Question Number 211560 by MrGaster last updated on 13/Sep/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{certificate}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \sqrt{\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \boldsymbol{\mathrm{cos}}\left(\boldsymbol{{x}}\right)−\boldsymbol{\mathrm{lncos}}\left(\boldsymbol{{x}}\right)\boldsymbol{{dx}}}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\sqrt{\mathrm{2}\boldsymbol{\mathrm{ln}}\mathrm{2}} \\ $$$$ \\ $$ Terms of Service…
Question Number 211546 by MrGaster last updated on 12/Sep/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\frac{\boldsymbol{{dx}}}{\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{{x}}+\mathrm{13}}}=? \\ $$ Answered by Frix last updated on 12/Sep/24 $$\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{13}}}\:\overset{\left[{t}=\frac{{x}−\mathrm{2}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{13}}}{\mathrm{3}}\right]}…
Question Number 210786 by zhou0429 last updated on 19/Aug/24 Answered by Berbere last updated on 19/Aug/24 $${x}^{\mathrm{9}} +\mathrm{1}=\underset{{k}=\mathrm{0}} {\overset{\mathrm{8}} {\prod}}\left({x}−{e}^{{i}\pi\left(\frac{\mathrm{1}+\mathrm{2}{k}}{\mathrm{9}}\right)} \right)={p}\left({x}\right) \\ $$$$=\Sigma\int\frac{\mathrm{1}}{\left({x}−{e}^{{i}\pi\left(\frac{\mathrm{1}+\mathrm{2}{k}}{\mathrm{9}}\right)} \right){p}'\left({e}^{{i}\pi\left(\frac{\mathrm{1}+\mathrm{2}{k}}{\mathrm{9}}\right)} \right)}{dx}…
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Question Number 208569 by pticantor last updated on 18/Jun/24 $$\boldsymbol{{help}}\:\boldsymbol{{me}}\:\boldsymbol{{to}}\:\boldsymbol{{solve}}\:\boldsymbol{{this}}\:\boldsymbol{{please}} \\ $$$$\:\:\boldsymbol{{y}}''−\sqrt{\mathrm{1}+\boldsymbol{{y}}'^{\mathrm{2}} }=\boldsymbol{{x}}^{\mathrm{2}} \\ $$$$\boldsymbol{{solve}}\:\boldsymbol{{this}}\:\boldsymbol{{differential}}\:\boldsymbol{{equation}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$ Terms of…
Question Number 208303 by Simurdiera last updated on 10/Jun/24 $${Resolver} \\ $$$$\frac{\partial^{\mathrm{2}} {u}}{\partial{y}^{\mathrm{2}} }\:−\:{x}^{\mathrm{2}} {u}\:=\:{xe}^{\mathrm{4}{y}} \\ $$ Answered by aleks041103 last updated on 12/Jun/24 $${u}={X}\left({x}\right){Y}\left({y}\right)…
Question Number 208264 by efronzo1 last updated on 09/Jun/24 $$\:\:\:\cancel{\underline{\underbrace{ }}} \\ $$ Answered by mr W last updated on 10/Jun/24 $${let}\:{y}'={p} \\ $$$${p}'−\mathrm{2}{p}=\mathrm{2}\left({e}^{{x}} \mathrm{cos}\:{x}−\mathrm{1}\right)…
Question Number 207991 by efronzo1 last updated on 02/Jun/24 $$\:\:\:\:\:\mathrm{y}\:\underline{\underbrace{\mathcal{W}}} \\ $$ Answered by mr W last updated on 02/Jun/24 $${particular}\:{solution}: \\ $$$${y}={A}\:\mathrm{sin}\:{x}+{B}\:\mathrm{cos}\:{x} \\ $$$${y}'={A}\:\mathrm{cos}\:{x}−{B}\:\mathrm{sin}\:{x}…
Question Number 207420 by Wuji last updated on 14/May/24 $$\mathrm{solve}\:\mathrm{the}\:\mathrm{Differential}\:\mathrm{equation} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\left(\frac{\mathrm{x}+\mathrm{3y}}{\mathrm{2x}}\right) \\ $$ Answered by mr W last updated on 14/May/24 $${let}\:{y}={tx} \\ $$$$\frac{{dy}}{{dx}}={t}+{x}\frac{{dt}}{{dx}}…
Question Number 207317 by mr W last updated on 11/May/24 $${solve}\:{for}\:{y} \\ $$$$\frac{\mathrm{1}}{{y}'}+\frac{\mathrm{1}}{{y}''}=\mathrm{1} \\ $$ Answered by Berbere last updated on 11/May/24 $${y}'={z} \\ $$$$\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{z}'}=\mathrm{1}\Rightarrow{z}'=\frac{{z}}{{z}−\mathrm{1}}\Rightarrow\left(\frac{{z}−\mathrm{1}}{{z}}\right){dz}={dx}…