Answer to Question #176283 in Differential Equations for Isaax Numoah Boateng

Question #176283

Suppose u(x) = cot x is an Integrating factor of the differential equation (sin x)? y" = 2y, find the general solution.


1
Expert's answer
2021-03-31T13:24:34-0400

Ans:- According to the question

integrating factor is

\Rightarrow I.F.=epdx=CotxI.F.=e^{\int{p}dx}=Cotx

\Rightarrow pdx=lnCotx\int{pdx}=ln{Cotx}

\Rightarrow p=2×Sec2xp=-2\times Sec2x

So then we arrange our question with respect to Integrating factor which has been given

As we see the question ? mark is coming (sinx)?y"=2y(sin x)? y" = 2y then instead of ? 2cosx2cosx written and yy'' written as yy' .

Thus the question is 2sinx×cosxy=2y2{sin x}\times{cosx} y'= 2y

y2y×Sec2x=0\Rightarrow y'-2y\times{Sec2x}=0

Integrating factor is CotxCotx

then y×Cotx=Cotx×0+cy\times{Cotx}=\int{Cotx}\times{0}+c

y×Cotx=c\Rightarrow y\times{Cotx}=c where c is an arbitrary constant


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