Question Number 96125 by john santu last updated on 30/May/20 $$\mathrm{4x}^{\mathrm{2}} \mathrm{y}''\:+\mathrm{12xy}'\:+\:\mathrm{3y}\:=\:\mathrm{0} \\ $$ Answered by john santu last updated on 30/May/20 $$\mathrm{cauchy}\:−\:\mathrm{euler}\: \\ $$$$\mathrm{4r}^{\mathrm{2}}…
Question Number 96117 by bobhans last updated on 30/May/20 $$\left(\mathrm{1}−\mathrm{2}{xy}\right)\:{dx}\:+\:\left(\mathrm{4}{y}^{\mathrm{3}} −{x}^{\mathrm{2}} \right)\:{dy}\:=\:\mathrm{0}\: \\ $$ Answered by john santu last updated on 30/May/20 Commented by john…
Question Number 96065 by john santu last updated on 29/May/20 $$\mathrm{y}'\:+\:\mathrm{y}\:=\:\mathrm{x}\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{y}^{\mathrm{2}} } \\ $$ Answered by bobhans last updated on 29/May/20 $$\mathrm{Bernoulli}\:\mathrm{eq}\: \\ $$$$\mathrm{v}\:=\:\mathrm{y}^{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}} \:=\:\mathrm{y}^{\frac{\mathrm{1}}{\mathrm{3}}}…
Question Number 96051 by i jagooll last updated on 29/May/20 $$\left(\mathrm{x}−\mathrm{y}\right)\:\mathrm{dx}\:+\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\:\mathrm{dy}\:=\:\mathrm{0}\: \\ $$ Commented by bobhans last updated on 29/May/20 $$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{y}−\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:.\:\mathrm{set}\:\mathrm{y}\:=\:\mathrm{vx}\:…
Question Number 95933 by john santu last updated on 28/May/20 $$\mathrm{y}'''+\mathrm{2y}'−\mathrm{3y}=\:\mathrm{e}^{\mathrm{x}} \:\left(\mathrm{x}+\mathrm{3}\right)\: \\ $$ Commented by Sourav mridha last updated on 29/May/20 $$\mathrm{are}\:\mathrm{you}\:\mathrm{sure}\:\mathrm{it}\:\mathrm{is}\:\mathrm{3rd}\:\mathrm{order}\:\mathrm{ODE},\mathrm{i}\:\mathrm{think} \\ $$$$\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{2nd}\:\mathrm{order}.…
Question Number 95801 by bobhans last updated on 27/May/20 $$\mathrm{2y}''−\mathrm{y}^{'} =\mathrm{1};\:\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{0}\:;\:\mathrm{y}'\left(\mathrm{0}\right)=\mathrm{1} \\ $$ Answered by mr W last updated on 27/May/20 $${u}={y}' \\ $$$${y}''=\frac{{du}}{{dx}} \\…
Question Number 95748 by bobhans last updated on 27/May/20 $$\mathrm{y}''\:=\:\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x} \\ $$ Answered by mathmax by abdo last updated on 27/May/20 $$\Rightarrow\mathrm{y}^{'} \:=\int\left(\mathrm{sinx}−\mathrm{cosx}\right)\mathrm{dx}\:=−\mathrm{cosx}−\mathrm{sinx}\:+\mathrm{c}\:\Rightarrow\mathrm{y}\:=\int\left(−\mathrm{cosx}−\mathrm{sinx}\:+\mathrm{c}\right)\mathrm{dx} \\ $$$$−\mathrm{sinx}\:+\mathrm{cosx}\:+\mathrm{cx}\:+\lambda…
Question Number 161272 by Ar Brandon last updated on 15/Dec/21 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{systeme}\:\left(\Sigma\right)\:\mathrm{below}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\Sigma\right)\begin{cases}{\overset{.} {{x}}\left({t}\right)={x}\left({t}\right)+\mathrm{2}{y}\left({t}\right)+{t}}\\{\overset{.} {{y}}\left({t}\right)=−\mathrm{4}{x}\left({t}\right)−\mathrm{3}{y}\left({t}\right)}\end{cases}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 95679 by Rio Michael last updated on 26/May/20 $$\mathrm{The}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{motion}\:\mathrm{for}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\mathrm{along}\:\mathrm{the}\:\mathrm{O}{X}\:\mathrm{axes}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{7}}\:\frac{{dt}}{{dx}}\:+\:\mathrm{4}{x}\:=\:\mathrm{0}. \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{motion}\:\mathrm{is}\:\mathrm{an}\:\mathrm{oscilatory}\:\mathrm{motion}\:\mathrm{hence}\:\mathrm{find} \\ $$$$\mathrm{its}\:\mathrm{period}. \\ $$ Commented by Tony Lin…
Question Number 95677 by Rio Michael last updated on 26/May/20 $$\:\mathrm{Given}\:{f}\left({x}\right)\:=\:\frac{\mathrm{ln}\:{x}}{{x}−\mathrm{1}} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{of}\:{f}. \\ $$$$\left(\mathrm{b}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{limts}\:\mathrm{of}\:{f}\:\mathrm{at}\:\mathrm{the}\:\mathrm{boundary}\:\mathrm{of}\:\mathrm{its}\:\mathrm{domain} \\ $$$$\mathrm{hence}\:\mathrm{state}\:\mathrm{the}\:\mathrm{asymptote}\:\mathrm{of}\:{y}\:=\:{f}\left({x}\right). \\ $$ Answered by mathmax by abdo last…