Question Number 107727 by Ar Brandon last updated on 12/Aug/20 $$\mathrm{Determine}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{solutions}\:\mathrm{to}\:\mathrm{the}\:\mathrm{equation}\:; \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}+\mathrm{t}^{\mathrm{3}} \right){x}'\left(\mathrm{t}\right)+\mathrm{t}^{\mathrm{2}} {x}\left(\mathrm{t}\right)+\mathrm{t}\left({x}\left(\mathrm{t}\right)\right)^{\mathrm{2}} =\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 107373 by Dwaipayan Shikari last updated on 10/Aug/20 $$\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\mathrm{n}} \\ $$ Answered by hgrocks last updated on 10/Aug/20 Commented by Dwaipayan…
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Question Number 107151 by john santu last updated on 09/Aug/20 $$\:\:\:\:\:\:\:\:\:@\mathrm{JS}@ \\ $$$$\left(\mathrm{D}^{\mathrm{2}} +\mathrm{7D}+\mathrm{12}\right)\mathrm{y}\:=\:\mathrm{e}^{\mathrm{x}} \:\mathrm{cos}\:\mathrm{2x}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 107108 by Ar Brandon last updated on 08/Aug/20 $$\mathrm{y}''+\mathrm{y}=\frac{\mathrm{1}}{\mathrm{cosx}} \\ $$ Commented by mohammad17 last updated on 08/Aug/20 $$\left({m}^{\mathrm{2}} −{i}^{\mathrm{2}} \right)=\mathrm{0}\Rightarrow{m}=\mp{i} \\ $$$${Yc}={c}_{\mathrm{1}}…
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Question Number 106775 by bemath last updated on 07/Aug/20 $$\:\:\:\:\:\:\:\:\:\:\:^{@\mathrm{bemath}@} \\ $$$$\:\left(\mathrm{1}\right)\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:−\mathrm{6}\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{9y}\:=\:\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} \\ $$$$\:\:\left(\mathrm{2}\right)\:\begin{cases}{\mathrm{x}^{\mathrm{3}} +\mathrm{3y}^{\mathrm{3}} \:=\:\mathrm{11}}\\{\mathrm{x}^{\mathrm{2}} \mathrm{y}\:+\mathrm{xy}^{\mathrm{2}} \:=\:\mathrm{6}}\end{cases}\: \\ $$ Answered by abdomathmax…
Question Number 106588 by bemath last updated on 06/Aug/20 $$\:\:\:\:\:\:\:@\mathrm{bemath}@ \\ $$$$\mathrm{x}\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:+\:\mathrm{3x}−\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{e}^{\mathrm{3x}} \: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 41028 by Choudharyvishal155@gmail.com last updated on 31/Jul/18 $${if}\:\mathrm{1}/{u}\:=\:\sqrt{\left({x}\mathrm{2}\:+\:{y}\mathrm{2}\:+{z}\mathrm{2}\right)} \\ $$$${then}\:{x}\partial{u}/\partial{x}\:+\:{y}\partial{u}/\partial{y}\:+\:{z}\partial{u}/\partial{z}\:=\:? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 31/Jul/18 $${u}=\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\:\:{so}\:\:\frac{\partial{u}}{\partial{x}}=\frac{−\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}}…
Question Number 40920 by ajfour last updated on 29/Jul/18 $$\frac{{d}^{\:\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }={a}−{b}\left(\mathrm{1}−\frac{{l}}{\:\sqrt{{x}^{\mathrm{2}} +{l}^{\mathrm{2}} }}\right){x}\: \\ $$$${Find}\:{x}\left({t}\right)\:{if}\:{x}\left(\mathrm{0}\right)={x}_{\mathrm{0}} \:,\:{x}'\left(\mathrm{0}\right)=\mathrm{0}\:. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on…