Question Number 95600 by Tony Lin last updated on 26/May/20 $${solve}\:{the}\:{differential}\:{equation} \\ $$$${L}\frac{{d}^{\mathrm{2}} {q}}{{dt}^{\mathrm{2}} }+{R}\frac{{dq}}{{dt}}+\frac{\mathrm{1}}{{C}}{q}=\varepsilon{cos}\omega{t} \\ $$$${which}\:{is}\:{in}\:{R}.{L}.{C}\:{circuit}\:{with}\:{forced}\:{oscillation} \\ $$$${where}\:{L}\:{is}\:{inductance} \\ $$$${R}\:{is}\:{resistance} \\ $$$${C}\:{is}\:{capacitanace} \\ $$$${q}\:{is}\:{charge}…
Question Number 160903 by blackmamba last updated on 08/Dec/21 $$\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}\:'=\:\mathrm{2}{xy}\:+\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} \: \\ $$ Answered by GuruBelakangPadang last updated on 09/Dec/21 $$\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}\:'=\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)'{y}\:+\left(\mathrm{1}+{x}^{\mathrm{2}}…
Question Number 95316 by i jagooll last updated on 24/May/20 $$\frac{{dy}}{{dx}}−{y}\:=\:{xy}^{\mathrm{5}} \: \\ $$ Answered by john santu last updated on 24/May/20 Terms of Service…
Question Number 95119 by Mr.D.N. last updated on 27/May/20 $$\:\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equations}:− \\ $$$$\bigstar.\left(\mathrm{x}\:\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:−\mathrm{x}\:\mathrm{cos}\:\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\:\mathrm{y}\:\mathrm{cos}\:\mathrm{x}=\mathrm{0}. \\ $$ Answered by niroj last updated on 23/May/20 $$\:\:\mathrm{2}.\:\left(\mathrm{x}\:\mathrm{sin}\:\mathrm{x}\:+\:\mathrm{cos}\:\mathrm{x}\:\right)\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}}…
Question Number 95098 by i jagooll last updated on 23/May/20 $$\left[\:\frac{\mathrm{y}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:+\:\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:\right]\:\mathrm{dx}\:+\:\left[\frac{\mathrm{y}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }−\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:\right]\mathrm{dy}=\mathrm{0} \\ $$ Answered by bobhans last…
Question Number 95068 by Mr.D.N. last updated on 23/May/20 $$\:\:\:\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equations}−: \\ $$$$\:\:\:\mathrm{1}.\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{sin}\left(\mathrm{x}+\mathrm{y}\right)+\:\mathrm{cos}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\: \\ $$ Commented by EmericGent last updated on 22/May/20 Would be easy with cos(x-y) instead of cos(x+y) Commented…
Question Number 94997 by i jagooll last updated on 22/May/20 Answered by bobhans last updated on 22/May/20 $$\mathrm{homogenous}\:\mathrm{solution} \\ $$$$\upsilon^{\mathrm{2}} −\mathrm{2}\upsilon+\mathrm{2}\:=\:\mathrm{0} \\ $$$$\upsilon\:=\:\mathrm{1}\pm\:{i}\:\Rightarrow\mathrm{y}_{\mathrm{h}} \:=\:\mathrm{Ae}^{\mathrm{x}} \mathrm{cos}\:\mathrm{x}\:+\mathrm{Be}^{\mathrm{x}}…
Question Number 160516 by peter frank last updated on 30/Nov/21 Answered by aleks041103 last updated on 06/Dec/21 $${u}={x}/{y}\Rightarrow{y}={x}/{u} \\ $$$$\Rightarrow{y}'=\frac{{u}−{u}'{x}}{{u}^{\mathrm{2}} }=\frac{\mathrm{1}}{{u}}−\frac{{u}'}{{u}^{\mathrm{2}} }{x} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{u}}−\frac{{x}}{{u}^{\mathrm{2}} }\:\frac{{du}}{{dx}}={e}^{{u}}…
Question Number 94969 by bobhans last updated on 22/May/20 $$\mathrm{y}''−\mathrm{3y}'+\mathrm{2y}=\mathrm{0}\:;\:\mathrm{y}=\mathrm{0},\mathrm{y}'=\mathrm{3}\:\mathrm{for}\:\mathrm{x}=\mathrm{0} \\ $$ Answered by bobhans last updated on 22/May/20 $$\mathrm{homogenous}\:\mathrm{solution} \\ $$$$\upsilon^{\mathrm{2}} −\mathrm{3}\upsilon+\mathrm{2}\:=\:\mathrm{0}\:\Rightarrow\left(\upsilon−\mathrm{2}\right)\left(\upsilon−\mathrm{1}\right)=\mathrm{0} \\ $$$$\begin{cases}{\upsilon=\mathrm{2}}\\{\upsilon=\mathrm{1}}\end{cases}\:\Rightarrow\mathrm{y}_{\mathrm{h}}…
Question Number 94956 by i jagooll last updated on 22/May/20 Commented by niroj last updated on 22/May/20 $$\:\mathrm{both}\:\mathrm{of}\:\mathrm{you}\:\mathrm{are}\:\mathrm{right}\:\mathrm{its}\:\mathrm{note}: \\ $$ Commented by i jagooll last…