Question Number 201837 by MrGHK last updated on 13/Dec/23 $${y}'''−{y}''+{y}'={sec}\left({t}\right),−\frac{\pi}{\mathrm{2}}<{t}<\frac{\pi}{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201832 by sonukgindia last updated on 13/Dec/23 Answered by aleks041103 last updated on 13/Dec/23 $${x}={e}^{−{t}} \:\Rightarrow\:{dx}=−{e}^{−{t}} {dt} \\ $$$$\Rightarrow{I}=\mathrm{32}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{5}} \left(−{t}\right)}{−{t}}{e}^{−{t}} {dt}=\mathrm{32}\int_{\mathrm{0}}…
Question Number 201833 by sebastian last updated on 13/Dec/23 $$ \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 201806 by sonukgindia last updated on 12/Dec/23 Commented by mr W last updated on 30/Dec/23 $${see}\:{Q}\mathrm{202543} \\ $$$${answer}\:{is}\:\mathrm{2}\left(\mathrm{4}^{\mathrm{3}} −\sqrt{\mathrm{1}}−\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−…−\sqrt{\mathrm{15}}\right) \\ $$ Commented by…
Question Number 201793 by sonukgindia last updated on 12/Dec/23 Answered by mr W last updated on 12/Dec/23 Commented by mr W last updated on 12/Dec/23…
Question Number 201740 by MathedUp last updated on 12/Dec/23 $${y}''\left({t}\right)+\frac{\mathrm{g}}{\ell}\centerdot\mathrm{sin}\left({y}\left({t}\right)\right)=\mathrm{0} \\ $$$$\boldsymbol{\mathcal{L}}_{{s}} \left\{{y}''\left({t}\right)\right\}+\frac{\mathrm{g}}{\ell}\boldsymbol{\mathcal{L}}_{{s}} \left\{\mathrm{sin}\left({y}\left({t}\right)\right)\right\}=\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} {y}''\left({t}\right){e}^{−{st}} \mathrm{d}{t}+\frac{\mathrm{g}}{\ell}\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{sin}\left({y}\left({t}\right)\right){e}^{−{st}} \mathrm{d}{t}=\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\:\infty}…
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Question Number 201712 by esmaeil last updated on 10/Dec/23 Commented by mr W last updated on 11/Dec/23 Commented by mr W last updated on 11/Dec/23…
Question Number 201654 by mokys last updated on 10/Dec/23 Answered by aleks041103 last updated on 10/Dec/23 $${First}\:{part}: \\ $$$${M}=\begin{pmatrix}{\mathrm{4}}&{\mathrm{3}}\\{\mathrm{1}}&{−\mathrm{2}}\end{pmatrix}\: \\ $$$${eigenvals}: \\ $$$${det}\left({M}−{xI}\right)=\begin{vmatrix}{\mathrm{4}−{x}}&{\mathrm{3}}\\{\mathrm{1}}&{−\mathrm{2}−{x}}\end{vmatrix}=\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{4}\right)\left({x}+\mathrm{2}\right)−\mathrm{3}=\mathrm{0}…