Question Number 122231 by fajri last updated on 15/Nov/20 Answered by mathmax by abdo last updated on 15/Nov/20 $$\mathrm{e}^{\mathrm{tA}} \:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{n}} \mathrm{A}^{\mathrm{n}} }{\mathrm{n}!}\:\:\mathrm{let}\:\mathrm{determine}\:\mathrm{A}^{\mathrm{n}} \\…
Question Number 122163 by benjo_mathlover last updated on 14/Nov/20 $$\:\left(\mathrm{2}{xy}−\mathrm{tan}\:{y}\right)\:{dx}\:+\:\left({x}^{\mathrm{2}} −{x}\:\mathrm{sec}\:^{\mathrm{2}} {y}\right)\:{dy}\:=\:\mathrm{0}\: \\ $$ Answered by liberty last updated on 14/Nov/20 $$\Rightarrow\mathrm{d}\left(\mathrm{x}^{\mathrm{2}} \mathrm{y}\right)−\mathrm{d}\left(\mathrm{xtan}\:\mathrm{y}\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{d}\left(\mathrm{x}^{\mathrm{2}}…
Question Number 56580 by subhankar10 last updated on 18/Mar/19 $$\left(\mathrm{D}^{\mathrm{3}} −\mathrm{2D}^{\mathrm{2}} +\mathrm{9D}−\mathrm{18}\right)\mathrm{y}=\mathrm{6cos3x} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 19/Mar/19 $${let}\:{y}={e}^{{mx}} \:{be}\:{a}\:{solution} \\ $$$${Dy}={me}^{{mx}}…
Question Number 122006 by bemath last updated on 13/Nov/20 $$\:\frac{{dy}}{{dx}}\:=\:\frac{{x}−{y}}{{x}+{y}}\:? \\ $$ Answered by bobhans last updated on 13/Nov/20 $${let}\:{y}\:=\:{zx}\:\Rightarrow\frac{{dy}}{{dx}}\:=\:{z}\:+\:{x}\:\frac{{dz}}{{dx}} \\ $$$$\Leftrightarrow\:{z}+{x}\:\frac{{dz}}{{dx}}\:=\:\frac{{x}−{zx}}{{x}+{zx}}\:=\:\frac{\mathrm{1}−{z}}{\mathrm{1}+{z}} \\ $$$$\Leftrightarrow\:{x}\:\frac{{dz}}{{dx}}\:=\:\frac{\mathrm{1}−{z}}{\mathrm{1}+{z}}\:−\:{z}\:=\:\frac{\mathrm{1}−\mathrm{2}{z}−{z}^{\mathrm{2}} }{\mathrm{1}+{z}}…
Question Number 121960 by bemath last updated on 13/Nov/20 $$\:\:\frac{{dy}}{{dx}}\:+\:\mathrm{3}{y}\:\mathrm{tan}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x}\: \\ $$ Answered by liberty last updated on 13/Nov/20 $$\:\mathrm{y}'\:+\:\mathrm{3y}\:\mathrm{tan}\:\mathrm{x}\:=\:\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x} \\ $$$$\:\mathrm{IF}\:\mathrm{u}\:=\:\mathrm{e}^{\int\:\mathrm{3}\:\mathrm{tan}\:\mathrm{x}\:\mathrm{dx}} \:=\:\mathrm{e}^{−\mathrm{3}\int\:\frac{\mathrm{d}\left(\mathrm{cos}\:\mathrm{x}\right)}{\mathrm{cos}\:\mathrm{x}}} =\:\mathrm{e}^{\mathrm{ln}\:\left(\mathrm{sec}\:^{\mathrm{3}} \mathrm{x}\:\right)}…
Question Number 121807 by abdelsalamalmukasabe last updated on 11/Nov/20 Answered by sukumara3@gmail.com last updated on 14/Nov/20 $${while}\:{cancelling}\:{step},{it}\:{is}\:{not}\:\mathrm{2}{yy}^{'} {x}.{But}\:{it}\:{is}\:\mathrm{2}{y}^{'} {y}^{'} {x}\:{only}. \\ $$ Terms of Service…
Question Number 121746 by KathleenannLabrador last updated on 11/Nov/20 $${what}\:{are}\:{the}\:{numerical}\:{coefficients}\:{in}\:{the}\:{quadration}\:{equation}\:−\mathrm{10}×\mathrm{2}−\mathrm{7}×+\mathrm{2}=\mathrm{0}?\:\:\:{a}.{a}=\mathrm{10}.{b}=.\mathrm{7}{c}=\mathrm{2}\:\:\:\:\:\:\:\:\:{b}.{a}=−\mathrm{2}.{b}=\mathrm{4}.{c}=−\mathrm{2}\:\:\:\:\:\:\:\:\:\:{c}.{a}=\mathrm{2}.{b}=\mathrm{7}.{c}=\mathrm{10}\:\:\:\:\:\:\:\:{d}.{a}=−\mathrm{10}.{b}=−\mathrm{7}.{c}=\mathrm{2} \\ $$ Answered by MJS_new last updated on 11/Nov/20 $$\mathrm{if}\:\mathrm{you}\:\mathrm{mean} \\ $$$$−\mathrm{10}{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\left({d}\right)\:{a}=−\mathrm{10};\:{b}=−\mathrm{7};\:{c}=\mathrm{2}…
Question Number 121687 by abdelsalamalmukasabe last updated on 10/Nov/20 Commented by bemath last updated on 11/Nov/20 $$\left(\frac{\mathrm{2}{y}^{\mathrm{4}} +\mathrm{1}}{{y}^{\mathrm{2}} }\right)\:{dx}\:+\:\left(\mathrm{4}{xy}\right)\:{dy}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}{y}^{\mathrm{4}} +\mathrm{1}\right)\:{dx}\:+\:\left(\mathrm{4}{xy}^{\mathrm{3}} \right)\:{dy}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\frac{{dy}}{{dx}}\:=\:\frac{−\left(\mathrm{2}{y}^{\mathrm{4}}…
Question Number 121633 by liberty last updated on 10/Nov/20 $$\mathrm{Given}\:\mathrm{implicit}\:\mathrm{expression}\:\begin{cases}{\mathrm{x}=\mathrm{tan}\:\alpha}\\{\mathrm{y}=\mathrm{tan}\:\beta}\end{cases} \\ $$$$\mathrm{where}\:\mathrm{x},\mathrm{y},\alpha\:\mathrm{and}\:\beta\:\mathrm{is}\:\mathrm{variable}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\frac{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{0}\:. \\ $$ Commented by Dwaipayan Shikari last updated on…
Question Number 121609 by benjo_mathlover last updated on 10/Nov/20 $$\:\mathrm{y}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:\mathrm{dx}\:−\mathrm{x}\left(\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\right)\:\mathrm{dy}\:=\:\mathrm{0}\: \\ $$$$ \\ $$ Answered by liberty last updated on 10/Nov/20…