$\mathrm{Find}\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:\:\mathrm{from}\:\mathrm{first}\:\mathrm{principle},\:\:\mathrm{if}\:\:\:\:\mathrm{y}\:=\:\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right) \\ $$ Commented by maxmathsup by imad last updated on 14/May/19 $$\frac{{dy}}{{dx}}\:=\mathrm{2}{sinx}\:{cosx}\:\:={sin}\left(\mathrm{2}{x}\right). \\ $$…
Question Number 59608 by Andrew Foxman last updated on 12/May/19 $${For}\:{your}\:{development}\:{solve}\:{this} \\ $$$$\frac{{d}^{\mathrm{2}} {r}}{{dt}^{\mathrm{2}} }=\frac{{A}}{{r}^{\mathrm{2}} \left({t}\right)}\:\:{where}\:{r}\left({t}\right)=\alpha{t}^{\beta} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 124923 by bemath last updated on 07/Dec/20 $$\:\frac{{dy}}{{dx}}\:−{y}\:=\:\mathrm{3cot}\:{x}\:.{e}^{\mathrm{sin}\:{x}} \: \\ $$$$ \\ $$ Commented by mohammad17 last updated on 07/Dec/20 $${P}\left({x}\right)=−\mathrm{1}\:\:\:\:\:,\:\:\:\:{Q}\left({x}\right)=\mathrm{3}{cotx}.{e}^{{sinx}} \\ $$$$…