Question Number 100766 by john santu last updated on 28/Jun/20 Answered by Rio Michael last updated on 28/Jun/20 $${AE}:\:{m}^{\mathrm{2}} \:+\mathrm{1}\:=\:\mathrm{0}\:\Rightarrow\:{m}\:=\pm{i} \\ $$$${y}_{{c}} \:=\:\left({A}\mathrm{cos}\:{x}\:+{B}\:\mathrm{sin}\:{x}\right) \\ $$$$\mathrm{let}\:{y}_{{p}}…
Question Number 100684 by bemath last updated on 28/Jun/20 $$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{xy}\right)\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{xy}\:+\:\mathrm{y}^{\mathrm{2}} \\ $$ Answered by bobhans last updated on 28/Jun/20 $$\mathrm{set}\:\mathrm{y}\:=\:\mathrm{sx}\:\Rightarrow\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{s}\:+\:\mathrm{x}\:\frac{\mathrm{ds}}{\mathrm{dx}} \\ $$$$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \mathrm{s}\right)\:\left(\mathrm{s}\:+\mathrm{x}\:\frac{\mathrm{ds}}{\mathrm{dx}}\right)\:=\:\mathrm{x}^{\mathrm{2}}…
Question Number 100594 by Coronavirus last updated on 27/Jun/20 $${solve}\:\:{the}\:{differential}\:\:{equations} \\ $$$$\mathrm{1}-\:\:{x}\mathrm{cos}\:\left(\mathrm{ln}\:\frac{{x}}{{y}}\right){dy}−{ydx}=\mathrm{0} \\ $$$$\mathrm{2}-\:\:{ydx}+\mathrm{2}{xdy}\:=\mathrm{2}{y}\frac{\sqrt{{x}}}{{cos}^{\mathrm{2}} \left({y}\right)}{dy}\:\:\:\:\:{y}\left(\mathrm{0}\right)=\pi \\ $$ Answered by smridha last updated on 27/Jun/20 $$\mathrm{2}.\left[\frac{\boldsymbol{{dx}}}{\mathrm{2}\sqrt{\boldsymbol{{x}}}}+\frac{\sqrt{\boldsymbol{{x}}}}{\boldsymbol{{y}}}\boldsymbol{{dy}}=\boldsymbol{{sec}}^{\mathrm{2}}…
Question Number 100585 by Rio Michael last updated on 27/Jun/20 $$\:\mathrm{Given}\:\mathrm{that}\:\:{G}\:=\:\left\{\mathrm{1},\left({x}\:+\:{yi}\right),\left({x}−{yi}\right)\right\}\:\mathrm{form}\:\mathrm{a}\:\mathrm{group} \\ $$$$\mathrm{under}\:\mathrm{complex}\:\mathrm{multiplication},\:\mathrm{describe}\:\mathrm{the}\:\mathrm{locus} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\left({x},{y}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 100565 by bemath last updated on 27/Jun/20 Commented by bobhans last updated on 28/Jun/20 $$\mathrm{let}\:\mathrm{z}\:=\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right) \\ $$$$\frac{\partial\mathrm{z}}{\partial\mathrm{x}}\:=\:\frac{\mathrm{y}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4y}^{\mathrm{2}} \right)−\left(\mathrm{2x}\right)\left(\mathrm{xy}−\mathrm{4y}\right)}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4y}^{\mathrm{2}} \right)}\:=\:\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}}…
Question Number 100344 by bemath last updated on 26/Jun/20 $$\mathrm{solve}\:\mathrm{y}''+\mathrm{y}\:=\:\mathrm{sin}\:\mathrm{x} \\ $$ Answered by bobhans last updated on 26/Jun/20 $$\mathrm{homogenous}\:\mathrm{solution} \\ $$$$\eta^{\mathrm{2}} +\mathrm{1}\:=\:\mathrm{0}\:\Rightarrow\eta\:=\pm\:\mathrm{i}\: \\ $$$$\mathrm{y}_{\mathrm{h}}…
Question Number 100337 by I want to learn more last updated on 26/Jun/20 $$\mathrm{Solve}:\:\:\:\left(\mathrm{1}\:+\:\:\mathrm{x}^{\mathrm{2}} \right)\mathrm{y}''\:\:−\:\:\mathrm{4xy}'\:\:+\:\:\mathrm{6y}\:\:\:=\:\:\mathrm{0} \\ $$ Answered by mathmax by abdo last updated on…
Question Number 100323 by bobhans last updated on 26/Jun/20 $$\mathbb{S}\mathrm{olve}\:\mathrm{x}^{\mathrm{2}} \mathrm{y}''+\mathrm{2xy}'−\mathrm{2y}=\mathrm{0} \\ $$ Commented by bemath last updated on 26/Jun/20 $$\mathrm{since}\:\mathrm{diff}\:\mathrm{a}\:\mathrm{power}\:\mathrm{pushes}\:\mathrm{down}\:\mathrm{the} \\ $$$$\mathrm{exponent}\:\mathrm{by}\:\mathrm{one}\:\mathrm{unit},\:\mathrm{the}\:\mathrm{form}\:\mathrm{of} \\ $$$$\mathrm{this}\:\mathrm{eq}\:\mathrm{suggest}\:\mathrm{that}\:\mathrm{we}\:\mathrm{look}\:\mathrm{for}…
Question Number 34746 by candre last updated on 10/May/18 $${y}+\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{\mathrm{1}}{{x}}\left({x}\mathrm{ln}\:{x}−\frac{{dy}}{{dx}}+\mathrm{cos}\:{x}\right) \\ $$$${y}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 100184 by bobhans last updated on 25/Jun/20 $$\frac{\mathrm{ydx}\:+\:\mathrm{xdy}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} }\:+\:\mathrm{xdx}\:=\:\mathrm{0} \\ $$ Answered by maths mind last updated on 26/Jun/20 $${u}={xy}\Rightarrow{du}={ydx}+{xdy} \\ $$$$\Leftrightarrow\frac{{du}}{\mathrm{1}−{u}^{\mathrm{2}}…