Question Number 46609 by maxmathsup by imad last updated on 29/Oct/18 $${solve}\:\:\:\:{x}\:{y}^{''} \:−{e}^{−{x}} {y}^{'} \:\:\:={x}\:{sinx} \\ $$ Commented by maxmathsup by imad last updated on…
Question Number 46608 by maxmathsup by imad last updated on 29/Oct/18 $${let}\:{the}\:{d}.{e}\:\:{xy}^{''} \:+\left({x}^{\mathrm{2}} −{x}\right){y}^{'} \:+\mathrm{2}{y}\:=\mathrm{0} \\ $$$${find}\:{a}\:{solution}\:{developpable}\:{at}\:{integr}\:{serie}. \\ $$$$ \\ $$ Terms of Service Privacy…
Question Number 112114 by mnjuly1970 last updated on 06/Sep/20 $$\:\:{solution}\:{of} \\ $$$$\Phi=\int_{\mathrm{0}} ^{\mathrm{1}} {xH}_{{x}} {dx}\:\overset{\gamma+\psi\left({x}+\mathrm{1}\right)} {=}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}\left(\gamma+\psi\left({x}+\mathrm{1}\right)\right){dx}\: \\ $$$$\:\overset{\psi\left({x}+\mathrm{1}\right)=\frac{\mathrm{1}}{{x}}+\psi\left({x}\right)} {=}\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}\left(\gamma+\frac{\mathrm{1}}{{x}}+\psi\left({x}\right)\right){dx} \\ $$$$=\frac{\gamma}{\mathrm{2}}+\mathrm{1}+\:\int_{\mathrm{0}}…
Question Number 111965 by mnjuly1970 last updated on 05/Sep/20 $$\:\:\:\:\:\:….{advanced}\:\:{calculus}… \\ $$$${evaluate}: \\ $$$$ \\ $$$$\:\:\:\:{i}::\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {xH}_{{x}} {dx}\:=???\:\: \\ $$$$\:\:\:\:\:{ii}::\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{{n}^{\mathrm{2}} \mathrm{2}^{{n}\:}…
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Question Number 177456 by cortano1 last updated on 05/Oct/22 Answered by a.lgnaoui last updated on 05/Oct/22 $${posons}\:\:{x}=\mathrm{sin}\:{t}\:\:\:\:{alors}\:{y}=\mathrm{cos}\:{t} \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\:{t}}+\frac{\mathrm{4}}{\mathrm{cos}\:{t}}=\frac{\mathrm{cos}\:{t}+\mathrm{4sin}\:{t}}{\mathrm{sin}\:{t}\mathrm{cos}\:{t}} \\ $$$$\left.{le}\:{rapoort}\:\:{est}\:{minimale}\:\:{si}\:\:\:\mathrm{scos}\:{t}+\mathrm{4sin}\:{t}\:{mnimale}\right) \\ $$$$\begin{cases}{\mathrm{cos}\:{t}+\mathrm{4sin}\:{t}=\mathrm{0}}\\{\mathrm{sin}\:{t}\mathrm{cos}\:{t}\neq\mathrm{0}}\end{cases} \\ $$$$\mathrm{4sin}\:{t}=−\mathrm{cos}\:{t}\:\:\:\mathrm{tan}\:{t}=−\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{sin}\:{t}}{\mathrm{cos}\:{t}}…
Question Number 177320 by mnjuly1970 last updated on 03/Oct/22 Answered by a.lgnaoui last updated on 04/Oct/22 $$\sqrt{\mathrm{2}}\:+\left(\mathrm{1}+\sqrt{\mathrm{2}}\:\right){x}+{x}^{\mathrm{2}} =\left({x}+\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} =\left({x}+\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}=−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\left[\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{1}+\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}+\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{1}+\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}\right] \\ $$$$\int_{\mathrm{0}} ^{\infty}…
Question Number 111584 by ajfour last updated on 04/Sep/20 Answered by ajfour last updated on 04/Sep/20 $${C}\left(\mathrm{0},{y}\right)\:\:\:;\:{A}\left({x},\mathrm{0}\right)\:\:;\:\:{B}\left({z},{z}^{\mathrm{2}} \right) \\ $$$$\sqrt{\left({x}−{z}\right)^{\mathrm{2}} +{z}^{\mathrm{4}} }+\sqrt{{z}^{\mathrm{2}} +\left({y}−{z}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\:{a}…
Question Number 177071 by cortano1 last updated on 30/Sep/22 Commented by Frix last updated on 30/Sep/22 $$\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist} \\ $$ Commented by Frix last updated on…
Question Number 45994 by peter frank last updated on 19/Oct/18 Answered by MrW3 last updated on 20/Oct/18 $${we}\:{know}\:\frac{{x}+{y}}{\mathrm{2}}\geqslant\sqrt{{xy}}\:{when}\:{x},{y}\geqslant\mathrm{0}. \\ $$$$ \\ $$$$\mathrm{2}\left({a}+{b}\right)={L}\Rightarrow{a}+{b}=\frac{{L}}{\mathrm{2}} \\ $$$${A}_{{rectangular}} ={ab}\leqslant\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}}…