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Category: Differential Equation

Question-57571

Question Number 57571 by cesar.marval.larez@gmail.com last updated on 07/Apr/19 Commented by MJS last updated on 07/Apr/19 $$\mathrm{what}\:\mathrm{is}\:“\mathrm{sen}''? \\ $$$$\mathrm{if}\:\mathrm{it}'\mathrm{s}\:\mathrm{sinus}\:\mathrm{or}\:\mathrm{secans}\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{it}\: \\ $$ Terms of Service Privacy…

Question-187858

Question Number 187858 by normans last updated on 23/Feb/23 Answered by som(math1967) last updated on 23/Feb/23 $$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{a}+{b}+{c}} \\ $$$$\Rightarrow\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right)−{abc}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} {b}+{abc}+{a}^{\mathrm{2}} {c}+{ab}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}+{abc}+\cancel{{abc}}…

dy-dx-y-4y-3-3y-2-x-with-y-1-1-

Question Number 122281 by benjo_mathlover last updated on 15/Nov/20 $$\:\:\frac{{dy}}{{dx}}\:=\:\frac{−{y}}{\mathrm{4}{y}^{\mathrm{3}} +\mathrm{3}{y}^{\mathrm{2}} +{x}}\:{with}\:{y}\left(\mathrm{1}\right)=−\mathrm{1} \\ $$ Answered by liberty last updated on 15/Nov/20 $$\left(\mathrm{4y}^{\mathrm{3}} +\mathrm{3y}^{\mathrm{2}} +\mathrm{x}\right)\mathrm{dy}\:+\:\mathrm{ydx}\:=\:\mathrm{0} \\…

2xy-tan-y-dx-x-2-x-sec-2-y-dy-0-

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}}…

dy-dx-x-y-x-y-

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}}…

dy-dx-3y-tan-x-1-2-sin-2x-

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)}…