Question Number 88492 by jagoll last updated on 11/Apr/20 $$\mathrm{y}''\:−\mathrm{4y}'+\mathrm{5y}\:=\:\mathrm{1}+\mathrm{8cos}\:\mathrm{x}+\mathrm{e}^{\mathrm{2x}} \\ $$ Commented by niroj last updated on 11/Apr/20 $$\:\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }−\mathrm{4}\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}+\mathrm{5}\boldsymbol{\mathrm{y}}=\:\mathrm{1}+\mathrm{8}\boldsymbol{\mathrm{cos}}\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{e}}^{\mathrm{2}\boldsymbol{\mathrm{x}}} \\ $$$$\:\:\left(\mathrm{D}^{\mathrm{2}} −\mathrm{4D}+\mathrm{5}\right)\mathrm{y}=\:\mathrm{1}+\mathrm{8cos}\:\mathrm{x}\:+\mathrm{e}^{\mathrm{2x}}…
Question Number 88238 by M±th+et£s last updated on 09/Apr/20 $${solve}\: \\ $$$$\left(\mathrm{3}{x}^{\mathrm{5}} {y}^{\mathrm{4}} +\mathrm{4}{y}\right){dx}+\left(\mathrm{2}{x}^{\mathrm{6}} {y}^{\mathrm{3}} +\mathrm{3}{x}\right){dy}=\mathrm{0} \\ $$ Answered by ajfour last updated on 09/Apr/20…
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Question Number 22535 by Arjun Daniel last updated on 19/Oct/17 $$\mathrm{H}{ow}\:{to}\:{solve}\:{this}\:{homogeneous}\:{equation}.\:{Can}\:{u}\:{help}\:{me}\:{plz} \\ $$$${Q}.\:{solve}\:\left({x}\:{sin}\:\frac{{y}}{{x}}\right){dy}\:−\left({y}\:{sin}^{−\mathrm{1}} \:\frac{{y}}{{x}}\right){dx}=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 153542 by EDWIN88 last updated on 08/Sep/21 $$\:{y}'''+{y}'=\mathrm{sec}\:{x}\: \\ $$ Answered by puissant last updated on 08/Sep/21 $${y}'''+{y}'={secx} \\ $$$$\Rightarrow\:{y}''+{y}={ln}\left({tan}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)\right)+{C}_{\mathrm{1}} \\ $$$$\Rightarrow\:\left({D}^{\mathrm{2}} +{D}\right){y}={ln}\left({tan}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)\right)+{C}_{\mathrm{1}}…
Question Number 87944 by jagoll last updated on 07/Apr/20 $$\mathrm{y}\:'\:=\:\frac{\mathrm{2xy}}{\mathrm{y}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} } \\ $$ Commented by john santu last updated on 07/Apr/20 $$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{2xy}}{\mathrm{y}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }…
Question Number 87905 by john santu last updated on 07/Apr/20 $$\mathrm{y}'\:=\:\mathrm{2}^{\mathrm{y}} \\ $$ Commented by john santu last updated on 07/Apr/20 $$\mathrm{2}^{−\mathrm{y}} .\mathrm{y}'\:=\:\mathrm{1} \\ $$$$\mathrm{e}^{−\mathrm{y}\:\mathrm{ln}\:\mathrm{2}}…
Question Number 87904 by john santu last updated on 07/Apr/20 $$\mathrm{y}\:''\:−\mathrm{3y}'\:+\mathrm{2y}\:=\:\mathrm{10sin}\:\mathrm{x}\:+\:\mathrm{2cos}\:\mathrm{2x} \\ $$ Commented by niroj last updated on 07/Apr/20 $$\:\:\mathrm{y}^{''} −\mathrm{3y}^{'} +\mathrm{2y}=\mathrm{10sin}\:\mathrm{x}+\mathrm{2cos}\:\mathrm{2x} \\ $$$$\:\:\:\left(\mathrm{D}^{\mathrm{2}}…
Question Number 87789 by john santu last updated on 06/Apr/20 $$\left(\mathrm{D}^{\mathrm{3}} −\mathrm{D}^{\mathrm{2}} \right)\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} +\mathrm{1}\:,\mathrm{y}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\mathrm{y}\:'\left(\mathrm{0}\right)=−\mathrm{1}\:,\mathrm{y}\:''\left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$ Commented by mathmax by abdo last updated…
Question Number 87504 by M±th+et£s last updated on 04/Apr/20 $${solve}\: \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}\left(\frac{\mathrm{2}+{y}}{\mathrm{1}+{x}+{y}}\right)^{\mathrm{2}} \\ $$ Answered by TANMAY PANACEA. last updated on 04/Apr/20 $${i}\:{think} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}\left(\frac{{x}+{y}}{\mathrm{1}+{x}+{y}}\right)^{\mathrm{2}}…