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} \\ $$$$\mathrm{The}\:\mathrm{condition}\:\mathrm{Cauchy}−\mathrm{Rieman}\:\mathrm{for} \\ $$$$\mathrm{the}\:\mathrm{analytical}\:\mathrm{is}\:\frac{\partial\mathrm{u}}{\partial\mathrm{x}}=\frac{\partial\mathrm{v}}{\partial\mathrm{y}}\left(\mathrm{1}\right),\frac{\partial\mathrm{u}}{\partial\mathrm{y}}=−\frac{\partial\mathrm{v}}{\partial\mathrm{x}}\left(\mathrm{2}\right) \\ $$$$\mathrm{We}\:\mathrm{have}\:\frac{\partial\mathrm{u}}{\partial\mathrm{x}}=\mathrm{2x},\frac{\partial\mathrm{u}}{\partial\mathrm{y}}=\mathrm{0},\frac{\partial\mathrm{v}}{\partial\mathrm{x}}=\mathrm{0},\frac{\partial\mathrm{v}}{\partial\mathrm{y}}=\mathrm{2y}…
Question Number 115869 by bobhans last updated on 29/Sep/20 $$\:\frac{{dy}}{{dx}}\:+\mathrm{2}{y}\:=\:{x}^{\mathrm{3}} .{e}^{−\mathrm{2}{y}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 115771 by bemath last updated on 28/Sep/20 $${y}''−{y}'−\mathrm{2}{y}={e}^{\mathrm{2}{x}} .\mathrm{cos}\:^{\mathrm{2}} {x} \\ $$ Answered by TANMAY PANACEA last updated on 28/Sep/20 Answered by mathmax…
Question Number 115721 by bemath last updated on 28/Sep/20 $$\left({D}^{\mathrm{2}} −\mathrm{6}{D}+\mathrm{9}\right){y}\:=\:\frac{{e}^{\mathrm{3}{x}} }{{x}^{\mathrm{2}} } \\ $$ Commented by mohammad17 last updated on 28/Sep/20 $${m}^{\mathrm{2}} −\mathrm{6}{m}+\mathrm{9}=\mathrm{0}\Rightarrow\left({m}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{m}_{\mathrm{1}}…
Question Number 115706 by Rio Michael last updated on 27/Sep/20 $$\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square},\:{A}\left({t}\right)\:\mathrm{is}\:\mathrm{increased} \\ $$$$\mathrm{at}\:\mathrm{a}\:\mathrm{rate}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{its}\:\mathrm{perimeter} \\ $$$$\mathrm{write}\:\mathrm{a}\:\mathrm{differential}\:\mathrm{equation}\:\mathrm{that}\:{A}\left({t}\right) \\ $$$$\mathrm{satisfy}\:,\:\mathrm{starting}\:\mathrm{from}\:\frac{{dA}}{{dt}}\:= \\ $$ Commented by Dwaipayan Shikari last updated…
Question Number 115539 by bemath last updated on 26/Sep/20 $$\frac{{dy}}{{dx}}\:=\:\frac{{e}^{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)} \:−{y}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$ Answered by bobhans last updated on 26/Sep/20 $$\frac{{dy}}{{dx}}\:+\:\frac{{y}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\:\frac{{e}^{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}…