Question Number 68926 by ramirez105 last updated on 16/Sep/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 68925 by ramirez105 last updated on 16/Sep/19 Commented by kaivan.ahmadi last updated on 16/Sep/19 $$\frac{\partial{u}}{\partial{x}}=\left({x}+\mathrm{3}\right)^{−\mathrm{1}} {cosy}\Rightarrow{u}\left({x},{y}\right)={cosyln}\left({x}+\mathrm{3}\right)+{h}\left({y}\right) \\ $$$$\frac{\partial{u}}{\partial{y}}={y}^{−\mathrm{1}} −{sinyln}\left(\mathrm{5}{x}+\mathrm{15}\right)=−{sinyln}\left({x}+\mathrm{3}\right)+\frac{{dh}}{{dy}}\Rightarrow \\ $$$${y}^{−\mathrm{1}} −{ln}\mathrm{5}=\frac{{dh}}{{dy}}\Rightarrow{h}={lny}−{yln}\mathrm{5} \\…
Question Number 68924 by ramirez105 last updated on 16/Sep/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 68923 by ramirez105 last updated on 16/Sep/19 Commented by kaivan.ahmadi last updated on 16/Sep/19 $$\frac{\partial{u}}{\partial{x}}=\frac{\mathrm{1}}{{y}}\Rightarrow{u}\left({x},{y}\right)=\int\frac{\mathrm{1}}{{y}}{dx}=\frac{{x}}{{y}}+{h}\left({y}\right) \\ $$$$\frac{\partial{u}}{\partial{y}}=−\frac{{x}}{{y}^{\mathrm{2}} }=−\frac{{x}}{{y}^{\mathrm{2}} }+\frac{{dh}}{{dy}}\Rightarrow\frac{{dh}}{{dy}}=\mathrm{0}\Rightarrow{h}={c}' \\ $$$$ \\ $$$$\Rightarrow{u}\left({x},{y}\right)={c}\Rightarrow\frac{{x}}{{y}}={c}\Rightarrow{x}−{cy}=\mathrm{0}…
Question Number 68860 by ramirez105 last updated on 16/Sep/19 Commented by kaivan.ahmadi last updated on 16/Sep/19 $${set}\:{M}={x}+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}\:\:{and}\:\:\:{N}=−{y}+\frac{{xy}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\frac{\partial{M}}{\partial{y}}=\frac{{y}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}}=\frac{\partial{N}}{\partial{x}} \\ $$$$\begin{cases}{\frac{\partial{u}}{\partial{x}}={x}+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}}\\{\frac{\partial{u}}{\partial{y}}=−{y}+\frac{{xy}}{\:\sqrt{{y}^{\mathrm{2}}…
Question Number 68855 by ramirez105 last updated on 16/Sep/19 Commented by kaivan.ahmadi last updated on 16/Sep/19 $${set}\:{M}=\mathrm{6}{x}+{y}^{\mathrm{2}} \:\:{and}\:{N}={y}\left(\mathrm{2}{x}−\mathrm{3}{y}\right) \\ $$$$\frac{\partial{M}}{\partial{y}}=\mathrm{2}{y}\:=\frac{\partial{N}}{\partial{x}}\Rightarrow{the}\:{equation}\:{is}\:{exact} \\ $$$$\frac{\partial{u}}{\partial{x}}=\mathrm{6}{x}+{y}^{\mathrm{2}} \:\:\:{and}\:\:\frac{\partial{u}}{\partial{y}}={y}\left(\mathrm{2}{x}−\mathrm{3}{y}\right)=\mathrm{2}{xy}−\mathrm{3}{y}^{\mathrm{2}} \\ $$$$\Rightarrow{u}\left({x},{y}\right)=\int\left(\mathrm{6}{x}+{y}^{\mathrm{2}}…
Question Number 68508 by oyemi kemewari last updated on 12/Sep/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 68506 by oyemi kemewari last updated on 12/Sep/19 $${y}'=\mathrm{4}{y}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{1} \\ $$$${what}\:{the}\:{primitive}\:{solution} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 133889 by EDWIN88 last updated on 25/Feb/21 $$\:\mathrm{y}'−\mathrm{y}.\mathrm{tan}\:\mathrm{x}\:=\:\mathrm{cos}\:\mathrm{x}−\mathrm{2x}\:\mathrm{sin}\:\mathrm{x} \\ $$$$\mathrm{y}\left(\frac{\pi}{\mathrm{6}}\right)\:=\:\mathrm{0}\:\Rightarrow\mathrm{y}\left(\mathrm{x}\right)=? \\ $$ Answered by bemath last updated on 25/Feb/21 Answered by Ñï= last…
Question Number 133887 by bemath last updated on 25/Feb/21 $$\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:;\:\mathrm{y}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com