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}: \\ $$ Commented by i jagooll last…
Question Number 94954 by Mr.D.N. last updated on 22/May/20 $$\:\:\boldsymbol{\mathrm{For}}\:\boldsymbol{\mathrm{any}}\:\boldsymbol{\mathrm{curve}},\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}: \\ $$$$\:\:\:\left(\frac{\mathrm{d}^{\mathrm{2}} \mathrm{x}}{\mathrm{ds}^{\mathrm{2}} }\right)^{\mathrm{2}} +\left(\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\right)^{\mathrm{2}} =\:\frac{\mathrm{1}}{\rho^{\mathrm{2}} } \\ $$$$\:\: \\ $$ Answered by…
Question Number 94934 by i jagooll last updated on 22/May/20 Commented by i jagooll last updated on 22/May/20 $$\mathrm{thank}\:\mathrm{you}\:\mathrm{both} \\ $$ Answered by john santu…
Question Number 160451 by Ar Brandon last updated on 29/Nov/21 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{system}\:\mathrm{below}: \\ $$$$\begin{cases}{{y}_{\mathrm{1}} '=\mathrm{2}{y}_{\mathrm{1}} +{y}_{\mathrm{2}} +{y}_{\mathrm{3}} }\\{{y}_{\mathrm{2}} '=−\mathrm{2}{y}_{\mathrm{1}} −{y}_{\mathrm{3}} }\\{{y}_{\mathrm{3}} '=\mathrm{2}{y}_{\mathrm{1}} +{y}_{\mathrm{2}} +\mathrm{2}{y}_{\mathrm{3}} }\end{cases} \\…
Question Number 29347 by sorour87 last updated on 07/Feb/18 $${y}^{\left(\mathrm{1}\right)} +\left(\frac{\mathrm{2}}{{x}\mathrm{ln}\:{x}}+\frac{\left(\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} \mathrm{ln}\:{x}\right)}{{x}\left(\mathrm{1}−\mathrm{2}{x}\mathrm{ln}\:{x}\right)}\right){y}=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 94847 by i jagooll last updated on 21/May/20 Answered by john santu last updated on 21/May/20 $$\mathrm{auxilarry}\:\mathrm{equation}\: \\ $$$$\lambda^{\mathrm{3}} −\lambda^{\mathrm{2}} +\lambda−\mathrm{1}\:=\:\mathrm{0}\: \\ $$$$\left(\lambda−\mathrm{1}\right)\left(\lambda^{\mathrm{2}}…