Question Number 103008 by bobhans last updated on 12/Jul/20

(dy/dx) − y.tan x = e^x .sec x

Answered by bemath last updated on 12/Jul/20

IF u(x) = e^(−∫ tan x dx ) = e^(∫ ((d(cos x))/(cos x))) = cos x. ⇔cos x (dy/dx) −y.sin x = e^x ((d(y.cos x))/dx) = d(e^x ) ⇒∫ ((d(y.cos x))/dx) = ∫ d(e^x ) ∴ y.cos x = e^x + C

Answered by mathmax by abdo last updated on 12/Jul/20

y^′ −y tanx =(e^x /(cosx)) (he)→y^′ =y tanx ⇒(y^′ /y) =tanx ⇒ln∣y∣ =∫ ((sinx)/(cosx))dx =−ln∣cosx∣ +c ⇒ y =(k/(∣cosx∣)) solution on { x/ cosx>0} mvc method →y^′ =(k^′ /(cosx)) +k((sinx)/(cos^2 x)) e ⇒(k^′ /(cosx)) +k ((sinx)/(cos^2 x)) −(k/(cosx))×((sinx)/(cosx)) =(e^x /(cosx)) ⇒k^′ =e^x ⇒k =e^x +λ ⇒ y(x) =(1/(cosx)){e^x +λ}

Commented bybemath last updated on 12/Jul/20

what (he) sir?

Commented byabdomsup last updated on 12/Jul/20

homogen equation