Question Number 181799 by amin96 last updated on 30/Nov/22 $$\:\boldsymbol{{solve}}\:\boldsymbol{{the}}\:\boldsymbol{{difgerential}}\:\boldsymbol{{equation}} \\ $$$$\boldsymbol{{y}}\sqrt{\mathrm{1}+\left(\boldsymbol{{y}}'\right)^{\mathrm{2}} }=\boldsymbol{{y}}'\: \\ $$ Answered by mr W last updated on 30/Nov/22 $${y}^{\mathrm{2}} \left(\mathrm{1}+\left({y}'\right)^{\mathrm{2}}…
Question Number 181795 by ali009 last updated on 30/Nov/22 $${use}\:{laplace}\:{transform}\:{to}\:{slove}\: \\ $$$${y}''+\mathrm{4}{y}=\mathrm{5}{u}\left({t}−\mathrm{1}\right),{y}\left(\mathrm{0}\right)=\mathrm{0},{y}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 116105 by ShakaLaka last updated on 30/Sep/20 $${solve}\:{the}\:{Cauchy}-{Euler}\: \\ $$$${Differential}\:{Equation}\:{by} \\ $$$${substituting}\:{x}={e}^{{t}} \\ $$$${x}^{\mathrm{3}} \:\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }\:+\:\mathrm{2}{x}^{\mathrm{2}} \:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{2}{y}\:=\:\mathrm{10}{x}\:+\:\frac{\mathrm{10}}{{x}} \\ $$$$ \\…
Question Number 181605 by ibe222 last updated on 27/Nov/22 Commented by Mastermind last updated on 27/Nov/22 $$\mathrm{We}\:\mathrm{should}\:\mathrm{Differentiate}? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 116054 by Study last updated on 30/Sep/20 $$\left({x}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{1}\right){dx}+\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{1}\right){dy}=\mathrm{0} \\ $$$${y}=? \\ $$ Commented by mohammad17 last updated on 30/Sep/20 $${M}\left({x},{y}\right)={x}^{\mathrm{2}}…
Question Number 116052 by Study last updated on 30/Sep/20 $$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{25}{y}=\mathrm{0}\:\:\:\:\:\:\:{y}=? \\ $$ Commented by mohammad17 last updated on 30/Sep/20 $$\left({D}^{\mathrm{2}} +\mathrm{25}\right){y}=\mathrm{0} \\ $$$$…
Question Number 116055 by Study last updated on 30/Sep/20 $$\mathrm{3}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{4}\frac{{dy}}{{dx}}+\mathrm{5}{y}=\mathrm{0}\:\:\:\:\:\:{y}=? \\ $$ Commented by mohammad17 last updated on 30/Sep/20 $$\left(\mathrm{3}{D}^{\mathrm{2}} +\mathrm{4}{D}+\mathrm{5}\right){y}=\mathrm{0} \\ $$$$…
Question Number 116053 by Study last updated on 30/Sep/20 $${y}\frac{{dy}}{{dx}}=\mathrm{1}+{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:{y}=? \\ $$ Answered by Dwaipayan Shikari last updated on 30/Sep/20 $$\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{1}+\mathrm{x}^{\mathrm{2}} \\ $$$$\int\mathrm{ydy}=\int\mathrm{1}+\mathrm{x}^{\mathrm{2}} \mathrm{dx}…
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Question Number 115968 by Engr_Jidda last updated on 29/Sep/20 Answered by 1549442205PVT last updated on 30/Sep/20 $$\mathrm{Put}\:\mathrm{u}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{1};\mathrm{v}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{y}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{The}\:\mathrm{condition}\:\mathrm{Cauchy}−\mathrm{Rieman}\:\mathrm{for} \\ $$$$\mathrm{the}\:\mathrm{analytical}\:\mathrm{is}\:\frac{\partial\mathrm{u}}{\partial\mathrm{x}}=\frac{\partial\mathrm{v}}{\partial\mathrm{y}}\left(\mathrm{1}\right),\frac{\partial\mathrm{u}}{\partial\mathrm{y}}=−\frac{\partial\mathrm{v}}{\partial\mathrm{x}}\left(\mathrm{2}\right) \\ $$$$\mathrm{We}\:\mathrm{have}\:\frac{\partial\mathrm{u}}{\partial\mathrm{x}}=\mathrm{2x},\frac{\partial\mathrm{u}}{\partial\mathrm{y}}=\mathrm{0},\frac{\partial\mathrm{v}}{\partial\mathrm{x}}=\mathrm{0},\frac{\partial\mathrm{v}}{\partial\mathrm{y}}=\mathrm{2y}…