Question Number 102653 by Sontsaronald last updated on 10/Jul/20

Answered by bemath last updated on 10/Jul/20

y′−ytan x=sin x  IF u(x)=e^(−∫tan x dx )  =  e^(ln(cosx)) =cos x  y(x)=((∫cos xsin x dx +C)/(cos x))  y(x)=sec x(−(1/4)cos 2x+C)

Answered by Ar Brandon last updated on 10/Jul/20

y′=ytan(x)+sin(x)    ⇒{y′−ytan(x)=sin(x)}.cos(x)    ⇒y′cos(x)−ysin(x)=sin(x)cos(x)    ⇒((d(ycos(x)))/dx)=sin(x)cos(x)  ⇒ycos(x)=∫sin(x)cos(x)dx=((sin^2 (x))/2)+C  ⇒y=((sin^2 (x)+K)/(2cos(x)))  ,  K=2C

Commented bySontsaronald last updated on 10/Jul/20

merci beaucoup

Commented bySontsaronald last updated on 10/Jul/20

il vous plait quelle est donc la bonne reponse?

Commented byprakash jain last updated on 10/Jul/20

sin^2 x=((1−cos 2x)/2)  ((sin^2 x+K)/(2cos x))=(1/(cosx))(((1−cos 2x+K)/(2×2)))  =sec x(−((cos 2x)/4)+constants)  both are equal

Commented byAr Brandon last updated on 10/Jul/20

Je vous en prie 👨 Les deux réponses sont les mêmes, comme vient de le prouver ce monsieur.😃

Commented byAr Brandon last updated on 10/Jul/20

👍😃

Commented bySontsaronald last updated on 10/Jul/20

merci beaucoup ohhh

Commented byAr Brandon last updated on 10/Jul/20

Cette expression ! 😯 Ça me rappelle de quelqu'un.

Answered by mathmax by abdo last updated on 11/Jul/20

y^′ −ytanx =sinx  he →y^′ −ytanx =0 ⇒y^′  =ytanx ⇒(y^′ /y)=tanx ⇒ln∣y∣ =∫((sinx)/(cosx)) dx =−ln∣cosx∣ +c  ⇒y =(k/(∣cosx∣))  let solve on d={x /cosx >0} ⇒y =(k/(cosx))  lagrange method ⇒y^′  =(k^′ /(cosx)) +k (((sinx)/(cos^2 x)))  e ⇒(k^′ /(cosx)) +((ksinx)/(cos^2 x)) −(k/(cosx))×((sinx)/(cosx)) =sinx ⇒(k^′ /(cosx)) =sinx ⇒k^′  =(1/2)sin(2x) ⇒  k(x) =(1/2)∫ sin(2x)dx =−(1/4)cos(2x) +λ ⇒  y(x) =(1/(cosx)){−(1/4)cos(2x)+λ}