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…
Question Number 95600 by Tony Lin last updated on 26/May/20 $${solve}\:{the}\:{differential}\:{equation} \\ $$$${L}\frac{{d}^{\mathrm{2}} {q}}{{dt}^{\mathrm{2}} }+{R}\frac{{dq}}{{dt}}+\frac{\mathrm{1}}{{C}}{q}=\varepsilon{cos}\omega{t} \\ $$$${which}\:{is}\:{in}\:{R}.{L}.{C}\:{circuit}\:{with}\:{forced}\:{oscillation} \\ $$$${where}\:{L}\:{is}\:{inductance} \\ $$$${R}\:{is}\:{resistance} \\ $$$${C}\:{is}\:{capacitanace} \\ $$$${q}\:{is}\:{charge}…
Question Number 160903 by blackmamba last updated on 08/Dec/21 $$\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}\:'=\:\mathrm{2}{xy}\:+\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} \: \\ $$ Answered by GuruBelakangPadang last updated on 09/Dec/21 $$\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}\:'=\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)'{y}\:+\left(\mathrm{1}+{x}^{\mathrm{2}}…
Question Number 95316 by i jagooll last updated on 24/May/20 $$\frac{{dy}}{{dx}}−{y}\:=\:{xy}^{\mathrm{5}} \: \\ $$ Answered by john santu last updated on 24/May/20 Terms of Service…
Question Number 95119 by Mr.D.N. last updated on 27/May/20 $$\:\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equations}:− \\ $$$$\bigstar.\left(\mathrm{x}\:\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:−\mathrm{x}\:\mathrm{cos}\:\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\:\mathrm{y}\:\mathrm{cos}\:\mathrm{x}=\mathrm{0}. \\ $$ Answered by niroj last updated on 23/May/20 $$\:\:\mathrm{2}.\:\left(\mathrm{x}\:\mathrm{sin}\:\mathrm{x}\:+\:\mathrm{cos}\:\mathrm{x}\:\right)\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}}…
Question Number 95098 by i jagooll last updated on 23/May/20 $$\left[\:\frac{\mathrm{y}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:+\:\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:\right]\:\mathrm{dx}\:+\:\left[\frac{\mathrm{y}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }−\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:\right]\mathrm{dy}=\mathrm{0} \\ $$ Answered by bobhans last…