Answer to Question #140080 in Differential Equations for ysabelle tagaruma

Question #140080
(y−(cosx)^2)dx+cosxdy=0
1
Expert's answer
2020-10-26T16:37:41-0400

"\\mathbf{Given:\\;(y-(cos\\;x)^2)dx+cos\\;x\\;dy=0}"


"\\mathbf{We\\;have\\;:\\dfrac{(y-cos^2x)}{cos\\;x}+\\dfrac{dy}{dx}=0}"


"\\implies \\mathbf{\\dfrac{dy}{dx}+\\dfrac{y}{cos\\;x}=cos\\;x}\\\\ \\\\"


"\\therefore\\;\\mathbf{This\\;is\\;in\\;the\\;form\\;of\\;Bernoulli's\\;equation,so\\;-}\\\\ \\\\\n\\mathbf{Integrating\\;factor=I.F.=exp\\left(\\int\\dfrac{1}{cos\\;x}dx\\right)=exp\\left(\\int{sec\\;x}dx\\right)}"

"=\\mathbf{exp(ln|sec\\;x+tan\\;x|)}"

"=|\\mathbf{sec\\;x+tan\\;x}|"


"\\therefore\\mathbf{Solution\\;is-}"


"\\mathbf{y(I.F.)=\\int Q(x)(I.F.)dx+C,\\;(here\\;Q(x)=cos\\;x)}"


"\\mathbf{So,\\;we\\;have\\;solution-}"


"\\mathbf{y(|sec\\;x+tan\\;x|)=\\int cos\\;x(|sec\\;x+tan\\;x|)dx+C}"


"\\implies \\mathbf{y(|sec\\;x+tan\\;x|)=\\int (1+sin\\;x)dx+C}"


"\\implies \\mathbf{y(|sec\\;x+tan\\;x|)= x-cos\\;x+C\\;\\;\\;\\;............Ans.}"




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