Question Number 104659 by bobhans last updated on 23/Jul/20 $${what}\:{is}\:{remainder}\:{of}\:\mathrm{12}!\:{in}\:{mod}\:\mathrm{17} \\ $$ Answered by bramlex last updated on 23/Jul/20 $${we}\:{can}\:{use}\:{Wilson}'{s}\:{theorem} \\ $$$${by}\:{which}\:\mathrm{16}!\:\equiv\:−\mathrm{1}\:\left({mod}\:\mathrm{17}\right) \\ $$$${Now}\:{use}\:{that}\: \\…
Question Number 104651 by Ar Brandon last updated on 23/Jul/20 $$\mathrm{Consider}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}; \\ $$$$\mathrm{x}''\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)=\mathrm{t}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{2t}\right). \\ $$$$\mathrm{Knowing}\:\mathrm{that}\:\mathrm{the}\:\mathrm{Complementary}\:\mathrm{Function}\:\mathrm{is}; \\ $$$$\mathrm{y}_{\mathrm{CF}} =\mathrm{acos}\left(\mathrm{t}\right)+\mathrm{bsin}\left(\mathrm{t}\right), \\ $$$$\mathrm{In}\:\mathrm{what}\:\mathrm{form}\:\mathrm{can}\:\mathrm{the}\:\mathrm{particular}\:\mathrm{integral}\:\mathrm{be}\:\mathrm{expressed} \\ $$$$\mathrm{so}\:\mathrm{as}\:\mathrm{to}\:\mathrm{obtain}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{given}\: \\ $$$$\mathrm{differential}\:\mathrm{equation}?…
Question Number 104599 by Ar Brandon last updated on 22/Jul/20 $$\left(\mathrm{1}+\mathrm{t}^{\mathrm{3}} \right)\mathrm{x}'\left(\mathrm{t}\right)+\mathrm{t}^{\mathrm{2}} \mathrm{x}\left(\mathrm{t}\right)+\mathrm{2}\left(\mathrm{x}\left(\mathrm{t}\right)\right)^{\mathrm{2}} =\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 104595 by Ar Brandon last updated on 22/Jul/20 $$\mathrm{Solve}\:\mathrm{the}\:\mathcal{DE} \\ $$$$\mathrm{x}''\left(\mathrm{t}\right)+\mathrm{2x}'\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)=\mathrm{te}^{−\mathrm{t}} \\ $$ Answered by mathmax by abdo last updated on 22/Jul/20 $$\left.\mathrm{h}\right)\rightarrow\mathrm{r}^{\mathrm{2}}…
Question Number 104592 by bemath last updated on 22/Jul/20 $$\left(\mathrm{2}{xy}−\mathrm{sec}\:^{\mathrm{2}} {x}\right)\:{dx}\:+\:\left({x}^{\mathrm{2}} +\mathrm{2}{y}\right){dy}\:=\:\mathrm{0} \\ $$ Answered by john santu last updated on 28/Jul/20 $${This}\:{is}\:{exact}\:{diff}\:{eq}\: \\ $$$${Here}\:{M}\left({x},{y}\right)=\:\mathrm{2}{xy}−\mathrm{sec}\:^{\mathrm{2}}…
Question Number 170116 by ali009 last updated on 16/May/22 $${solve}\:{the}\:{D}.{E}. \\ $$$$\left({x}+\mathrm{2}{y}−\mathrm{4}\right){dx}+\left(\mathrm{2}{x}+{y}−\mathrm{5}\right){dy}=\mathrm{0} \\ $$ Commented by mr W last updated on 17/May/22 $${see}\:{Q}\mathrm{169558} \\ $$…
Question Number 38876 by tawa tawa last updated on 30/Jun/18 $$\mathrm{solve}:\:\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:\:+\:\:\mathrm{2x}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:\:+\:\mathrm{5y}\:=\:\mathrm{0} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18 $${it}\:{is}\:{realy}\:{very}\:{good}\:{problem}…{in}\:{search}..{and} \\ $$$${finding}\:{way}\:{to}\:{solve}\:{it}……
Question Number 169913 by ali009 last updated on 12/May/22 $${solve}\:{the}\:{D}.{E} \\ $$$${dx}+\left(−{sin}\left({y}\right)+\frac{{x}}{{y}}\right){dy}=\mathrm{0} \\ $$ Commented by cortano1 last updated on 12/May/22 $$\:\mathrm{sin}\:{y}\:{dy}\:−\frac{{x}}{{y}}\:{dy}\:=\:{dx}\: \\ $$$$\:{y}\:\mathrm{sin}\:{y}\:{dy}\:=\:{x}\:{dy}\:+{y}\:{dx}\: \\…
Question Number 104357 by bemath last updated on 21/Jul/20 $$\left({x}+{y}+\mathrm{1}\right)\:\frac{{dy}}{{dx}}\:=\:\mathrm{1}\: \\ $$ Answered by john santu last updated on 21/Jul/20 $${let}\:{z}\:=\:{x}+{y}+\mathrm{1} \\ $$$$\frac{{dz}}{{dx}}\:=\:\mathrm{1}+\:\frac{{dy}}{{dx}}\:\Rightarrow\frac{{dy}}{{dx}}\:=\:\frac{{dz}}{{dx}}−\mathrm{1} \\ $$$$\left(\rightarrow\right)\:{z}.\left(\frac{{dz}}{{dx}}−\mathrm{1}\right)\:=\:\mathrm{1}\:…
Question Number 38802 by tawa tawa last updated on 30/Jun/18 $$\mathrm{solve}:\:\:\:\mathrm{y}''\left(\mathrm{1}\:+\:\mathrm{4x}^{\mathrm{2}} \right)\:−\:\mathrm{8y}\:=\:\mathrm{0} \\ $$ Answered by MrW3 last updated on 30/Jun/18 $${y}={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${y}'=\mathrm{2}{ax}+{b}…