Answer to Question #140080 in Differential Equations for ysabelle tagaruma

Question

(y−(cosx)^2)dx+cosxdy=0

Answer 1

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

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

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

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

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

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

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

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

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

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

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

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