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Category: Differential Equation

Two-small-charged-spheresn-Two-small-charged-spheresn-cotain-charges-q1-and-q2r-espectively-A-charge-da-ism-reoved-from-sphere-carryingh-carge-q1-and-is-transferredtoh-te-other-Find-charge-

Question Number 167917 by guddupari last updated on 29/Mar/22 $$\:_{} \: \\ $$$$\mathrm{Two}\:\mathrm{small}\:\mathrm{charged}\:\mathrm{spheresn} \\ $$$$\mathrm{Two}\:\mathrm{small}\:\mathrm{charged}\:\mathrm{spheresn} \\ $$$$\mathrm{cotain}\:\mathrm{charges}\:+\:\mathrm{q1}\:\mathrm{and}\:+\:\mathrm{q2r} \\ $$$$\mathrm{espectively}.\:\mathrm{A}\:\mathrm{charge}\:\mathrm{da}\:\mathrm{ism} \\ $$$$\mathrm{reoved}\:\mathrm{from}\:\mathrm{sphere}\:\mathrm{carryingh} \\ $$$$\mathrm{carge}\:\mathrm{q1}\:\mathrm{and}\:\mathrm{is}\:\mathrm{transferredtoh} \\ $$$$\mathrm{te}\:\mathrm{other}.\:\mathrm{Find}\:\mathrm{charge}\:\mathrm{on}\:\mathrm{eachs}…

sinx-dy-dx-2ycosx-3sinx-

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)}…

cos-x-dy-dx-y-sin-x-2x-cos-2-x-y-pi-4-15pi-2-2-32-

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}…

xy-y-y-2-

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-101846

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}}…