Question Number 101846 by bemath last updated on 05/Jul/20 Answered by bramlex last updated on 05/Jul/20 $$\Leftrightarrow\:\frac{{dy}}{{dx}}\:−\mathrm{2}{x}^{−\mathrm{1}} {y}\:=\:{x}^{\mathrm{2}} \:\mathrm{cos}\:{x} \\ $$$${integrating}\:{factor}\:\:{u}\:=\:{e}^{\int−\mathrm{2}{x}^{−\mathrm{1}} \:{dx}} \\ $$$${u}\:=\:{e}\:^{−\mathrm{2}\:\mathrm{ln}\left({x}\right)} \:=\:{x}^{−\mathrm{2}}…
Question Number 101777 by 1549442205 last updated on 04/Jul/20 $$\mathrm{Find}\:\mathrm{a}\:\mathrm{curve}\:\mathrm{passing}\:\mathrm{through}\:\mathrm{pointA}\left(\mathrm{0};\mathrm{1}\right) \\ $$$$\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{formed}\:\mathrm{by}\:\mathrm{the}\:\mathrm{axis}\:\mathrm{Oy} \\ $$$$,\mathrm{tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{at}\:\mathrm{its}\:\mathrm{arbitrary}\:\mathrm{point} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{radius}−\mathrm{vector}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of} \\ $$$$\mathrm{contact},\mathrm{issosceles}\left(\mathrm{and}\:\mathrm{base}\:\mathrm{is}\:\mathrm{the}\:\mathrm{segment}\right. \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{from}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{contact} \\ $$$$\left.\mathrm{to}\:\mathrm{the}\:\mathrm{axis}\:\mathrm{Oy}\right) \\ $$ Terms…
Question Number 101756 by bramlex last updated on 04/Jul/20 $$\int\:\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{11}{x}+\mathrm{6}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}−\mathrm{5}\right)}\:{dx}\: \\ $$ Answered by john santu last updated on 04/Jul/20 $$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{11}{x}+\mathrm{6}\:=\:\left({x}−\mathrm{5}\right)\left({ax}+{b}\right)+{c}\left({x}^{\mathrm{2}} +\mathrm{1}\right)…
Question Number 101680 by john santu last updated on 04/Jul/20 Answered by bemath last updated on 04/Jul/20 $$\mathrm{let}\:\mathrm{point}\:\mathrm{R}\left(\mathrm{x},\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\mathrm{area}\:\mathrm{of}\:\Delta\mathrm{PQR}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\begin{vmatrix}{\:\:\:\mathrm{2}\:\:\:\:\mathrm{4}\:\:\:\:\mathrm{1}}\\{−\mathrm{1}\:\:\:\mathrm{1}\:\:\:\mathrm{1}}\\{\:\:\:\mathrm{x}\:\:\:\:\mathrm{x}^{\mathrm{2}} \:\:\mathrm{1}}\end{vmatrix} \\ $$$$\mathrm{A}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{2}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)−\mathrm{4}\left(−\mathrm{1}−\mathrm{x}\right)+\mathrm{1}\left(−\mathrm{x}^{\mathrm{2}}…
Question Number 101419 by bemath last updated on 02/Jul/20 $$\mathrm{y}''−\mathrm{4y}'+\mathrm{3y}\:=\:\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{1}+\mathrm{e}^{\mathrm{x}} } \\ $$ Answered by john santu last updated on 02/Jul/20 $$\mathrm{AE}:\:\lambda^{\mathrm{2}} −\mathrm{4}\lambda+\mathrm{3}\:=\:\mathrm{0} \\…
Question Number 101362 by bobhans last updated on 02/Jul/20 Commented by bemath last updated on 02/Jul/20 $$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3y}^{\mathrm{2}} \right)\mathrm{dx}+\left(\mathrm{2xy}−\mathrm{x}\right)\mathrm{dy}\:=\mathrm{0} \\ $$$$\mathrm{M}=\:\mathrm{x}^{\mathrm{2}} +\mathrm{3y}^{\mathrm{2}} \rightarrow\:\frac{\partial\mathrm{M}}{\partial\mathrm{y}}\:=\:\mathrm{6y} \\ $$$$\mathrm{N}\:=\:\mathrm{2xy}−\mathrm{x}\rightarrow\frac{\partial\mathrm{N}}{\partial\mathrm{x}}\:=\:\mathrm{2y}−\mathrm{1}…
Question Number 101350 by bobhans last updated on 02/Jul/20 Answered by bemath last updated on 02/Jul/20 $$\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}\: \\…