Question Number 102295 by 1549442205 last updated on 08/Jul/20 $$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{equation}}\:: \\ $$$$\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{lny}}−\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{y}}'=\boldsymbol{\mathrm{y}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 102283 by Ar Brandon last updated on 08/Jul/20 $$\mathrm{sinx}\frac{\mathrm{dy}}{\mathrm{dx}}−\mathrm{2ycosx}=\mathrm{3sinx} \\ $$ Answered by john santu last updated on 08/Jul/20 $$\frac{{dy}}{{dx}}\:−\mathrm{2cot}\:{x}.{y}\:=\:\mathrm{3} \\ $$$${IF}\:\Rightarrow{u}\left({x}\right)={e}^{\int\:−\mathrm{2cot}\:{x}\:{dx}\:} =\:{e}^{−\mathrm{2ln}\left(\mathrm{sin}\:{x}\right)}…
Question Number 101848 by bramlex last updated on 05/Jul/20 $$\left(\mathrm{cos}\:\mathrm{x}\right)\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\mathrm{y}\:\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{2x}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:, \\ $$$$\mathrm{y}\left(\frac{\pi}{\mathrm{4}}\right)\:=\:\frac{−\mathrm{15}\pi^{\mathrm{2}} \sqrt{\mathrm{2}}}{\mathrm{32}} \\ $$ Answered by john santu last updated on 05/Jul/20 $$\frac{{dy}}{{dx}}\:+\:{y}.\mathrm{tan}\:\left({x}\right)=\:\mathrm{2}{x}\:\mathrm{cos}\:{x}…
Question Number 101841 by bemath last updated on 05/Jul/20 $${xy}'\:+\:{y}\:=\:{y}^{\mathrm{2}} \\ $$ Answered by bobhans last updated on 05/Jul/20 $${xy}'\:=\:{y}^{\mathrm{2}} −{y}\:\Rightarrow\:\frac{{dy}}{{y}\left({y}−\mathrm{1}\right)}\:=\:\frac{{dx}}{{x}} \\ $$$$\int\:\frac{{dy}}{{y}−\mathrm{1}}\:−\:\int\:\frac{{dy}}{{y}}\:=\:\frac{{dx}}{{x}} \\ $$$$\mathrm{ln}\left(\frac{{y}−\mathrm{1}}{{y}}\right)\:=\:\mathrm{ln}\mid{Cx}\mid\:\Rightarrow\:\mathrm{1}−\frac{\mathrm{1}}{{y}}\:=\:\mid{Cx}\mid\:…
Question Number 101846 by bemath last updated on 05/Jul/20 Answered by bramlex last updated on 05/Jul/20 $$\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}…