Question Number 101350 by bobhans last updated on 02/Jul/20 Answered by bemath last updated on 02/Jul/20 $$\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{1}−{y}^{\mathrm{2}} \right){dx}−{xy}\:{dy}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{1}−{y}^{\mathrm{2}} \right)\:{dx}\:=\:{xy}\:{dy}\: \\ $$$$\frac{\left({x}^{\mathrm{2}}…
Question Number 166764 by amin96 last updated on 27/Feb/22 $$\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}=\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{x}}\right)\:\: \\ $$ Answered by mr W last updated on 27/Feb/22 $${let}\:{u}={y}−{x} \\ $$$${y}={u}+{x} \\ $$$$\frac{{dy}}{{dx}}=\frac{{du}}{{dx}}+\mathrm{1}…
Question Number 100902 by bemath last updated on 29/Jun/20 $$\mathrm{solve}\:\mathrm{y}''−\mathrm{4y}'+\mathrm{4y}=\mathrm{0}\: \\ $$$$\mathrm{with}\:\mathrm{variation}\:\mathrm{method} \\ $$ Commented by bramlex last updated on 29/Jun/20 $${AE}\::\:\lambda^{\mathrm{2}} −\mathrm{4}\lambda+\mathrm{4}\:=\mathrm{0} \\ $$$$\left(\lambda−\mathrm{2}\right)^{\mathrm{2}}…
Question Number 100891 by bemath last updated on 29/Jun/20 $$\mathrm{u}_{\mathrm{tt}} \:=\:\mathrm{u}_{\mathrm{xx}} \:−\:\mathrm{6x}\:;\:\mathrm{0}\leqslant\mathrm{x}<\pi\:,\:\mathrm{t}>\mathrm{0} \\ $$$$\mathrm{u}_{\left(\mathrm{0},\mathrm{t}\right)} \:=\:\mathrm{0}\:;\:\mathrm{u}_{\left(\pi,\mathrm{t}\right)} \:=\:\pi^{\mathrm{3}} +\mathrm{3}\pi \\ $$$$\mathrm{u}_{\left(\mathrm{x},\mathrm{0}\right)} \:=\:\mathrm{x}^{\mathrm{3}} +\mathrm{3x}+\mathrm{3sin}\:\mathrm{x} \\ $$$$\mathrm{u}_{\mathrm{t}} \left(\mathrm{x},\mathrm{0}\right)\:=\:\mathrm{0}\: \\…
Question Number 100766 by john santu last updated on 28/Jun/20 Answered by Rio Michael last updated on 28/Jun/20 $${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}}…