Question Number 124050 by john_santu last updated on 30/Nov/20 $$\left(\mathrm{6}{x}+{y}^{\mathrm{2}} \right){dx}\:+{y}\left(\mathrm{2}{xy}−\mathrm{3}{y}^{\mathrm{2}} \right){dy}\:=\:\mathrm{0}\: \\ $$ Answered by mohammad17 last updated on 30/Nov/20 $${M}=\mathrm{6}{x}+{y}^{\mathrm{2}} \rightarrow{M}_{{y}} =\mathrm{2}{y}\rightarrow\left(\ast\right) \\…
Question Number 123982 by Ar Brandon last updated on 29/Nov/20 $$\mathrm{Consider}\:\mathrm{the}\:\mathrm{D}.\mathrm{E}\:\left(\mathrm{E}\right):\:\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{y}'−\mathrm{2y}=\mathrm{x}^{\mathrm{3}} \left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} \mathrm{e}^{−\mathrm{x}} \\ $$$$\mathrm{a}\backslash\:\mathrm{Resolve}\:\mathrm{the}\:\mathrm{DE}\:\mathrm{x}\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)\mathrm{y}'−\mathrm{2y}=\mathrm{0} \\ $$$$\mathrm{b}\backslash\mathrm{We}\:\mathrm{wish}\:\mathrm{to}\:\mathrm{find}\:\mathrm{a}\:\mathrm{function}\:\mathrm{g}\left(\mathrm{x}\right)\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function} \\ $$$$\mathrm{h}\left(\mathrm{x}\right)\:\mathrm{defined}\:\mathrm{by}\:\mathrm{h}\left(\mathrm{x}\right)=\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{particular}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\left(\mathrm{E}\right)…
Question Number 123983 by Ar Brandon last updated on 29/Nov/20 $$\mathrm{Consider}\:\mathrm{the}\:\mathrm{real}\:\mathrm{number}\:\mathrm{c}\:\mathrm{and}\:\mathrm{the}\:\mathrm{DE}\:\left(\mathrm{H}\right)\:\mathrm{below} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}''+\mathrm{2y}'+\mathrm{cy}=\mathrm{0} \\ $$$$\mathrm{a}\backslash\mathrm{Given}\:\mathrm{c}=\mathrm{3}.\:\mathrm{Rewrite}\:\mathrm{and}\:\mathrm{resolve}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{the}\:\mathrm{equation}\:\left(\mathrm{H}\right) \\ $$$$\mathrm{b}\backslash\mathrm{for}\:\mathrm{c}=\mathrm{1},\:\mathrm{Rewrite}\:\mathrm{and}\:\mathrm{resolve}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{the}\:\mathrm{equation}\:\left(\mathrm{H}\right) \\ $$$$\mathrm{c}\backslash\mathrm{for}\:\mathrm{c}=\mathrm{10}; \\ $$$$\:\:\:\:\mathrm{i}.\:\mathrm{Rewrite}\:\mathrm{and}\:\mathrm{resolve}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{the}\:\mathrm{equation}\:\left(\mathrm{H}\right) \\ $$$$\:\:\:\mathrm{ii}.\:\mathrm{Determine}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{at}\:+\infty\:\mathrm{of}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{this}\:\mathrm{DE}. \\ $$$$\:\:\mathrm{iii}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{verifying}\:\mathrm{the}\:\mathrm{initial}\:\mathrm{conditions}:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{0}…
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Question Number 123858 by Dwaipayan Shikari last updated on 28/Nov/20 $$\overset{..} {\theta}+\frac{{g}}{{l}}{sin}\theta=\mathrm{0}\:\:\:\:\:\left({Find}\:{the}\:{exact}\:{solution}\right) \\ $$ Answered by Olaf last updated on 29/Nov/20 $$\overset{\bullet\bullet} {\theta}+\frac{{g}}{{l}}\mathrm{sin}\theta\:=\:\mathrm{0}\:\left(\mathrm{1}\right) \\ $$$$\left(\mathrm{1}\right)×\overset{\bullet}…
Question Number 123716 by Engr_Jidda last updated on 02/Jan/21 Commented by Dwaipayan Shikari last updated on 27/Nov/20 $${What}\:{is}\:{B}_{{n}} \:{here}? \\ $$ Commented by Engr_Jidda last…
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Question Number 123715 by Engr_Jidda last updated on 27/Nov/20 Answered by Dwaipayan Shikari last updated on 27/Nov/20 $$\int_{−\infty} ^{\infty} \frac{{e}^{\mathrm{2}{x}} }{{e}^{\mathrm{3}{x}} +\mathrm{1}}{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty}…
Question Number 123342 by bemath last updated on 25/Nov/20 $$\:\:\frac{{y}+\frac{{dy}}{{dx}}}{{y}}\:=\:{ay}\:,\:{for}\:{a}\:{some}\:{constant} \\ $$ Answered by liberty last updated on 25/Nov/20 $$\Rightarrow\:{y}\:+\:\frac{{dy}}{{dx}}\:=\:{ay}^{\mathrm{2}} \:;\:\frac{{dy}}{{dx}}\:=\:{ay}^{\mathrm{2}} −{y} \\ $$$$\:\Rightarrow\frac{{dy}}{{y}\left({ay}−\mathrm{1}\right)}\:=\:{dx}\: \\…
Question Number 123207 by benjo_mathlover last updated on 24/Nov/20 $$\:\:{x}^{\mathrm{2}} \:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:−\mathrm{3}{x}\:\frac{{dy}}{{dx}}\:+\:\mathrm{4}{y}\:=\:\ell{n}^{\mathrm{2}} \left({x}\right)−\ell{n}\left({x}\right) \\ $$ Commented by mnjuly1970 last updated on 24/Nov/20 $$\:\:\:\:\:{cauchy}\:{differential}\:{equation}. \\…