Question Number 104893 by bramlex last updated on 24/Jul/20

(d^2 y/dx^2 ) + tan x (dy/dx) = sec x + cot x

Answered by mathmax by abdo last updated on 24/Jul/20

y^(′′)  +tanx y^′  =((1+sinx)/(cosx))  let y^′  =z so e ⇒z^′  +tanx z =((1+sinx)/(cosx))  h→z^′  =−tanx z ⇒(z^′ /z) =−tanx ⇒ ln∣z∣ =−∫ tanx dx +c  =−∫ ((sinx)/(cosx))dx +c =ln∣cosx∣ +c ⇒z =k ∣cosx∣ let determine solution on  { x /cosx >0}  mvc method ⇒z^′  =k^′  cosx −ksinx  e ⇒k^′  cosx−ksinx + tamx ×k cosx =((1+sinx)/(cosx)) ⇒k^(′ )  =((1+sinx)/(cos^2 x)) ⇒  k =∫  ((1+sinx)/(cos^2 x))dx =∫  (dx/(cos^2 x)) +∫ ((sinx)/(cos^2 x))dx =(1/(cosx)) +∫ (dx/(cos^2 x))  we have  ∫ (dx/(cos^2 x)) =∫ (1+tan^2 x)dx =tanx  +c ⇒k (x) =(1/(cosx)) +tanx +λ ⇒  z(x) =((1/(cosx)) +tanx +λ)cosx =1+sinx +λcosx  y^′  =z ⇒y^′  =1+sinx +λ cosx ⇒y(x) =∫ (1+sinx +λ cosx)dx  y(x)=x −cosx + λ sinx

Commented bymathmax by abdo last updated on 24/Jul/20

sorry i have solved y^(′′)  +tanx y^′  =secx +tanx but the way is the same...

Commented bybramlex last updated on 24/Jul/20

ok sir . thank you

Commented bymathmax by abdo last updated on 24/Jul/20

you are welcome