Question Number 117228 by bemath last updated on 10/Oct/20 $$\left(\frac{{x}+{y}}{{y}−\mathrm{1}}\right)\:{dx}\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}+\mathrm{1}}{{y}−\mathrm{1}}\right)^{\mathrm{2}} {dy}\:=\:\mathrm{0} \\ $$ Answered by TANMAY PANACEA last updated on 10/Oct/20 $$\left(\frac{{x}+{y}}{{y}−\mathrm{1}}\right){dx}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}+{y}}{{y}−\mathrm{1}}−\mathrm{1}\right)^{\mathrm{2}} {dy}=\mathrm{0} \\ $$$$\left(\frac{{x}+{y}}{{y}−\mathrm{1}}\right){dx}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}+{y}}{{y}−\mathrm{1}}\right)^{\mathrm{2}}…
Question Number 51494 by Tawa1 last updated on 27/Dec/18 $$\mathrm{Solve}:\:\:\:\:\:\:\:\:\left(\mathrm{t}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:\frac{\mathrm{dp}}{\mathrm{dt}}\:\:=\:\:\mathrm{p}^{\mathrm{t}} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 27/Dec/18 $$\left({t}^{\mathrm{2}} +\mathrm{1}\right)\frac{{d}^{\mathrm{2}} {p}}{{dt}^{\mathrm{2}} }+\frac{{dp}}{{dt}}×\mathrm{2}{t}=\frac{{d}}{{dt}}\left({p}^{{t}} \right)…
Question Number 117029 by bemath last updated on 09/Oct/20 $$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2y}^{\mathrm{2}} \right)\:\mathrm{dx}\:+\:\left(\mathrm{4xy}−\mathrm{y}^{\mathrm{2}} \right)\:\mathrm{dy}\:=\:\mathrm{0} \\ $$ Answered by bemath last updated on 09/Oct/20 $$\mathrm{letting}\:\mathrm{y}\:=\:\varphi\mathrm{x}\:\Rightarrow\mathrm{dy}\:=\:\varphi\:\mathrm{dx}+\mathrm{x}\:\mathrm{d}\varphi \\ $$$$\Rightarrow\left(\mathrm{x}^{\mathrm{2}}…
Question Number 116614 by bemath last updated on 05/Oct/20 $$\:\left(\mathrm{x}^{\mathrm{2}} \:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{y}}\right)−\mathrm{y}\right)\mathrm{dx}\:+\:\mathrm{x}\:\mathrm{dy}\:=\:\mathrm{0} \\ $$$$\mathrm{where}\:\mathrm{y}\left(\mathrm{1}\right)=\:\frac{\pi}{\mathrm{4}} \\ $$ Commented by TANMAY PANACEA last updated on 05/Oct/20 $${i}\:{think}\:{it}\:{should}\:{be}\:{cos}^{\mathrm{2}}…
Question Number 182120 by amin96 last updated on 04/Dec/22 Answered by SEKRET last updated on 04/Dec/22 $$\:\:\:\sqrt[{\mathrm{4}}]{\mathrm{5}+\frac{\mathrm{3}}{\:\sqrt[{\mathrm{6}}]{\mathrm{13}+\frac{\mathrm{8}}{….}}}\:}\:\:\:\:=\:\:\boldsymbol{\mathrm{a}} \\ $$$$\:\:\:\:\:\:\mathrm{5}\:+\frac{\mathrm{3}}{\:\sqrt[{\mathrm{6}}]{\mathrm{13}+\frac{\mathrm{8}}{\boldsymbol{\mathrm{a}}}}}\:\:=\:\boldsymbol{\mathrm{a}}^{\mathrm{4}} \\ $$$$\:\:\:\left(\mathrm{13}+\frac{\mathrm{8}}{\boldsymbol{\mathrm{a}}}\right)=\:\frac{\mathrm{3}^{\mathrm{6}} }{\left(\boldsymbol{\mathrm{a}}^{\mathrm{4}} −\mathrm{1}\right)^{\mathrm{6}} } \\…
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Question Number 182003 by MWSuSon last updated on 03/Dec/22 $$\mathrm{u}_{\mathrm{xx}} −\mathrm{u}_{\mathrm{x}} \mathrm{u}_{\mathrm{y}} −\mathrm{u}_{\mathrm{yy}} +\mathrm{2u}_{\mathrm{y}} −\mathrm{2u}_{\mathrm{x}} =\mathrm{e}^{\mathrm{2x}+\mathrm{3y}} +\mathrm{sin}\left(\mathrm{2x}+\mathrm{y}\right)+\mathrm{xy} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 116457 by bobhans last updated on 04/Oct/20 $$\mathrm{Use}\:\mathrm{Laplace}\:\mathrm{transform}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{the}\: \\ $$$$\mathrm{initial}\:\mathrm{value}\:\mathrm{problem}\:\mathrm{ty}''+\left(\mathrm{4t}−\mathrm{2}\right)\mathrm{y}'+\left(\mathrm{13t}−\mathrm{4}\right)\mathrm{y}=\mathrm{0} \\ $$$$\mathrm{where}\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{0}\:\mathrm{and}\:\mathrm{y}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$ Answered by Olaf last updated on 04/Oct/20 $$\mathrm{L}\left({y}''\right)\:=\:\mathrm{p}^{\mathrm{2}} \mathrm{L}\left(\mathrm{p}\right)−\mathrm{p}{y}\left(\mathrm{0}\right)−{y}'\left(\mathrm{0}\right)…
Question Number 116425 by bemath last updated on 04/Oct/20 $$\:\mathrm{y}'−\mathrm{y}\:=\:−\mathrm{2xy}^{\mathrm{3}} \\ $$ Commented by bemath last updated on 04/Oct/20 $$\mathrm{thank}\:\mathrm{you}\:\mathrm{mr}\:\mathrm{Bob}\:\mathrm{and}\:\mathrm{mr}\:\mathrm{Olaf} \\ $$ Answered by bobhans…
Question Number 116329 by bemath last updated on 03/Oct/20 $$\mathrm{solve}\:\mathrm{the}\:\mathrm{diff}\:\mathrm{equation}\:\: \\ $$$$\left(\mathrm{1}\right)\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{x}+\mathrm{3y}−\mathrm{5}}{\mathrm{x}−\mathrm{y}−\mathrm{1}} \\ $$$$\left(\mathrm{2}\right)\:\left(\mathrm{3y}−\mathrm{7x}−\mathrm{3}\right)\mathrm{dx}+\left(\mathrm{7y}−\mathrm{3x}−\mathrm{7}\right)\mathrm{dy}=\mathrm{0} \\ $$ Answered by mr W last updated on 03/Oct/20 $$\left(\mathrm{1}\right)\:\:\frac{{dy}}{{dx}}=\frac{{x}+\mathrm{3}{y}−\mathrm{5}}{{x}−{y}−\mathrm{1}}…