Question Number 103008 by bobhans last updated on 12/Jul/20 $$\frac{{dy}}{{dx}}\:−\:{y}.\mathrm{tan}\:{x}\:=\:{e}^{{x}} .\mathrm{sec}\:{x} \\ $$ Answered by bemath last updated on 12/Jul/20 $${IF}\:{u}\left({x}\right)\:=\:{e}^{−\int\:{tan}\:{x}\:{dx}\:} =\:{e}^{\int\:\frac{{d}\left({cos}\:{x}\right)}{{cos}\:{x}}} = \\ $$$$\mathrm{cos}\:{x}.\:\Leftrightarrow\mathrm{cos}\:{x}\:\frac{{dy}}{{dx}}\:−{y}.\mathrm{sin}\:{x}\:=\:{e}^{{x}}…
Question Number 101303 by Dwaipayan Shikari last updated on 01/Jul/20 $${i}^{{i}^{{i}.\infty} } =? \\ $$ Commented by Dwaipayan Shikari last updated on 01/Jul/20 $${If}\:{there}\:{any}\:{possible}\:{solutions}? \\…
Question Number 100327 by bobhans last updated on 26/Jun/20 $$\mathcal{S}\mathrm{olve}\:\mathrm{x}^{\mathrm{2}} \mathrm{y}''−\mathrm{3xy}'−\mathrm{5y}=\mathrm{0} \\ $$ Commented by bobhans last updated on 26/Jun/20 $$\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)−\mathrm{3n}−\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{n}^{\mathrm{2}} −\mathrm{4n}−\mathrm{5}=\mathrm{0},\:\left(\mathrm{n}−\mathrm{5}\right)\left(\mathrm{n}+\mathrm{1}\right)=\mathrm{0} \\…
Question Number 96138 by john santu last updated on 30/May/20 $${xy}'\:+\:{y}^{\mathrm{2}} \:=\:{x}^{\mathrm{2}} {e}^{{x}} \: \\ $$ Commented by Sourav mridha last updated on 30/May/20 $$\boldsymbol{{it}}'\mathrm{s}\:\mathrm{wrong},\mathrm{it}\:\boldsymbol{{should}}\:\boldsymbol{{be}}…
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Question Number 154561 by amin96 last updated on 19/Sep/21 Answered by liberty last updated on 19/Sep/21 $$\frac{\mathrm{3}}{\mathrm{4}} \\ $$ Commented by amin96 last updated on…
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Question Number 78333 by Lontum Hans last updated on 16/Jan/20 $$\mathrm{prove}\:\mathrm{by}\:\mathrm{contradiction}\:\mathrm{that}\:\sqrt{\mathrm{2}\:}\:\mathrm{is}\:\mathrm{irrational}. \\ $$ Answered by MJS last updated on 16/Jan/20 $$\sqrt{\mathrm{2}}\in\mathbb{Q}\wedge\sqrt{\mathrm{2}}>\mathrm{0}\:\Rightarrow\:\sqrt{\mathrm{2}}=\frac{{p}}{{q}} \\ $$$${p},\:{q}\:\in\mathbb{N}\wedge\mathrm{gcd}\:\left({p},\:{q}\right)\:=\mathrm{1}\:\Leftrightarrow\:{p}\nmid{q}\wedge{q}\nmid{p} \\ $$$$\left(\sqrt{\mathrm{2}}=\frac{{p}}{{q}}\right)^{\mathrm{2}}…
Question Number 77330 by TawaTawa last updated on 05/Jan/20 Commented by $@ty@m123 last updated on 05/Jan/20 Commented by $@ty@m123 last updated on 05/Jan/20 $${In}\:\bigtriangleup{PDC}\:\&\bigtriangleup{QAP} \\…