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Question Number 126122 by benjo_mathlover last updated on 17/Dec/20 $$\:\:{y}.{y}'+\mathrm{3}{y}=\mathrm{2}{x}\: \\ $$ Answered by Dwaipayan Shikari last updated on 17/Dec/20 $${y}'+\mathrm{3}=\frac{\mathrm{2}{x}}{{y}}\:\:\:\:\:\:\:\:{y}={vx}\Rightarrow{y}'={v}+{xv}' \\ $$$$\Rightarrow{xv}'+{v}+\mathrm{3}=\frac{\mathrm{2}}{{v}}\Rightarrow{xv}'=\frac{\mathrm{2}−{v}^{\mathrm{2}} −\mathrm{3}{v}}{{v}} \\…
Question Number 126076 by liberty last updated on 17/Dec/20 $$\:\:\:{y}\left(\mathrm{ln}\:\left(\frac{{y}}{{x}}\right)+\mathrm{1}\right){dx}−{xdy}\:=\:\mathrm{0} \\ $$ Answered by benjo_mathlover last updated on 17/Dec/20 Terms of Service Privacy Policy Contact:…
Question Number 125991 by bramlexs22 last updated on 16/Dec/20 $${If}\:{a}\:{and}\:{b}\:{arbitary}\:{constants},\: \\ $$$${find}\:{a}\:{second}\:−\:{order}\:{equation} \\ $$$${which}\:{has}\:{y}\:=\:{ae}^{{x}} +{b}\:\mathrm{cos}\:{x}\:{as}\:{a} \\ $$$${general}\:{solution}. \\ $$ Answered by liberty last updated on…
Question Number 125958 by bramlexs22 last updated on 15/Dec/20 $$\:\:\:\:{y}\:{dx}\:+{x}\left(\mathrm{ln}\:{x}\:−\mathrm{ln}\:{y}−\mathrm{1}\right){dy}=\mathrm{0} \\ $$$$\:\:\:\:{where}\:{y}\left(\mathrm{1}\right)=\mathrm{0} \\ $$ Answered by Olaf last updated on 16/Dec/20 $${ydx}+{x}\left(\mathrm{ln}{x}−\mathrm{ln}{y}−\mathrm{1}\right){dy}\:=\:\mathrm{0} \\ $$$$\mathrm{Let}\:{y}\:=\:{xu} \\…
Question Number 60367 by bhanukumarb2@gmail.com last updated on 20/May/19 Commented by bhanukumarb2@gmail.com last updated on 20/May/19 $$\mathrm{1}\: \\ $$ Commented by Mr X pcx last…
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Question Number 60351 by bhanukumarb2@gmail.com last updated on 20/May/19 Commented by bhanukumarb2@gmail.com last updated on 20/May/19 $${without}\:{applying}\:{geomtry}\:{plz}\:{solve}\:{it}….. \\ $$ Commented by bhanukumarb2@gmail.com last updated on…
Question Number 125721 by bramlexs22 last updated on 13/Dec/20 $$\:\left(\mathrm{3}{xy}+\mathrm{2}{x}^{\mathrm{2}} \right){dy}\:+\:\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{y}\right){dx}=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 125693 by fajri last updated on 13/Dec/20 $${find}\:: \\ $$$$ \\ $$$$\frac{{d}^{\mathrm{4}} {y}}{{dx}^{\mathrm{4}} }\:+\:\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }\:−\:\mathrm{7}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:−\:\frac{{dy}}{{dx}\:\:}\:+\:\mathrm{6}{y}\:=\:\mathrm{0} \\ $$$$ \\ $$$${for}\:{y}\left(\mathrm{0}\right)\:=\:\mathrm{1},\:{y}'\left(\mathrm{0}\right)\:=\:\mathrm{0},\:{y}''\left(\mathrm{0}\right)\:=\:−\mathrm{2}\:,\:{y}'''\left(\mathrm{0}\right)\:=\:−\mathrm{1} \\…