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Question-113330

Question Number 113330 by mathdave last updated on 12/Sep/20 Answered by MJS_new last updated on 12/Sep/20 $$\mathrm{use}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{half}\:\mathrm{derivate} \\ $$$$\frac{{d}^{{q}} }{{dx}^{{q}} }{x}^{{p}} =\frac{\Gamma\:\left({p}+\mathrm{1}\right)}{\Gamma\:\left({p}−{q}+\mathrm{1}\right)}{x}^{{p}−{q}} \\ $$$$\Rightarrow \\…

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Question Number 113271 by mathdave last updated on 12/Sep/20 $${solve}\:{the}\:{initial}\:{boundary}\:{value} \\ $$$${problem}\:{of}\:{wave}\:{equation} \\ $$$$\frac{\partial^{\mathrm{2}} {u}\left({x},{t}\right)}{\partial{t}^{\mathrm{2}} }=\mathrm{9}\frac{\partial^{\mathrm{2}} {u}\left({x},{t}\right)}{\partial{x}^{\mathrm{2}} },\mathrm{0}<{x}<\mathrm{2},{t}>\mathrm{0} \\ $$$${u}\left(\mathrm{0},{t}\right)=\mathrm{1},{u}\left(\mathrm{2},{t}\right)=\mathrm{3},{t}>\mathrm{0} \\ $$$${u}\left({x},\mathrm{0}\right)=\mathrm{2},\mathrm{0}<{x}<\mathrm{2} \\ $$$$\frac{\partial{u}}{\partial{t}}\left({x},\mathrm{0}\right)=\mathrm{sin2}{x},\mathrm{0}<{x}<\mathrm{2} \\…

Let-a-be-a-positive-number-Then-a-a-1-a-i-a-1-i-a-i-2-a-Where-is-the-mistake-

Question Number 178725 by ComplexPrime last updated on 20/Oct/22 $$\mathrm{Let}\:{a}\:\mathrm{be}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{number}.\:\mathrm{Then} \\ $$$$\sqrt{{a}}=\sqrt{\left(−{a}\right)\left(−\mathrm{1}\right)}=\sqrt{−{a}}\centerdot{i}=\sqrt{{a}\left(−\mathrm{1}\right)}\centerdot{i}=\sqrt{{a}}\centerdot{i}^{\mathrm{2}} =−\sqrt{{a}} \\ $$$$\mathrm{Where}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mistake}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com